Class 10: Heights and Distances

$\displaystyle \textbf{Question 1: }\text{A tower stands vertically on the ground. From a point on the ground, }20\text{ m}$
$\displaystyle \text{away from the foot of the tower, the angle of elevation of the top of the tower is }60^\circ.$
$\displaystyle \text{What is the height of the tower?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower }=h\text{ m.}$
$\displaystyle \text{Distance of the point from the foot of the tower }=20\text{ m.}$
$\displaystyle \tan60^\circ=\frac{h}{20}$
$\displaystyle \sqrt{3}=\frac{h}{20}$
$\displaystyle h=20\sqrt{3}\text{ m}$
$\displaystyle \therefore \text{The height of the tower is }20\sqrt{3}\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{The angle of elevation of a ladder leaning against a wall is }60^\circ\text{ and the}$
$\displaystyle \text{foot of the ladder is }9.5\text{ m away from the wall. Find the length of the ladder.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the length of the ladder }=l\text{ m.}$
$\displaystyle \text{Distance of the foot of the} \\ \text{ladder from the wall }=9.5\text{ m.}$
$\displaystyle \cos60^\circ=\frac{9.5}{l}$
$\displaystyle \frac{1}{2}=\frac{9.5}{l}$
$\displaystyle l=2\times9.5=19\text{ m}$
$\displaystyle \therefore \text{The length of the ladder is }19\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{A ladder }15\text{ metres long just reaches the top of a vertical wall. If the ladder}$
$\displaystyle \text{makes an angle of }60^\circ\text{ with the wall, find the height of the wall.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the wall }=h\text{ m.}$
$\displaystyle \text{Length of the ladder }=15\text{ m} \\ \text{and the angle between the ladder and the wall }=60^\circ.$
$\displaystyle \cos60^\circ=\frac{h}{15}$
$\displaystyle \frac{1}{2}=\frac{h}{15}$
$\displaystyle h=\frac{15}{2}=7.5\text{ m}$
$\displaystyle \therefore \text{The height of the wall is }7.5\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{A vertical tower stands on a horizontal plane and is surmounted by a vertical}$
$\displaystyle \text{flag-staff. At a point on the plane }70\text{ metres away from the tower, an observer notices}$
$\displaystyle \text{that the angles of elevation of the top and the bottom of the flag-staff are respectively }$
$\displaystyle 60^\circ\text{ and }45^\circ. \ \text{Find the height of the flag-staff and that of the tower.}\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower }=h\text{ m and the height of the flag-staff }=x\text{ m.}$
$\displaystyle \text{Distance of the observation point from the tower }=70\text{ m.}$
$\displaystyle \tan45^\circ=\frac{h}{70}$
$\displaystyle 1=\frac{h}{70}$
$\displaystyle h=70\text{ m}$
$\displaystyle \text{Now, the total height of the} \\ \text{tower and the flag-staff }=(h+x)\text{ m.}$
$\displaystyle \tan60^\circ=\frac{h+x}{70}$
$\displaystyle \sqrt{3}=\frac{70+x}{70}$
$\displaystyle 70\sqrt{3}=70+x$
$\displaystyle x=70(\sqrt{3}-1)\text{ m}$
$\displaystyle \therefore \text{The height of the flag-staff is }70(\sqrt{3}-1)\text{ m and the height of the tower is }70\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Two points }A\text{ and }B\text{ are on the same side of a tower and in the}$
$\displaystyle \text{same straight line with its base. The angles of depression of these points from the top}$
$\displaystyle \text{of the tower are }60^\circ\text{ and }45^\circ\text{ respectively. If the height of the tower is }15\text{ m, then}$
$\displaystyle \text{find the distance between these points.}\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distances of points }A\text{ and }B\text{ from the foot of the tower be }x\text{ m and }y\text{ m respectively.}$
$\displaystyle \text{The corresponding angles of elevation at }A\text{ and }B\text{ are }60^\circ\text{ and }45^\circ.$
$\displaystyle \tan60^\circ=\frac{15}{x}$
$\displaystyle \sqrt{3}=\frac{15}{x}$
$\displaystyle x=\frac{15}{\sqrt{3}}=5\sqrt{3}\text{ m}$
$\displaystyle \tan45^\circ=\frac{15}{y}$
$\displaystyle 1=\frac{15}{y}$
$\displaystyle y=15\text{ m}$
$\displaystyle \therefore \text{Distance between the two points}=y-x$
$\displaystyle =15-5\sqrt{3}=5(3-\sqrt{3})\text{ m}$
$\displaystyle \therefore \text{The distance between the two points is }5(3-\sqrt{3})\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{A vertical tower stands on a horizontal plane and is surmounted by a vertical}$
$\displaystyle \text{flag-staff of height }5\text{ metres. At a point on the plane, the angles of elevation of the bottom}$
$\displaystyle \text{and the top of the flag-staff are respectively }30^\circ\text{ and }60^\circ.\text{ Find the height of the}$
$\displaystyle \text{tower.}\hfill\text{[CBSE 2015, 16, 19, 20]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower }=h\text{ m} \\ \text{and the distance of the observation point from the tower }=x\text{ m.}$
$\displaystyle \tan30^\circ=\frac{h}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{x}$
$\displaystyle x=\sqrt{3}h$
$\displaystyle \text{The total height of the tower and} \\ \text{the flag-staff }=(h+5)\text{ m.}$
$\displaystyle \tan60^\circ=\frac{h+5}{x}$
$\displaystyle \sqrt{3}=\frac{h+5}{\sqrt{3}h}$
$\displaystyle 3h=h+5$
$\displaystyle 2h=5$
$\displaystyle h=\frac{5}{2}=2.5\text{ m}$
$\displaystyle \therefore \text{The height of the tower is }2.5\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{The angle of elevation of a tower from a point on the same level as the foot}$
$\displaystyle \text{of the tower is }30^\circ.\text{ On advancing }150\text{ metres towards the foot of the tower, the angle of}$
$\displaystyle \text{elevation of the tower becomes }60^\circ.\text{ Show that the height of the tower is }129.9\text{ metres.}$
$\displaystyle \text{(Use }\sqrt{3}=1.732\text{).}\hfill\text{[CBSE 2006]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distance of the nearer point from the foot of the tower }=x\text{ m.}$
$\displaystyle \therefore \text{Distance of the farther point from the foot of the tower }=(x+150)\text{ m.}$
$\displaystyle \text{Let the height of the tower }=h\text{ m.}$
$\displaystyle \tan60^\circ=\frac{h}{x}$
$\displaystyle \sqrt{3}=\frac{h}{x}$
$\displaystyle h=\sqrt{3}x \qquad ...(i)$
$\displaystyle \tan30^\circ=\frac{h}{x+150}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{x+150}$
$\displaystyle h=\frac{x+150}{\sqrt{3}} \qquad ...(ii)$
$\displaystyle \text{From (i) and (ii),}$
$\displaystyle \sqrt{3}x=\frac{x+150}{\sqrt{3}}$
$\displaystyle 3x=x+150$
$\displaystyle 2x=150$
$\displaystyle x=75$
$\displaystyle \therefore h=75\sqrt{3}=75\times1.732=129.9\text{ m}$
$\displaystyle \therefore \text{The height of the tower is }129.9\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{The angle of elevation of the top of a tower as observed from a point in a}$
$\displaystyle \text{horizontal plane through the foot of the tower is }32^\circ.\text{ When the observer moves towards}$
$\displaystyle \text{the tower a distance of }100\text{ m, he finds the angle of elevation of the top to be }63^\circ.$
$\displaystyle \text{Find the height of the tower and the distance of the first position from the tower.}$
$\displaystyle \text{[Take }\tan32^\circ=0.6248\text{ and }\tan63^\circ=1.9626\text{].}\hfill\text{[CBSE 2001C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distance of the first position from the foot of the tower }=x\text{ m.}$
$\displaystyle \therefore \text{Distance after moving }100\text{ m towards the tower }=(x-100)\text{ m.}$
