\displaystyle \textbf{Question 1. }\text{A }30\text{ m long rope is tightly stretched and tied from the top of pole to the ground.}
\displaystyle \text{If the rope makes an angle of }60^{\circ}\text{ with the ground, the height of the pole is:} \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{(a) }10\sqrt{3}\text{ m} \qquad \text{(b) }30\sqrt{3}\text{ m}
\displaystyle \text{(c) }15\text{ m} \qquad \text{(d) }15\sqrt{3}\text{ m}
\displaystyle \text{Answer:}
\displaystyle \textbf{(d)}
\displaystyle \theta=60^\circ
\displaystyle \frac{h}{30}=\sin60^\circ
\displaystyle \frac{h}{30}=\frac{\sqrt{3}}{2}
\displaystyle \Rightarrow h=15\sqrt{3}\text{ m}
\\

\displaystyle \textbf{Question 2. }\text{A drone was used to facilitate movement of an ambulance on the straight highway}
\displaystyle \text{to a point }P\text{ on the ground where there was an accident. The ambulance was travelling at}
\displaystyle \text{the speed of }60\text{ km/h. The drone stopped at a point }Q,\ 100\text{ m vertically above the point }P.
\displaystyle \text{The angle of depression of the ambulance was found to be }30^{\circ}\text{ at a particular instant.} \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{Based on above information, answer the following questions:}
\displaystyle \text{(i) Represent the above situation with the help of a diagram.}
\displaystyle \text{(ii) Find the distance between the ambulance and the site of accident }(P)
\displaystyle \text{at the particular instant. (Use }\sqrt{3}=1.73\text{)}
\displaystyle \text{(iii) (a) Find the time (in seconds) in which the angle of depression changes from }30^{\circ}
\displaystyle \text{to }45^{\circ}.
\displaystyle \text{OR}
\displaystyle \text{(iii) (b) How long (in seconds) will the ambulance take to reach point }P\text{ from a point }T
\displaystyle \text{on the highway such that angle of depression of the ambulance at }T\text{ is }60^{\circ}\text{ from the drone?}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle \text{In }\triangle PQR,
\displaystyle \frac{PQ}{PR}=\tan30^\circ
\displaystyle \Rightarrow \frac{100}{PR}=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow PR=100\sqrt{3}=100\times1.73
\displaystyle =173\text{ m}

\displaystyle \textbf{(ii)(a)}
\displaystyle \text{In }\triangle PQS,
\displaystyle \frac{PQ}{PS}=\tan45^\circ=1
\displaystyle \Rightarrow PQ=PS=100\text{ m}
\displaystyle \text{Now,}
\displaystyle PR=PS+SR=173
\displaystyle \Rightarrow SR=173-100=73
\displaystyle \text{As,}
\displaystyle \text{speed}=\frac{\text{Distance}}{\text{time}}
\displaystyle \Rightarrow \text{time}=\frac{\text{Distance}}{\text{speed}}
\displaystyle \Rightarrow \text{time}=\frac{73\times18}{60\times5}
\displaystyle =\frac{219}{50}=\frac{43.8}{10}
\displaystyle =4.38\text{ sec}
\displaystyle \therefore \text{Time }30^\circ\text{ to }45^\circ=4.38\text{ sec}

\displaystyle \textbf{(iii)(b)}
\displaystyle \text{In }\triangle QPT,
\displaystyle \frac{PQ}{PT}=\tan60^\circ
\displaystyle \Rightarrow \frac{100}{PT}=\sqrt{3}
\displaystyle \Rightarrow PT=\frac{100}{\sqrt{3}}=\frac{100}{3}\times\sqrt{3}
\displaystyle =57.667\text{ m}
\displaystyle \text{As, speed}=\frac{\text{Distance}}{\text{time}}
\displaystyle \Rightarrow \frac{60\times\frac{5}{18}}{\text{time}}=57.667
\displaystyle \Rightarrow \text{time}=\frac{57667\times18}{1000\times5\times60}
\displaystyle =\frac{173001}{10000\times5}
\displaystyle =\frac{173000}{10000\times5}
\displaystyle \approx\frac{34.6}{10}
\displaystyle =3.46\text{ sec}
\displaystyle \therefore \text{Time taken by ambulance from }T\text{ to point }P
\displaystyle =3.46\text{ sec (approx)}
\\