$\displaystyle \text{Let the height of the tower }=h\text{ m.}$
$\displaystyle \tan32^\circ=\frac{h}{x}$
$\displaystyle 0.6248=\frac{h}{x}$
$\displaystyle h=0.6248x \qquad ...(i)$
$\displaystyle \tan63^\circ=\frac{h}{x-100}$
$\displaystyle 1.9626=\frac{h}{x-100}$
$\displaystyle h=1.9626(x-100) \qquad ...(ii)$
$\displaystyle \text{From (i) and (ii),}$
$\displaystyle 0.6248x=1.9626(x-100)$
$\displaystyle 0.6248x=1.9626x-196.26$
$\displaystyle 1.3378x=196.26$
$\displaystyle x=\frac{196.26}{1.3378}=146.70\text{ m (approx.).}$
$\displaystyle \text{From (i),}$
$\displaystyle h=0.6248\times146.70=91.66\text{ m (approx.).}$
$\displaystyle \therefore \text{The height of the tower is }91.66\text{ m and the distance of the first position from the tower}$
$\displaystyle \text{is }146.70\text{ m (approx.).}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{The angle of elevation of the top of a tower from a point }A\text{ on the}$
$\displaystyle \text{ground is }30^\circ.\text{ On moving a distance of }20\text{ metres towards the foot of the tower to a}$
$\displaystyle \text{point }B,\text{ the angle of elevation increases to }60^\circ.\text{ Find the height of the tower and the}$
$\displaystyle \text{distance of the tower from point }A.\hfill\text{[CBSE 2002, 15, 17]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distance of point }B\text{ from the foot of the tower }=x\text{ m.}$
$\displaystyle \therefore \text{Distance of point }A\text{ from the foot of the tower }=(x+20)\text{ m.}$
$\displaystyle \text{Let the height of the tower }=h\text{ m.}$
$\displaystyle \tan60^\circ=\frac{h}{x}$
$\displaystyle \sqrt{3}=\frac{h}{x}$
$\displaystyle h=\sqrt{3}x \qquad ...(i)$
$\displaystyle \tan30^\circ=\frac{h}{x+20}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{x+20}$
$\displaystyle h=\frac{x+20}{\sqrt{3}} \qquad ...(ii)$
$\displaystyle \text{From (i) and (ii),}$
$\displaystyle \sqrt{3}x=\frac{x+20}{\sqrt{3}}$
$\displaystyle 3x=x+20$
$\displaystyle 2x=20$
$\displaystyle x=10$
$\displaystyle \therefore h=10\sqrt{3}\text{ m}$
$\displaystyle \text{Distance of the tower from point }A=x+20=10+20=30\text{ m.}$
$\displaystyle \therefore \text{The height of the tower is }10\sqrt{3}\text{ m and its distance from point }A\text{ is }30\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{From the top of a building }15\text{ m high, the angle of elevation of the}$
$\displaystyle \text{top of a tower is }30^\circ.\text{ From the bottom of the same building, the angle of elevation of the}$
$\displaystyle \text{top of the tower is }60^\circ.\text{ Find the height of the tower and the distance between the tower}$
$\displaystyle \text{and the building.}\hfill\text{[CBSE 2002]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower }=h\text{ m and the distance between the tower and building }=x\text{ m.}$
$\displaystyle \text{From the bottom of the building,}$
$\displaystyle \tan60^\circ=\frac{h}{x}$
$\displaystyle \sqrt{3}=\frac{h}{x}$
$\displaystyle h=\sqrt{3}x \qquad ...(i)$
$\displaystyle \text{From the top of the building, the vertical} \\ \text{height above its top }=(h-15)\text{ m.}$
$\displaystyle \tan30^\circ=\frac{h-15}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h-15}{x}$
$\displaystyle h-15=\frac{x}{\sqrt{3}} \qquad ...(ii)$
$\displaystyle \text{Substituting }h=\sqrt{3}x\text{ from (i) in (ii),}$
$\displaystyle \sqrt{3}x-15=\frac{x}{\sqrt{3}}$
$\displaystyle 3x-15\sqrt{3}=x$
$\displaystyle 2x=15\sqrt{3}$
$\displaystyle x=\frac{15\sqrt{3}}{2}\text{ m}$
$\displaystyle \text{From (i),}$
$\displaystyle h=\sqrt{3}\times\frac{15\sqrt{3}}{2}=\frac{45}{2}=22.5\text{ m}$
$\displaystyle \therefore \text{The height of the tower is }22.5\text{ m and the distance between the tower and the building is}$
$\displaystyle \frac{15\sqrt{3}}{2}\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{On a horizontal plane there is a vertical tower with a flag pole on the}$
$\displaystyle \text{top of the tower. At a point }9\text{ metres away from the foot of the tower, the angles of}$
$\displaystyle \text{elevation of the top and bottom of the flag pole are }60^\circ\text{ and }30^\circ\text{ respectively. Find the}$
$\displaystyle \text{height of the tower and the flag pole mounted on it.}\hfill\text{[CBSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower }=h\text{ m and the height of the flag pole }=x\text{ m.}$
$\displaystyle \text{Distance of the observation point from the foot of the tower }=9\text{ m.}$
$\displaystyle \tan30^\circ=\frac{h}{9}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{9}$
$\displaystyle h=\frac{9}{\sqrt{3}}=3\sqrt{3}\text{ m}$
$\displaystyle \text{The total height of the} \\ \text{tower and the flag pole }=(h+x)\text{ m.}$
$\displaystyle \tan60^\circ=\frac{h+x}{9}$
$\displaystyle \sqrt{3}=\frac{h+x}{9}$
$\displaystyle h+x=9\sqrt{3}$
$\displaystyle x=9\sqrt{3}-3\sqrt{3}=6\sqrt{3}\text{ m}$
$\displaystyle \therefore \text{The height of the tower is }3\sqrt{3}\text{ m and the height of the flag pole is }6\sqrt{3}\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{An observer, }1.5\text{ m tall, is }28.5\text{ m away from a tower }30\text{ m high.}$
$\displaystyle \text{Determine the angle of elevation of the top of the tower from his eye.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the angle of elevation of the top of the tower from the observer's eye be }\theta.$
$\displaystyle \text{Vertical height of the top of the tower above the observer's eye}=30-1.5=28.5\text{ m.}$
$\displaystyle \text{Horizontal distance of the observer from the tower }=28.5\text{ m.}$
$\displaystyle \tan\theta=\frac{28.5}{28.5}=1$
$\displaystyle \tan\theta=\tan45^\circ$
$\displaystyle \therefore \theta=45^\circ$
$\displaystyle \therefore \text{The angle of elevation of the top of the tower from the observer's eye is }45^\circ.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{A }1.5\text{ m tall boy is standing at some distance from a }30\text{ m tall}$
$\displaystyle \text{building. The angle of elevation from his eyes to the top of the building increases from }30^\circ$
$\displaystyle \text{to }60^\circ\text{ as he walks towards the building. Find the distance he walked towards the}$
$\displaystyle \text{building.}\hfill\text{[CBSE 2014, 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Height of the building above the boy's eye level}=30-1.5=28.5\text{ m.}$
$\displaystyle \text{Let the initial distance of the boy from the building }=x\text{ m.}$
$\displaystyle \text{Let the final distance of the boy from the building }=y\text{ m.}$
$\displaystyle \tan30^\circ=\frac{28.5}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{28.5}{x}$
$\displaystyle x=28.5\sqrt{3}\text{ m}$
$\displaystyle \tan60^\circ=\frac{28.5}{y}$
$\displaystyle \sqrt{3}=\frac{28.5}{y}$
$\displaystyle y=\frac{28.5}{\sqrt{3}}=9.5\sqrt{3}\text{ m}$
$\displaystyle \therefore \text{Distance walked}=x-y$
$\displaystyle =28.5\sqrt{3}-9.5\sqrt{3}=19\sqrt{3}\text{ m}$
$\displaystyle \therefore \text{The boy walked }19\sqrt{3}\text{ m, i.e. approximately }32.91\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{The shadow of a tower standing on a level ground is found to be }40\text{ m}$
$\displaystyle \text{longer when Sun's altitude is }30^\circ\text{ than when it was }60^\circ.\text{ Find the height of the}$
$\displaystyle \text{tower.}\hfill\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower }=h\text{ m.}$
$\displaystyle \text{Let the lengths of its shadows at }30^\circ\text{ and }60^\circ\text{ be }x\text{ m and }y\text{ m respectively.}$
$\displaystyle \tan30^\circ=\frac{h}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{x}$
$\displaystyle x=h\sqrt{3}$
$\displaystyle \tan60^\circ=\frac{h}{y}$
$\displaystyle \sqrt{3}=\frac{h}{y}$
$\displaystyle y=\frac{h}{\sqrt{3}}$
$\displaystyle \text{Given, }x-y=40$
$\displaystyle h\sqrt{3}-\frac{h}{\sqrt{3}}=40$
$\displaystyle \frac{3h-h}{\sqrt{3}}=40$
$\displaystyle \frac{2h}{\sqrt{3}}=40$
$\displaystyle h=20\sqrt{3}\text{ m}$
$\displaystyle \therefore \text{The height of the tower is }20\sqrt{3}\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{From a point on the ground, the angles of elevation of the bottom and top}$
$\displaystyle \text{of a transmission tower fixed at the top of a }20\text{ m high building are }45^\circ\text{ and }60^\circ$
$\displaystyle \text{respectively. Find the height of the transmission tower.}\hfill\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the transmission tower }=h\text{ m.}$
$\displaystyle \text{Let the distance of the observation} \\ \text{point from the building }=x\text{ m.}$
$\displaystyle \tan45^\circ=\frac{20}{x}$
$\displaystyle 1=\frac{20}{x}$
$\displaystyle x=20\text{ m}$
$\displaystyle \text{Total height of the building and the} \\ \text{transmission tower }=(20+h)\text{ m.}$
$\displaystyle \tan60^\circ=\frac{20+h}{20}$
$\displaystyle \sqrt{3}=\frac{20+h}{20}$
$\displaystyle 20\sqrt{3}=20+h$
$\displaystyle h=20\sqrt{3}-20=20(\sqrt{3}-1)\text{ m}$
$\displaystyle \therefore \text{The height of the transmission tower is }20(\sqrt{3}-1)\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{The angles of depression of the top and bottom of an }8\text{ m tall}$
$\displaystyle \text{building from the top of a multistoried building are }30^\circ\text{ and }45^\circ\text{ respectively. Find}$
$\displaystyle \text{the height of the multistoried building and the distance between the two buildings.}$
$\displaystyle \hfill\text{[CBSE 2009, 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the multistoried building }=h\text{ m.}$
$\displaystyle \text{Let the distance between the two buildings }=x\text{ m.}$
$\displaystyle \text{The corresponding angles of elevation from the bottom and top of the }8\text{ m building are }45^\circ\text{ and }30^\circ.$
$\displaystyle \tan45^\circ=\frac{h}{x}$
$\displaystyle 1=\frac{h}{x}$
$\displaystyle x=h \qquad ...(i)$
$\displaystyle \tan30^\circ=\frac{h-8}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h-8}{x}$
$\displaystyle \text{Using }x=h\text{ from (i),}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h-8}{h}$
$\displaystyle \sqrt{3}(h-8)=h$
$\displaystyle h(\sqrt{3}-1)=8\sqrt{3}$
$\displaystyle h=\frac{8\sqrt{3}}{\sqrt{3}-1}$
$\displaystyle =\frac{8\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$
$\displaystyle =4\sqrt{3}(\sqrt{3}+1)=12+4\sqrt{3}=4(3+\sqrt{3})\text{ m}$
$\displaystyle \therefore x=h=4(3+\sqrt{3})\text{ m}$
$\displaystyle \therefore \text{The height of the multistoried building and the distance between the two buildings are each}$
$\displaystyle 4(3+\sqrt{3})\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{A statue }1.6\text{ m tall stands on the top of a pedestal. From a point on the}$
$\displaystyle \text{ground, the angle of elevation of the top of the statue is }60^\circ\text{ and from the same point,}$
$\displaystyle \text{the angle of elevation of the top of the pedestal is }45^\circ.\text{ Find the height of the}$
$\displaystyle \text{pedestal.}\hfill\text{[CBSE 2008, 14]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the pedestal }=h\text{ m and the distance of the observation point from it }=x\text{ m.}$
$\displaystyle \tan45^\circ=\frac{h}{x}$
$\displaystyle 1=\frac{h}{x}$
$\displaystyle x=h \qquad ...(i)$
$\displaystyle \text{Total height of the pedestal and} \\ \text{the statue }=(h+1.6)\text{ m.}$
$\displaystyle \tan60^\circ=\frac{h+1.6}{x}$
$\displaystyle \sqrt{3}=\frac{h+1.6}{h}\qquad [\text{Using (i)}]$
$\displaystyle \sqrt{3}h=h+1.6$
$\displaystyle h(\sqrt{3}-1)=1.6$
$\displaystyle h=\frac{1.6}{\sqrt{3}-1}$
$\displaystyle =\frac{1.6(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$
$\displaystyle =0.8(\sqrt{3}+1)\text{ m}$
$\displaystyle \therefore \text{The height of the pedestal is }0.8(\sqrt{3}+1)\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{From the top of a }120\text{ m high tower, a man observes two cars on the}$
$\displaystyle \text{opposite sides of the tower and in a straight line with the base of the tower, with angles of}$
$\displaystyle \text{depression }60^\circ\text{ and }45^\circ.\text{ Find the distance between the cars.}$
$\displaystyle \text{(Take }\sqrt{3}=1.732\text{)}\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distances of the two cars from the foot of the tower be }x\text{ m and }y\text{ m respectively.}$
$\displaystyle \text{The corresponding angles of elevation of the top of the tower are }60^\circ\text{ and }45^\circ.$
$\displaystyle \tan60^\circ=\frac{120}{x}$
$\displaystyle \sqrt{3}=\frac{120}{x}$
$\displaystyle x=\frac{120}{\sqrt{3}}=40\sqrt{3}\text{ m}$
$\displaystyle \tan45^\circ=\frac{120}{y}$
$\displaystyle 1=\frac{120}{y}$
$\displaystyle y=120\text{ m}$
$\displaystyle \text{Since the cars are on opposite sides of the tower,}$
$\displaystyle \text{Distance between the cars}=x+y$
$\displaystyle =40\sqrt{3}+120$
$\displaystyle =40(1.732)+120=69.28+120=189.28\text{ m}$
$\displaystyle \therefore \text{The distance between the two cars is }189.28\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{From the top of a }7\text{ m high building, the angle of elevation of the top of a}$
$\displaystyle \text{cable tower is }60^\circ\text{ and the angle of depression of its foot is }45^\circ.\text{ Determine the height}$
$\displaystyle \text{of the tower.}\hfill\text{[CBSE 2014, 17, 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distance between the building and the tower }=x\text{ m.}$
$\displaystyle \text{Since the angle of depression of the foot of the tower is }45^\circ,$
$\displaystyle \tan45^\circ=\frac{7}{x}$
$\displaystyle 1=\frac{7}{x}$
$\displaystyle x=7\text{ m}$
$\displaystyle \text{Let the height of the tower above} \\ \text{the level of the building }=h\text{ m.}$
$\displaystyle \tan60^\circ=\frac{h}{7}$
$\displaystyle \sqrt{3}=\frac{h}{7}$
$\displaystyle h=7\sqrt{3}\text{ m}$
$\displaystyle \therefore \text{Height of the tower}=7+7\sqrt{3}=7(\sqrt{3}+1)\text{ m.}$
$\displaystyle \therefore \text{The height of the tower is }7(\sqrt{3}+1)\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{As observed from the top of a }75\text{ m tall lighthouse, the angles of}$
$\displaystyle \text{depression of two ships are }30^\circ\text{ and }45^\circ.\text{ If one ship is exactly behind the other on the}$
$\displaystyle \text{same side of the lighthouse, find the distance between the two ships.}$
$\displaystyle \hfill\text{[CBSE 2013, 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distances of the ships corresponding to angles }30^\circ\text{ and }45^\circ\text{ be }x\text{ m and }y\text{ m respectively.}$
$\displaystyle \tan30^\circ=\frac{75}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{75}{x}$
$\displaystyle x=75\sqrt{3}\text{ m}$
$\displaystyle \tan45^\circ=\frac{75}{y}$
$\displaystyle 1=\frac{75}{y}$
$\displaystyle y=75\text{ m}$
$\displaystyle \text{Since the ships are on the same side of the lighthouse,}$