\displaystyle \textbf{Question 3. }\text{A pole }6\text{ m high is fixed on the top of a tower. The angle of elevation}
\displaystyle \text{of the top of the pole observed from a point }P\text{ on the ground is }60^{\circ}\text{ and the}
\displaystyle \text{angle of depression of the point }P\text{ from the top of the tower is }45^{\circ}. \text{Find the height}
\displaystyle \text{of the tower and the distance of point }P\text{ from the foot of the tower. (Use }\sqrt{3}=1.73\text{)} \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AB\text{ be the tower and }BC\text{ be the pole.}
\displaystyle \text{Let }AB=h\text{ m. Now, }CB=6\text{ m.}
\displaystyle \text{Also, }\angle CPA=60^\circ\text{ and }\angle BPA=45^\circ
\displaystyle \text{In }\triangle CAP,
\displaystyle \tan60^\circ=\frac{CA}{AP}
\displaystyle \Rightarrow \sqrt{3}=\frac{h+6}{AP}
\displaystyle \Rightarrow AP=\frac{h+6}{\sqrt{3}}\qquad (i)
\displaystyle \text{In }\triangle BAP,
\displaystyle \tan45^\circ=\frac{AB}{AP}
\displaystyle \Rightarrow 1=\frac{h}{AP}
\displaystyle \Rightarrow AP=h\qquad (ii)
\displaystyle \text{From }(i)\text{ and }(ii),\text{ we get}
\displaystyle h=\frac{h+6}{\sqrt{3}}
\displaystyle \Rightarrow h=\frac{6}{\sqrt{3}-1}
\displaystyle \Rightarrow h=\frac{6}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}
\displaystyle \Rightarrow h=\frac{6(\sqrt{3}+1)}{2}
\displaystyle \Rightarrow h=3(\sqrt{3}+1)
\displaystyle \Rightarrow h=3(1.73+1)=3\times2.73=8.19\text{ m}
\displaystyle \therefore \text{Height of tower is }8.19\text{ m.}
\displaystyle \text{Also, }AP=h=8.19\text{ m}
\displaystyle \therefore \text{Distance of point }P\text{ from foot of tower is }8.19\text{ m.}
\\

\displaystyle \textbf{Question 4. }\text{Due to short circuit, a fire has broken out in New Home Complex. Two buildings,}
\displaystyle \text{namely }X\text{ and }Y\text{ have mainly been affected. The fire engine has arrived and it has been}
\displaystyle \text{stationed at a point which is in between the two buildings. A ladder at point }O\text{ is fixed}
\displaystyle \text{in front of the fire engine. The ladder inclined at an angle }60^{\circ}\text{ to the horizontal is leaning}
\displaystyle \text{against the wall of the terrace (top) of the building }Y. \text{The foot of the ladder is kept fixed}
\displaystyle \text{and after some time it is made to lean against the terrace (top) of the opposite building }X
\displaystyle \text{at an angle of }45^{\circ}\text{ with the ground. Both the buildings along with the foot of the ladder,}
\displaystyle \text{fixed at }O\text{ are in a straight line.} \hspace{0.2cm}\text{[CBSE 2024(C)]} \displaystyle \text{Based on the above given information, answer the following questions:}
\displaystyle \text{(i) Find the length of the ladder.}
\displaystyle \text{(ii) Find the distance of the building }Y\text{ from point }O,\text{ i.e. }OA.
\displaystyle \text{(iii) (a) Find the horizontal distance between the two buildings.}
\displaystyle \text{OR}
\displaystyle \text{(b) Find the height of the building }X.
\displaystyle \text{Answer:}

\displaystyle \textbf{(i)}
\displaystyle \text{In }\triangle OAP,
\displaystyle \frac{OP}{12\sqrt{3}}=\mathrm{cosec}\,60^\circ=\frac{2}{\sqrt{3}}
\displaystyle \Rightarrow OP=24\text{ m}
\displaystyle \therefore \text{Length of ladder is }24\text{ m}