$\displaystyle \text{Distance between the ships}=x-y$
$\displaystyle =75\sqrt{3}-75=75(\sqrt{3}-1)\text{ m}$
$\displaystyle \therefore \text{The distance between the two ships is }75(\sqrt{3}-1)\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{The angle of elevation of the top of the building from the foot of the}$
$\displaystyle \text{tower is }30^\circ\text{ and the angle of elevation of the top of the tower from the foot of the}$
$\displaystyle \text{building is }60^\circ.\text{ If the tower is }50\text{ m high, find the height of the building.}$
$\displaystyle \hfill\text{[CBSE 2012, 15, 17]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distance between the tower and the building }=x\text{ m.}$
$\displaystyle \text{Let the height of the building }=h\text{ m.}$
$\displaystyle \tan60^\circ=\frac{50}{x}$
$\displaystyle \sqrt{3}=\frac{50}{x}$
$\displaystyle x=\frac{50}{\sqrt{3}}\text{ m}$
$\displaystyle \tan30^\circ=\frac{h}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{50/\sqrt{3}}$
$\displaystyle h=\frac{50}{3}\text{ m}$
$\displaystyle \therefore \text{The height of the building is }\frac{50}{3}\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{From a point on a bridge across a river, the angles of depression of the}$
$\displaystyle \text{banks on opposite sides of the river are }30^\circ\text{ and }45^\circ\text{ respectively. If the bridge is at a}$
$\displaystyle \text{height of }30\text{ m from the banks, find the width of the river.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the horizontal distances of the two banks from the point vertically below the observer be }x\text{ m and }y\text{ m.}$
$\displaystyle \text{The corresponding angles of elevation from the banks are }30^\circ\text{ and }45^\circ.$
$\displaystyle \tan30^\circ=\frac{30}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{30}{x}$
$\displaystyle x=30\sqrt{3}\text{ m}$
$\displaystyle \tan45^\circ=\frac{30}{y}$
$\displaystyle 1=\frac{30}{y}$
$\displaystyle y=30\text{ m}$
$\displaystyle \text{Since the banks are on opposite sides,}$
$\displaystyle \text{Width of the river}=x+y$
$\displaystyle =30\sqrt{3}+30=30(\sqrt{3}+1)\text{ m}$
$\displaystyle \therefore \text{The width of the river is }30(\sqrt{3}+1)\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{The angle of elevation of the top of a hill at the foot of a tower is }60^\circ$
$\displaystyle \text{and the angle of elevation of the top of the tower from the foot of the hill is }30^\circ.\text{ If the}$
$\displaystyle \text{tower is }50\text{ m high, what is the height of the hill?}\hfill\text{[CBSE 2006C, 13]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distance between the foot of the tower and the foot of the hill }=x\text{ m.}$
$\displaystyle \text{Let the height of the hill }=h\text{ m.}$
$\displaystyle \tan30^\circ=\frac{50}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{50}{x}$
$\displaystyle x=50\sqrt{3}\text{ m}$
$\displaystyle \tan60^\circ=\frac{h}{x}$
$\displaystyle \sqrt{3}=\frac{h}{50\sqrt{3}}$
$\displaystyle h=50\sqrt{3}\times\sqrt{3}=150\text{ m}$
$\displaystyle \therefore \text{The height of the hill is }150\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{(i) Two boats approach a light house in mid-sea from opposite directions.}$
$\displaystyle \text{The angles of elevation of the top of the light house from two boats are }30^\circ\text{ and }45^\circ$
$\displaystyle \text{respectively. If the distance between two boats is }100\text{ m, find the height of the light house.}$
$\displaystyle \hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the light house }=h\text{ m.}$
$\displaystyle \text{Let the distances of the boats from the foot of the light house be }x\text{ m and }y\text{ m respectively.}$
$\displaystyle \tan30^\circ=\frac{h}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{x}$
$\displaystyle x=h\sqrt{3}$
$\displaystyle \tan45^\circ=\frac{h}{y}$
$\displaystyle 1=\frac{h}{y}$
$\displaystyle y=h$
$\displaystyle \text{Since the boats are on opposite sides of the light house,}$
$\displaystyle x+y=100$
$\displaystyle h\sqrt{3}+h=100$
$\displaystyle h(\sqrt{3}+1)=100$
$\displaystyle h=\frac{100}{\sqrt{3}+1}$
$\displaystyle =\frac{100(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=50(\sqrt{3}-1)\text{ m}$
$\displaystyle \therefore \text{The height of the light house is }50(\sqrt{3}-1)\text{ m.}$
$\displaystyle \\$

$\displaystyle \text{(ii) From the top of a }45\text{ m high lighthouse, the angles of depression of two ships on}$
$\displaystyle \text{the opposite sides of it are observed to be }30^\circ\text{ and }60^\circ.\text{ If the line joining the ships}$
$\displaystyle \text{passes through the foot of the lighthouse, find the distance between the ships.}$
$\displaystyle \text{(Use }\sqrt{3}=1.73\text{)}\hfill\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distances of the ships from the foot of the lighthouse be }x\text{ m and }y\text{ m respectively.}$
$\displaystyle \text{The corresponding angles of elevation of the top of the lighthouse are }30^\circ\text{ and }60^\circ.$
$\displaystyle \tan30^\circ=\frac{45}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{45}{x}$
$\displaystyle x=45\sqrt{3}\text{ m}$
$\displaystyle \tan60^\circ=\frac{45}{y}$
$\displaystyle \sqrt{3}=\frac{45}{y}$
$\displaystyle y=\frac{45}{\sqrt{3}}=15\sqrt{3}\text{ m}$
$\displaystyle \text{Since the ships are on opposite sides of the lighthouse,}$
$\displaystyle \text{Distance between the ships}=x+y$
$\displaystyle =45\sqrt{3}+15\sqrt{3}=60\sqrt{3}$
$\displaystyle =60\times1.73=103.8\text{ m}$
$\displaystyle \therefore \text{The distance between the two ships is }103.8\text{ m.}$

$\displaystyle \text{(iii) Two ships are sailing in the sea on either side of a lighthouse. The angles of depression}$
$\displaystyle \text{to the two ships as observed from the top of the lighthouse are }60^\circ\text{ and }45^\circ\text{ respectively.}$
$\displaystyle \text{If the distance between the ships is }100\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)\text{ m, then find the height of the}$
$\displaystyle \text{lighthouse.}\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the lighthouse }=h\text{ m.}$
$\displaystyle \text{Let the distances of the ships from the foot of the lighthouse be }x\text{ m and }y\text{ m respectively.}$
$\displaystyle \tan60^\circ=\frac{h}{x}$
$\displaystyle \sqrt{3}=\frac{h}{x}$
$\displaystyle x=\frac{h}{\sqrt{3}}$
$\displaystyle \tan45^\circ=\frac{h}{y}$
$\displaystyle 1=\frac{h}{y}$
$\displaystyle y=h$
$\displaystyle \text{Since the ships are on either side of the lighthouse,}$
$\displaystyle x+y=100\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)$
$\displaystyle \frac{h}{\sqrt{3}}+h=100\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)$
$\displaystyle \frac{h(1+\sqrt{3})}{\sqrt{3}}=\frac{100(1+\sqrt{3})}{\sqrt{3}}$
$\displaystyle h=100\text{ m}$
$\displaystyle \therefore \text{The height of the lighthouse is }100\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{Two men on either side of a cliff }80\text{ m high observe the angles of}$
$\displaystyle \text{elevation of the top of the cliff to be }30^\circ\text{ and }60^\circ\text{ respectively. Find the distance}$
$\displaystyle \text{between the two men.}\hfill\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distances of the two men from the foot of the cliff be }x\text{ m and }y\text{ m respectively.}$
$\displaystyle \tan30^\circ=\frac{80}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{80}{x}$
$\displaystyle x=80\sqrt{3}\text{ m}$
$\displaystyle \tan60^\circ=\frac{80}{y}$
$\displaystyle \sqrt{3}=\frac{80}{y}$
$\displaystyle y=\frac{80}{\sqrt{3}}\text{ m}$
$\displaystyle \text{Since the two men are on either side of the cliff,}$
$\displaystyle \text{Distance between the two men}=x+y$
$\displaystyle =80\sqrt{3}+\frac{80}{\sqrt{3}}$
$\displaystyle =\frac{240+80}{\sqrt{3}}=\frac{320}{\sqrt{3}}=\frac{320\sqrt{3}}{3}\text{ m}$
$\displaystyle \therefore \text{The distance between the two men is }\frac{320\sqrt{3}}{3}\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{An aeroplane is flying at a height of }210\text{ m. Flying at this height, at some}$
$\displaystyle \text{instant the angles of depression of two points in a line in opposite directions on both the}$
$\displaystyle \text{banks of the river are }45^\circ\text{ and }60^\circ.\text{ Find the width of the river.}$
$\displaystyle \text{(Use }\sqrt{3}=1.73\text{).}\hfill\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the horizontal distances of the two banks from the point vertically below the aeroplane be }x\text{ m and }y\text{ m.}$
$\displaystyle \text{The corresponding angles of elevation of the aeroplane from the banks are }45^\circ\text{ and }60^\circ.$
$\displaystyle \tan45^\circ=\frac{210}{x}$
$\displaystyle 1=\frac{210}{x}$
$\displaystyle x=210\text{ m}$
$\displaystyle \tan60^\circ=\frac{210}{y}$
$\displaystyle \sqrt{3}=\frac{210}{y}$
$\displaystyle y=\frac{210}{\sqrt{3}}=70\sqrt{3}\text{ m}$
$\displaystyle \text{Since the two banks are in opposite directions,}$
$\displaystyle \text{Width of the river}=x+y$
$\displaystyle =210+70\sqrt{3}$
$\displaystyle =210+70(1.73)=210+121.1=331.1\text{ m}$
$\displaystyle \therefore \text{The width of the river is }331.1\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{(i) A flag-staff stands on the top of a }5\text{ m high tower. From a point on}$
$\displaystyle \text{the ground, the angle of elevation of the top of the flag-staff is }60^\circ\text{ and from the same}$
$\displaystyle \text{point, the angle of elevation of the top of the tower is }45^\circ.\text{ Find the height of the}$
$\displaystyle \text{flag-staff.}\hfill\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the flag-staff }=h\text{ m and the distance of the point from the tower }=x\text{ m.}$
$\displaystyle \tan45^\circ=\frac{5}{x}$
$\displaystyle 1=\frac{5}{x}$
$\displaystyle x=5\text{ m}$
$\displaystyle \text{Total height of the tower} \\ \text{and the flag-staff }=(5+h)\text{ m.}$
$\displaystyle \tan60^\circ=\frac{5+h}{5}$
$\displaystyle \sqrt{3}=\frac{5+h}{5}$
$\displaystyle 5\sqrt{3}=5+h$
$\displaystyle h=5(\sqrt{3}-1)\text{ m}$
$\displaystyle \therefore \text{The height of the flag-staff is }5(\sqrt{3}-1)\text{ m.}$

$\displaystyle \text{(ii) A pole }6\text{ m high is fixed on the top of a tower. The angle of elevation of the top of}$
$\displaystyle \text{the pole observed from a point }P\text{ on the ground is }60^\circ\text{ and the angle of depression of}$
$\displaystyle \text{the point }P\text{ from the top of the tower is }45^\circ.\text{ Find the height of the tower and the}$
$\displaystyle \text{distance of point }P\text{ from the foot of the tower. (Use }\sqrt{3}=1.73\text{).}\hfill\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower }=h\text{ m and the distance of point }P\text{ from its foot }=x\text{ m.}$
$\displaystyle \text{The angle of elevation of the top of the tower from }P=45^\circ.$
$\displaystyle \tan45^\circ=\frac{h}{x}$
$\displaystyle 1=\frac{h}{x}$
$\displaystyle x=h \qquad ...(i)$
$\displaystyle \text{Total height of the tower} \\ \text{and the pole }=(h+6)\text{ m.}$
$\displaystyle \tan60^\circ=\frac{h+6}{x}$
$\displaystyle \sqrt{3}=\frac{h+6}{h}\qquad [\text{Using (i)}]$
$\displaystyle \sqrt{3}h=h+6$
$\displaystyle h(\sqrt{3}-1)=6$
$\displaystyle h=\frac{6}{\sqrt{3}-1}$
$\displaystyle =\frac{6(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=3(\sqrt{3}+1)\text{ m}$
$\displaystyle h=3(1.73+1)=8.19\text{ m}$
$\displaystyle \therefore x=h=8.19\text{ m}$
$\displaystyle \therefore \text{The height of the tower and the distance of point }P\text{ from its foot are each }8.19\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{The horizontal distance between two poles is }15\text{ m. The angle of}$
$\displaystyle \text{depression of the top of the first pole as seen from the top of the second pole is }30^\circ.$
$\displaystyle \text{If the height of the second pole is }24\text{ m, find the height of the first pole.}$
$\displaystyle \text{(}\sqrt{3}=1.732\text{)}\hfill\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the first pole }=h\text{ m.}$
$\displaystyle \text{Difference in the heights} \\ \text{of the two poles }=(24-h)\text{ m.}$
$\displaystyle \tan30^\circ=\frac{24-h}{15}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{24-h}{15}$
$\displaystyle 24-h=\frac{15}{\sqrt{3}}=5\sqrt{3}$
$\displaystyle h=24-5\sqrt{3}$
$\displaystyle =24-5(1.732)=24-8.66=15.34\text{ m}$
$\displaystyle \therefore \text{The height of the first pole is }15.34\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{The angles of depression of two ships from the top of a light house}$
$\displaystyle \text{and on the same side of it are found to be }45^\circ\text{ and }30^\circ\text{ respectively. If the ships are}$
$\displaystyle 200\text{ m apart, find the height of the light house.}\hfill\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the light house }=h\text{ m.}$
$\displaystyle \text{Let the distances of the nearer and farther ships from its foot be }x\text{ m and }y\text{ m respectively.}$
$\displaystyle \tan45^\circ=\frac{h}{x}$
$\displaystyle 1=\frac{h}{x}$
$\displaystyle x=h$
$\displaystyle \tan30^\circ=\frac{h}{y}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{y}$
$\displaystyle y=h\sqrt{3}$
$\displaystyle \text{Since the ships are }200\text{ m apart,}$
$\displaystyle y-x=200$
$\displaystyle h\sqrt{3}-h=200$
$\displaystyle h(\sqrt{3}-1)=200$
$\displaystyle h=\frac{200}{\sqrt{3}-1}$
$\displaystyle =\frac{200(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$
$\displaystyle =100(\sqrt{3}+1)\text{ m}$
$\displaystyle \therefore \text{The height of the light house is }100(\sqrt{3}+1)\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{The angles of elevation of the top of a tower from two points at}$
$\displaystyle \text{a distance of }4\text{ m and }9\text{ m from the base of the tower and in the same straight line with it}$
$\displaystyle \text{are complementary. Prove that the height of the tower is }6\text{ m.}\hfill\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower }=h\text{ m.}$
$\displaystyle \text{Let the angle of elevation from the point }4\text{ m from the tower be }\theta.$
$\displaystyle \therefore \text{The angle of elevation from the point }9\text{ m from the tower is }(90^\circ-\theta).$
$\displaystyle \tan\theta=\frac{h}{4} \qquad ...(i)$
$\displaystyle \tan(90^\circ-\theta)=\frac{h}{9}$
$\displaystyle \cot\theta=\frac{h}{9} \qquad ...(ii)$
$\displaystyle \text{Since }\tan\theta\cdot\cot\theta=1,$
$\displaystyle \frac{h}{4}\times\frac{h}{9}=1$
$\displaystyle \frac{h^2}{36}=1$
$\displaystyle h^2=36$
$\displaystyle h=6\text{ m}\qquad [\text{Since height is positive}]$
$\displaystyle \therefore \text{The height of the tower is }6\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }\text{From a point on a bridge across a river, the angles of depression of}$