\displaystyle \textbf{(ii)}
\displaystyle \text{In }\triangle OAP,
\displaystyle \frac{OA}{12\sqrt{3}}=\cot60^\circ=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow OA=12\text{ m}
\displaystyle \therefore \text{The distance of the building }Y\text{ from point }O
\displaystyle \text{i.e. }OA\text{ is }12\text{ m}

\displaystyle \textbf{(iii)(a)}
\displaystyle OP=OR=24\text{ m}
\displaystyle \therefore \text{In }\triangle OCR,
\displaystyle \frac{OC}{24}=\cos45^\circ=\frac{1}{\sqrt{2}}
\displaystyle \Rightarrow OC=12\sqrt{2}\text{ m}
\displaystyle \therefore \text{Distance between two buildings}
\displaystyle =OA+OC
\displaystyle =(12+12\sqrt{2})\text{ m}
\displaystyle \text{or }12(1+\sqrt{2})\text{ m}

\displaystyle \textbf{32. (iii)(b)}
\displaystyle OP=OR=24\text{ m}
\displaystyle \therefore \text{In }\triangle OCR,
\displaystyle \frac{RC}{24}=\sin45^\circ=\frac{1}{\sqrt{2}}
\displaystyle \Rightarrow RC=12\sqrt{2}\text{ m}
\displaystyle \therefore \text{Height of building }X\text{ is }12\sqrt{2}\text{ m}
\\

\displaystyle \textbf{Question 5. }\text{A straight highway leads to the foot of a tower. A man standing on the top}
\displaystyle \text{of the }75\text{ m high tower observes two cars at angles of depression of }30^{\circ}\text{ and }60^{\circ},
\displaystyle \text{which are approaching the foot of the tower. If one car is exactly behind the other}
\displaystyle \text{on the same side of the tower, find the distance between the two cars. (use }\sqrt{3}=1.73\text{)} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABD,
\displaystyle \frac{AB}{BD}=\tan60^\circ
\displaystyle \Rightarrow \frac{75}{BD}=\sqrt{3}
\displaystyle \Rightarrow BD=\frac{75}{\sqrt{3}}\text{ m}
\displaystyle \text{In }\triangle ABC,
\displaystyle \frac{AB}{BC}=\tan30^\circ
\displaystyle \Rightarrow \frac{75}{BC}=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow BC=75\sqrt{3}\text{ m}
\displaystyle \text{Now, distance between two cars }=DC
\displaystyle DC=BC-BD
\displaystyle =75\sqrt{3}-\frac{75}{\sqrt{3}}
\displaystyle =50\sqrt{3}=50\times1.73\text{ m}
\displaystyle =86.5\text{ m}
\\

\displaystyle \textbf{Question 6. }\text{From the top of a }7\text{ m high building, the angle of elevation of the top}
\displaystyle \text{of a cable tower is }60^{\circ}\text{ and the angle of depression of its foot is }30^{\circ}.
\displaystyle \text{Determine the height of tower.} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ADC,
\displaystyle \frac{AD}{DC}=\tan30^\circ
\displaystyle \Rightarrow \frac{7}{y}=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow y=7\sqrt{3}\text{ m}
\displaystyle \therefore AE=DC
\displaystyle \Rightarrow AE=y=7\sqrt{3}\text{ m}
\displaystyle \text{In }\triangle BEA,
\displaystyle \frac{BE}{AE}=\tan60^\circ
\displaystyle \Rightarrow \frac{x}{y}=\sqrt{3}
\displaystyle \Rightarrow \frac{x}{7\sqrt{3}}=\sqrt{3}
\displaystyle \Rightarrow x=21\text{ m}
\displaystyle \therefore \text{Height of tower }BC=BE+EC
\displaystyle =21+7=28\text{ m}
\\

\displaystyle \textbf{Question 7. }\text{If a pole }6\text{ m high casts a shadow }2\sqrt{3}\text{ m long on the ground,}
\displaystyle \text{then sun's elevation is:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }60^{\circ} \qquad \text{(b) }45^{\circ}
\displaystyle \text{(c) }30^{\circ} \qquad \text{(d) }90^{\circ}
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle \text{Here, }\tan\theta=\frac{6}{2\sqrt{3}}=\sqrt{3}
\displaystyle \Rightarrow \theta=60^\circ
\\