$\displaystyle \text{the banks on opposite sides of the river are }30^\circ\text{ and }45^\circ.\text{ If the bridge is at a height}$
$\displaystyle \text{of }8\text{ m from the banks, then find the width of the river.}\hfill\text{[CBSE 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }AC=8\text{ m be the height of the bridge above the banks.}$
$\displaystyle \text{Let }BC=x\text{ m and }CD=y\text{ m.}$
$\displaystyle \text{The corresponding angles of elevation at }B\text{ and }D\text{ are }45^\circ\text{ and }30^\circ.$
$\displaystyle \tan45^\circ=\frac{AC}{BC}=\frac{8}{x}$
$\displaystyle 1=\frac{8}{x}$
$\displaystyle x=8\text{ m}$
$\displaystyle \tan30^\circ=\frac{AC}{CD}=\frac{8}{y}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{8}{y}$
$\displaystyle y=8\sqrt{3}\text{ m}$
$\displaystyle \text{Width of the river}=BC+CD$
$\displaystyle =8+8\sqrt{3}=8(1+\sqrt{3})\text{ m}$
$\displaystyle \therefore \text{The width of the river is }8(1+\sqrt{3})\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: }\text{The angle of elevation of the top of a chimney from the top of a}$
$\displaystyle \text{tower is }60^\circ\text{ and the angle of depression of the foot of the chimney from the top of the}$
$\displaystyle \text{tower is }30^\circ.\text{ If the height of the tower is }40\text{ m, find the height of the chimney.}$
$\displaystyle \text{According to pollution control norms, the minimum height of a smoke emitting chimney}$
$\displaystyle \text{should be }100\text{ m. State if the height of the above mentioned chimney meets the pollution}$
$\displaystyle \text{norms. What value is discussed in this question?}\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the horizontal distance between the tower and the chimney }=x\text{ m.}$
$\displaystyle \text{Height of the tower }=40\text{ m.}$
$\displaystyle \tan30^\circ=\frac{40}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{40}{x}$
$\displaystyle x=40\sqrt{3}\text{ m}$
$\displaystyle \text{Let the height of the chimney }=h\text{ m.}$
$\displaystyle \text{The height of the chimney above} \\ \text{the top of the tower }=(h-40)\text{ m.}$
$\displaystyle \tan60^\circ=\frac{h-40}{40\sqrt{3}}$
$\displaystyle \sqrt{3}=\frac{h-40}{40\sqrt{3}}$
$\displaystyle h-40=120$
$\displaystyle h=160\text{ m}$
$\displaystyle \therefore \text{The height of the chimney is }160\text{ m.}$
$\displaystyle \text{Since }160\text{ m}>100\text{ m, the chimney meets the prescribed pollution control norms.}$
$\displaystyle \therefore \text{The value discussed is environmental awareness and concern for pollution control.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 33: }\text{From the top of a building }AB,\ 60\text{ m high, the angles of depression}$
$\displaystyle \text{of the top and bottom of a vertical lamp post }CD\text{ are observed to be }30^\circ\text{ and }60^\circ$
$\displaystyle \text{respectively. Find}$
$\displaystyle \text{(i) the horizontal distance between }AB\text{ and }CD.$
$\displaystyle \text{(ii) the height of the lamp post.}$
$\displaystyle \text{(iii) the difference between the heights of the building and the lamp post.}$
$\displaystyle \text{(iv) the distance between the tops of the building and the lamp post.}$
$\displaystyle \hfill\text{[CBSE 2009, 2024]}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Let the horizontal distance between the building and the lamp post }=x\text{ m.}$
$\displaystyle \text{Height of the building }AB=60\text{ m.}$
$\displaystyle \text{The angle of elevation of the top of the building from }C\text{ is }60^\circ.$
$\displaystyle \tan60^\circ=\frac{60}{x}$
$\displaystyle \sqrt{3}=\frac{60}{x}$
$\displaystyle x=\frac{60}{\sqrt{3}}=20\sqrt{3}\text{ m}$
$\displaystyle \therefore \text{(i) The horizontal distance between }AB\text{ and }CD\text{ is }20\sqrt{3}\text{ m.}$

$\displaystyle \text{Let the height of the lamp post }CD=h\text{ m.}$
$\displaystyle \text{The vertical difference between the tops of the building and lamp post }=(60-h)\text{ m.}$
$\displaystyle \tan30^\circ=\frac{60-h}{20\sqrt{3}}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{60-h}{20\sqrt{3}}$
$\displaystyle 60-h=20$
$\displaystyle h=40\text{ m}$
$\displaystyle \therefore \text{(ii) The height of the lamp post is }40\text{ m.}$

$\displaystyle \text{Difference between the heights of the building and the lamp post}=60-40=20\text{ m.}$
$\displaystyle \therefore \text{(iii) The difference between their heights is }20\text{ m.}$

$\displaystyle \text{Let the distance between the tops of the building and the lamp post }=d\text{ m.}$
$\displaystyle \sin30^\circ=\frac{60-40}{d}$
$\displaystyle \frac{1}{2}=\frac{20}{d}$
$\displaystyle d=40\text{ m}$
$\displaystyle \therefore \text{(iv) The distance between the tops of the building and the lamp post is }40\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 34: }\text{A moving boat is observed from the top of a }150\text{ m high cliff, moving}$
$\displaystyle \text{away from the cliff. The angle of depression of the boat changes from }60^\circ\text{ to }45^\circ\text{ in}$
$\displaystyle 2\text{ minutes. Find the speed of the boat in m/h.}\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the initial and final distances of the boat from the foot of the cliff be }x\text{ m and }y\text{ m.}$
$\displaystyle \text{The corresponding angles of elevation of the top of the cliff are }60^\circ\text{ and }45^\circ.$
$\displaystyle \tan60^\circ=\frac{150}{x}$
$\displaystyle \sqrt{3}=\frac{150}{x}$
$\displaystyle x=\frac{150}{\sqrt{3}}=50\sqrt{3}\text{ m}$
$\displaystyle \tan45^\circ=\frac{150}{y}$
$\displaystyle 1=\frac{150}{y}$
$\displaystyle y=150\text{ m}$
$\displaystyle \text{Distance travelled by the boat}=y-x$
$\displaystyle =150-50\sqrt{3}\text{ m}$
$\displaystyle \text{Time taken}=2\text{ minutes}=\frac{2}{60}\text{ hour}=\frac{1}{30}\text{ hour}$
$\displaystyle \text{Speed of the boat}=\frac{150-50\sqrt{3}}{1/30}$
$\displaystyle =30(150-50\sqrt{3})=4500-1500\sqrt{3}\text{ m/h}$
$\displaystyle \approx 4500-1500(1.732)=1902\text{ m/h}$
$\displaystyle \therefore \text{The speed of the boat is }4500-1500\sqrt{3}\text{ m/h, i.e. approximately }1902\text{ m/h.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 35: }\text{A man in a boat rowing away from a light house }100\text{ m high takes}$
$\displaystyle 2\text{ minutes to change the angle of elevation of the top of the light house from }60^\circ\text{ to }30^\circ.$
$\displaystyle \text{Find the speed of the boat in metres per minute. (Use }\sqrt{3}=1.732\text{).}$
$\displaystyle \hfill\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the initial and final distances of the boat from the foot of the light house be }x\text{ m and }y\text{ m.}$
$\displaystyle \tan60^\circ=\frac{100}{x}$
$\displaystyle \sqrt{3}=\frac{100}{x}$
$\displaystyle x=\frac{100}{\sqrt{3}}\text{ m}$
$\displaystyle \tan30^\circ=\frac{100}{y}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{100}{y}$
$\displaystyle y=100\sqrt{3}\text{ m}$
$\displaystyle \text{Distance travelled by the boat}=y-x$
$\displaystyle =100\sqrt{3}-\frac{100}{\sqrt{3}}$
$\displaystyle =\frac{300-100}{\sqrt{3}}=\frac{200}{\sqrt{3}}\text{ m}$
$\displaystyle \text{Time taken}=2\text{ minutes}$
$\displaystyle \text{Speed of the boat}=\frac{200/\sqrt{3}}{2}=\frac{100}{\sqrt{3}}\text{ m/min}$
$\displaystyle =\frac{100\sqrt{3}}{3}=\frac{100\times1.732}{3}=57.74\text{ m/min (approx.).}$
$\displaystyle \therefore \text{The speed of the boat is }57.74\text{ m/min (approx.).}$
$\displaystyle \\$

$\displaystyle \textbf{Question 36: }\text{If the angle of elevation of a cloud from a point }h\text{ metres above a}$
$\displaystyle \text{lake is }\alpha\text{ and the angle of depression of its reflection in the lake is }\beta,\text{ prove that the}$
$\displaystyle \text{distance of the cloud from the point of observation is }\frac{2h\sec\alpha}{\tan\beta-\tan\alpha}.$
$\displaystyle \hfill\text{[CBSE 2004]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the cloud be }H\text{ metres above the surface of the lake.}$
$\displaystyle \text{Its reflection will be }H\text{ metres below the surface of the lake.}$
$\displaystyle \text{Let the horizontal distance of the cloud from the point vertically below the observer be }x\text{ metres.}$
$\displaystyle \text{The observer is }h\text{ metres above the surface of the lake.}$
$\displaystyle \therefore \tan\alpha=\frac{H-h}{x} \qquad ...(i)$
$\displaystyle \tan\beta=\frac{H+h}{x} \qquad ...(ii)$
$\displaystyle \text{Subtracting (i) from (ii),}$
$\displaystyle \tan\beta-\tan\alpha=\frac{H+h-H+h}{x}$
$\displaystyle \tan\beta-\tan\alpha=\frac{2h}{x}$
$\displaystyle x=\frac{2h}{\tan\beta-\tan\alpha} \qquad ...(iii)$
$\displaystyle \text{Let the distance of the cloud from the point of observation be }d.$
$\displaystyle \cos\alpha=\frac{x}{d}$
$\displaystyle d=x\sec\alpha$
$\displaystyle \text{Using (iii),}$
$\displaystyle d=\frac{2h\sec\alpha}{\tan\beta-\tan\alpha}$
$\displaystyle \therefore \text{The distance of the cloud from the point of observation is }\frac{2h\sec\alpha}{\tan\beta-\tan\alpha}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 37: }\text{From an aeroplane vertically above a straight horizontal road, the}$
$\displaystyle \text{angles of depression of two consecutive mile stones on opposite sides of the aeroplane are}$
$\displaystyle \text{observed to be }\alpha\text{ and }\beta.\text{ Show that the height in miles of the aeroplane above the road is}$
$\displaystyle \text{given by }\frac{\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}.\hfill\text{[CBSE 2004]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the aeroplane above the road }=h\text{ miles.}$
$\displaystyle \text{Let the horizontal distances of the two mile stones from the point vertically below the aeroplane}$
$\displaystyle \text{be }x\text{ miles and }y\text{ miles respectively.}$
$\displaystyle \text{Since the mile stones are consecutive and on opposite sides,}$
$\displaystyle x+y=1 \qquad ...(i)$
$\displaystyle \tan\alpha=\frac{h}{x}$
$\displaystyle x=\frac{h}{\tan\alpha} \qquad ...(ii)$
$\displaystyle \tan\beta=\frac{h}{y}$
$\displaystyle y=\frac{h}{\tan\beta} \qquad ...(iii)$
$\displaystyle \text{Substituting (ii) and (iii) in (i),}$
$\displaystyle \frac{h}{\tan\alpha}+\frac{h}{\tan\beta}=1$
$\displaystyle h\left(\frac{\tan\beta+\tan\alpha}{\tan\alpha\tan\beta}\right)=1$
$\displaystyle h=\frac{\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}$
$\displaystyle \therefore \text{The height of the aeroplane above the road is }\frac{\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}\text{ miles.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 38: }\text{A ladder rests against a wall at an angle }\alpha\text{ to the horizontal. Its}$
$\displaystyle \text{foot is pulled away from the wall through a distance }a,\text{ so that it slides a distance }b\text{ down}$
$\displaystyle \text{the wall making an angle }\beta\text{ with the horizontal. Show that}$
$\displaystyle \frac{a}{b}=\frac{\cos\alpha-\cos\beta}{\sin\beta-\sin\alpha}.$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Let the length of the ladder be }l.$
$\displaystyle \text{Initially, its horizontal distance from the wall}=l\cos\alpha.$
$\displaystyle \text{Finally, its horizontal distance from the wall}=l\cos\beta.$
$\displaystyle \therefore a=l\cos\beta-l\cos\alpha=l(\cos\beta-\cos\alpha) \qquad ...(i)$
$\displaystyle \text{Initially, the height reached by the ladder}=l\sin\alpha.$
$\displaystyle \text{Finally, the height reached by the ladder}=l\sin\beta.$
$\displaystyle \therefore b=l\sin\alpha-l\sin\beta=l(\sin\alpha-\sin\beta) \qquad ...(ii)$
$\displaystyle \text{Dividing (i) by (ii),}$
$\displaystyle \frac{a}{b}=\frac{l(\cos\beta-\cos\alpha)}{l(\sin\alpha-\sin\beta)}$
$\displaystyle =\frac{\cos\beta-\cos\alpha}{\sin\alpha-\sin\beta}$
$\displaystyle =\frac{\cos\alpha-\cos\beta}{\sin\beta-\sin\alpha}$
$\displaystyle \therefore \frac{a}{b}=\frac{\cos\alpha-\cos\beta}{\sin\beta-\sin\alpha}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 39: }\text{From the top of a light house, the angles of depression of two ships}$
$\displaystyle \text{on the opposite sides of it are observed to be }\alpha\text{ and }\beta.\text{ If the height of the light house}$
$\displaystyle \text{be }h\text{ metres and the line joining the ships passes through the foot of the light house, show}$
$\displaystyle \text{that the distance between the ships is }\frac{h(\tan\alpha+\tan\beta)}{\tan\alpha\tan\beta}\text{ metres.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distances of the two ships from the foot of the light house be }x\text{ m and }y\text{ m respectively.}$
$\displaystyle \text{The corresponding angles of elevation of the top of the light house are }\alpha\text{ and }\beta.$
$\displaystyle \tan\alpha=\frac{h}{x}$
$\displaystyle x=\frac{h}{\tan\alpha} \qquad ...(i)$
$\displaystyle \tan\beta=\frac{h}{y}$
$\displaystyle y=\frac{h}{\tan\beta} \qquad ...(ii)$
$\displaystyle \text{Since the ships are on opposite sides of the light house,}$
$\displaystyle \text{Distance between the ships}=x+y$
$\displaystyle =\frac{h}{\tan\alpha}+\frac{h}{\tan\beta}$
$\displaystyle =\frac{h(\tan\beta+\tan\alpha)}{\tan\alpha\tan\beta}$
$\displaystyle =\frac{h(\tan\alpha+\tan\beta)}{\tan\alpha\tan\beta}\text{ metres.}$
$\displaystyle \therefore \text{The distance between the ships is }\frac{h(\tan\alpha+\tan\beta)}{\tan\alpha\tan\beta}\text{ metres.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 40: }\text{From the top of a tower }h\text{ metre high, the angles of depression}$
$\displaystyle \text{of two objects, which are in the line with the foot of the tower, are }\alpha\text{ and }$
$\displaystyle \beta\ (\beta>\alpha) \ \text{Find the distance between the two objects.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distances of the objects corresponding to angles }\alpha\text{ and }\beta\text{ be }x\text{ m and }y\text{ m respectively.}$
$\displaystyle \text{The corresponding angles of elevation of the top of the tower are }\alpha\text{ and }\beta.$
$\displaystyle \tan\alpha=\frac{h}{x}$
$\displaystyle x=\frac{h}{\tan\alpha}=h\cot\alpha$
$\displaystyle \tan\beta=\frac{h}{y}$
$\displaystyle y=\frac{h}{\tan\beta}=h\cot\beta$
$\displaystyle \text{Since }\beta>\alpha,\text{ the object} \\ \text{corresponding to }\beta\text{ is nearer to the tower.}$
$\displaystyle \therefore \text{Distance between the two objects}=x-y$
$\displaystyle =h\cot\alpha-h\cot\beta$
$\displaystyle =h(\cot\alpha-\cot\beta)$
$\displaystyle \therefore \text{The distance between the two objects is }h(\cot\alpha-\cot\beta)\text{ metres.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 41: }\text{A window of a house is }h\text{ metre above the ground. From the}$
$\displaystyle \text{window, the angles of elevation and depression of the top and bottom of another house}$