\displaystyle \textbf{Question 8. }\text{A vertical pole }10\text{ m long casts a shadow of length }5\text{ m on the ground.}
\displaystyle \text{At the same time, a tower casts a shadow of length }12.5\text{ m on the ground.}
\displaystyle \text{The height of the tower is:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }20\text{ m} \qquad \text{(b) }22\text{ m}
\displaystyle \text{(c) }25\text{ m} \qquad \text{(d) }24\text{ m}
\displaystyle \text{Answer:}
\displaystyle \textbf{(c)}  \text{ Given,}\displaystyle \text{Let }h\text{ m be the height of the tower.}
\displaystyle \therefore \text{ from }\triangle ABC\text{ and }\triangle DEF,
\displaystyle \Rightarrow \frac{10}{5}=\frac{h}{12.5}
\displaystyle \Rightarrow h=\frac{125}{5}=25\text{ m}
\\

\displaystyle \textbf{Question 9. }\text{The height of a tower is }20\text{ m. The length of its shadow made on the}
\displaystyle \text{level ground when the Sun's altitude is }60^{\circ},\text{ is:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }\frac{20}{\sqrt{3}}\text{ m} \qquad \text{(b) }\frac{20}{3}\text{ m}
\displaystyle \text{(c) }20\sqrt{3}\text{ m} \qquad \text{(d) }20\text{ m}
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle \text{According to the question,}
\displaystyle \tan60^\circ=\frac{20}{\text{length of shadow}}
\displaystyle \therefore \text{length of the shadow}=\frac{20}{\sqrt{3}}\text{ m}
\\

\displaystyle \textbf{Question 10. }\text{From the top of a tower }50\text{ m high, the angles of depression of the top}
\displaystyle \text{and bottom of a pole are observed to be }45^{\circ}\text{ and }60^{\circ}\text{ respectively. Find the height}
\displaystyle \text{of pole, if the pole and tower stand on the same plane. (Use }\frac{1}{\sqrt{3}}=0.577\text{)} \hspace{0.2cm}\text{[CBSE 2023 (C)]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }h\text{ be the height of the pole.}
\displaystyle \text{In }\triangle ADE,
\displaystyle \tan45^\circ=\frac{AE}{DE}
\displaystyle \Rightarrow \tan45^\circ=\frac{50-h}{DE}
\displaystyle \Rightarrow DE=(50-h)\text{ m}
\displaystyle \text{Now, in }\triangle ABC,
\displaystyle \tan60^\circ=\frac{50}{BC}
\displaystyle \Rightarrow BC=\frac{50}{\sqrt{3}}
\displaystyle \text{Since }BC=DE
\displaystyle \Rightarrow \frac{50}{\sqrt{3}}=50-h\qquad \left[\because \frac{1}{\sqrt{3}}=0.577\right]
\displaystyle \Rightarrow 28.85=50-h
\displaystyle \Rightarrow h=50-28.85=21.15\text{ m}
\displaystyle \therefore \text{Height of the pole }=21.15\text{ m}
\\

\displaystyle \textbf{Question 11. }\text{In the given figure below }AB\text{ is tower of height }50\text{ m. A man standing}
\displaystyle \text{on its top, observes two cars on the opposite sides of the tower with angles of depression}
\displaystyle 30^{\circ}\text{ and }45^{\circ}\text{ respectively. Find the distance between the two cars.} \hspace{0.2cm}\text{[CBSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{Given,}
\displaystyle \text{In }\triangle ACB,
\displaystyle \tan30^\circ=\frac{AB}{BC}
\displaystyle \Rightarrow \frac{1}{\sqrt{3}}=\frac{50}{BC}
\displaystyle \Rightarrow BC=50\sqrt{3}\qquad (i)
\displaystyle \text{In }\triangle ABD,
\displaystyle \tan45^\circ=\frac{50}{BD}
\displaystyle \Rightarrow 1=\frac{50}{BD}
\displaystyle \Rightarrow BD=50\qquad (ii)
\displaystyle \text{Distance between cars }=BC+BD
\displaystyle =50\sqrt{3}+50
\displaystyle =50(\sqrt{3}+1)\text{ m}
\\