$\displaystyle \text{situated on the opposite side of the lane are found to be }\alpha\text{ and }\beta\text{ respectively. Prove}$
$\displaystyle \text{that the height of the house is }h(1+\tan\alpha\cot\beta)\text{ metres.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the horizontal distance} \\ \text{between the two houses }=x\text{ m.}$
$\displaystyle \text{The window is }h\text{ m above the ground.}$
$\displaystyle \tan\beta=\frac{h}{x}$
$\displaystyle x=\frac{h}{\tan\beta}=h\cot\beta \qquad ...(i)$
$\displaystyle \text{Let the height of the opposite} \\ \text{house above the level of the window }=y\text{ m.}$
$\displaystyle \tan\alpha=\frac{y}{x}$
$\displaystyle y=x\tan\alpha$
$\displaystyle =h\cot\beta\tan\alpha \qquad [\text{Using (i)}]$
$\displaystyle \therefore \text{Height of the opposite house}=h+y$
$\displaystyle =h+h\tan\alpha\cot\beta$
$\displaystyle =h(1+\tan\alpha\cot\beta)$
$\displaystyle \therefore \text{The height of the house is }h(1+\tan\alpha\cot\beta)\text{ metres.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 42: }\text{The lower window of a house is at a height of }2\text{ m above the ground}$
$\displaystyle \text{and its upper window is }4\text{ m vertically above the lower window. At a certain instant, the}$
$\displaystyle \text{angles of elevation of a balloon from these windows are observed to be }60^\circ\text{ and }30^\circ$
$\displaystyle \text{respectively. Find the height of the balloon above the ground.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Height of the lower window above the ground}=2\text{ m.}$
$\displaystyle \text{Height of the upper window above the ground}=2+4=6\text{ m.}$
$\displaystyle \text{Let the height of the balloon above the ground}=h\text{ m.}$
$\displaystyle \text{Let its horizontal distance from the house}=x\text{ m.}$
$\displaystyle \tan60^\circ=\frac{h-2}{x}$
$\displaystyle \sqrt{3}=\frac{h-2}{x}$
$\displaystyle x=\frac{h-2}{\sqrt{3}} \qquad ...(i)$
$\displaystyle \tan30^\circ=\frac{h-6}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h-6}{x}$
$\displaystyle x=\sqrt{3}(h-6) \qquad ...(ii)$
$\displaystyle \text{From (i) and (ii),}$
$\displaystyle \frac{h-2}{\sqrt{3}}=\sqrt{3}(h-6)$
$\displaystyle h-2=3(h-6)$
$\displaystyle h-2=3h-18$
$\displaystyle 2h=16$
$\displaystyle h=8\text{ m}$
$\displaystyle \therefore \text{The height of the balloon above the ground is }8\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 43: }\text{One observer estimates the angle of elevation to the basket of a hot air}$
$\displaystyle \text{balloon to be }60^\circ,\text{ while another observer }100\text{ m away estimates the angle of elevation}$
$\displaystyle \text{to be }30^\circ.\text{ Find:}$
$\displaystyle \text{(i) the height of the basket from the ground.}$
$\displaystyle \text{(ii) the distance of the basket from the first observer's eye.}$
$\displaystyle \text{(iii) the horizontal distance of the second observer from the basket.}$
$\displaystyle \hfill\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the horizontal distance of the first observer from the basket}=x\text{ m.}$
$\displaystyle \therefore \text{Horizontal distance of the second observer from the basket}=(x+100)\text{ m.}$
$\displaystyle \text{Let the height of the basket from the ground}=h\text{ m.}$
$\displaystyle \tan60^\circ=\frac{h}{x}$
$\displaystyle \sqrt{3}=\frac{h}{x}$
$\displaystyle h=\sqrt{3}x \qquad ...(i)$
$\displaystyle \tan30^\circ=\frac{h}{x+100}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{x+100}$
$\displaystyle h=\frac{x+100}{\sqrt{3}} \qquad ...(ii)$
$\displaystyle \text{From (i) and (ii),}$
$\displaystyle \sqrt{3}x=\frac{x+100}{\sqrt{3}}$
$\displaystyle 3x=x+100$
$\displaystyle 2x=100$
$\displaystyle x=50\text{ m}$
$\displaystyle h=50\sqrt{3}\text{ m}$
$\displaystyle \therefore \text{(i) The height of the basket from the ground is }50\sqrt{3}\text{ m.}$

$\displaystyle \text{Let the distance of the basket from the first observer's eye}=d\text{ m.}$
$\displaystyle \sin60^\circ=\frac{h}{d}$
$\displaystyle \frac{\sqrt{3}}{2}=\frac{50\sqrt{3}}{d}$
$\displaystyle d=100\text{ m}$
$\displaystyle \therefore \text{(ii) The distance of the basket from the first observer's eye is }100\text{ m.}$

$\displaystyle \text{Horizontal distance of the second observer from the basket}=x+100$
$\displaystyle =50+100=150\text{ m}$
$\displaystyle \therefore \text{(iii) The horizontal distance of the second observer from the basket is }150\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 44: }\text{The angle of elevation of the top of a tower }24\text{ m high from the}$
$\displaystyle \text{foot of another tower in the same plane is }60^\circ.\text{ The angle of elevation of the top of the}$
$\displaystyle \text{second tower from the foot of the first tower is }30^\circ.\text{ Find the distance between the two}$
$\displaystyle \text{towers and the height of the other tower. Also, find the length of the wire attached to the}$
$\displaystyle \text{tops of both the towers.}\hfill\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the distance between the two towers }=x\text{ m.}$
$\displaystyle \text{Height of the first tower }=24\text{ m.}$
$\displaystyle \tan60^\circ=\frac{24}{x}$
$\displaystyle \sqrt{3}=\frac{24}{x}$
$\displaystyle x=\frac{24}{\sqrt{3}}=8\sqrt{3}\text{ m}$
$\displaystyle \therefore \text{The distance between the two towers is }8\sqrt{3}\text{ m.}$

$\displaystyle \text{Let the height of the second tower }=h\text{ m.}$
$\displaystyle \tan30^\circ=\frac{h}{8\sqrt{3}}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{8\sqrt{3}}$
$\displaystyle h=8\text{ m}$
$\displaystyle \therefore \text{The height of the second tower is }8\text{ m.}$

$\displaystyle \text{Difference between the heights of the towers}=24-8=16\text{ m.}$
$\displaystyle \text{Let the length of the wire joining the tops of the towers}=l\text{ m.}$
$\displaystyle l^2=(8\sqrt{3})^2+16^2$
$\displaystyle =192+256=448$
$\displaystyle l=\sqrt{448}=8\sqrt{7}\text{ m}$
$\displaystyle \therefore \text{The length of the wire attached to the tops of both towers is }8\sqrt{7}\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 45: }\text{The angle of elevation of an airborne helicopter from a point }A\text{ on}$
$\displaystyle \text{the ground is }45^\circ.\text{ After a flight of }15\text{ seconds, the angle of elevation of the helicopter}$
$\displaystyle \text{changes to }30^\circ.\text{ If the helicopter is flying at a constant height of }2000\text{ m, find the speed}$
$\displaystyle \text{of the helicopter. (Take }\sqrt{3}=1.732\text{)}\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the initial and final horizontal distances of the helicopter from point }A\text{ be }x\text{ m and }y\text{ m.}$
$\displaystyle \tan45^\circ=\frac{2000}{x}$
$\displaystyle 1=\frac{2000}{x}$
$\displaystyle x=2000\text{ m}$
$\displaystyle \tan30^\circ=\frac{2000}{y}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{2000}{y}$
$\displaystyle y=2000\sqrt{3}\text{ m}$
$\displaystyle \text{Distance travelled by the helicopter}=y-x$
$\displaystyle =2000\sqrt{3}-2000=2000(\sqrt{3}-1)\text{ m}$
$\displaystyle =2000(1.732-1)=1464\text{ m}$
$\displaystyle \text{Time taken}=15\text{ seconds}$
$\displaystyle \text{Speed of the helicopter}=\frac{1464}{15}=97.6\text{ m/s}$
$\displaystyle =\frac{97.6\times18}{5}=351.36\text{ km/h}$
$\displaystyle \therefore \text{The speed of the helicopter is }97.6\text{ m/s, i.e. }351.36\text{ km/h.}$
$\displaystyle \\$