\displaystyle \textbf{Question 12. }\text{Kite festival is celebrated in many countries at different times of the year.}
\displaystyle \text{In India, every year }14\text{th January is celebrated as International Kite Day. On this day many}
\displaystyle \text{people visit India and participate in the festival by flying various kinds of kites.} \hspace{0.2cm}\text{[CBSE 2022]}
\displaystyle \text{The picture given below, shows three kites flying together.}
\displaystyle \text{In the above figure, the angles of elevation of two kites (Points }A\text{ and }B\text{) from the}
\displaystyle \text{hands of a man (Point }C\text{) are found to be }30^{\circ}\text{ and }60^{\circ}\text{ respectively. Taking}
\displaystyle AD=50\text{ m and }BE=60\text{ m, find}
\displaystyle \text{(i) the lengths of strings used (take them straight) for kites }A\text{ and }B\text{ as shown in the figure.}
\displaystyle \text{(ii) the distance }d\text{ between these two kites.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle \text{In }\triangle ADC,
\displaystyle \sin30^\circ=\frac{50}{AC}=\frac{1}{2}
\displaystyle \Rightarrow AC=100
\displaystyle \therefore \text{Length of string used for kite }c\\ A\text{ is }100\text{ m}
\displaystyle \text{In }\triangle BEC,
\displaystyle \sin60^\circ=\frac{60}{BC}
\displaystyle \Rightarrow \frac{\sqrt{3}}{2}=\frac{60}{BC}
\displaystyle \Rightarrow BC=\frac{120}{\sqrt{3}}=40\sqrt{3}
\displaystyle \therefore \text{Length of string used for kite }B\text{ is }40\sqrt{3}\text{ m}

\displaystyle \textbf{(ii)}
\displaystyle \angle ACB=180^\circ-\angle ACD-\angle BCE
\displaystyle =180^\circ-60^\circ-30^\circ=90^\circ
\displaystyle \text{So, }\triangle ACB\text{ is a right angled triangle.}
\displaystyle AB^{2}=AC^{2}+BC^{2}
\displaystyle d^{2}=10000+4800=14800
\displaystyle d=20\sqrt{37}\text{ m}
\\

\displaystyle \textbf{Question 13. }\text{The ratio of the length of a vertical rod and the length of its shadow}
\displaystyle \text{is }1:\sqrt{3}. \text{Find the angle of elevation of the Sun at that moment.} \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \tan\theta=\frac{x}{\sqrt{3}x}=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow \theta=30^\circ
\\

\displaystyle \textbf{Question 14. }\text{A vertical tower stands on a horizontal plane and is surmounted by a vertical}
\displaystyle \text{flag-staff of height }6\text{ m. At a point on the plane, the angle of elevation of the bottom}
\displaystyle \text{and top of the flag-staff are }30^{\circ}\text{ and }45^{\circ}\text{ respectively. Find the height of the tower.}
\displaystyle \text{(Take }\sqrt{3}=1.73\text{)} \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AB\to\text{ Tower of height }h\text{ m}
\displaystyle BC\to\text{ flag staff of height }6\text{ m}
\displaystyle O\text{ is point of observation, }\angle AOB=30^\circ,\ \angle AOC=45^\circ
\displaystyle \text{Let }OA=x\text{ m}
\displaystyle \text{In right-angled triangle }OAB,
\displaystyle \frac{AB}{OA}=\tan30^\circ
\displaystyle \Rightarrow \frac{h}{x}=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow x=\sqrt{3}h\qquad (i)
\displaystyle \text{In right-angled triangle }OAC,
\displaystyle \frac{AC}{OA}=\tan45^\circ
\displaystyle \Rightarrow \frac{h+6}{x}=1
\displaystyle \Rightarrow h+6=x
\displaystyle \Rightarrow h+6=\sqrt{3}h\qquad [\text{from }(i)]
\displaystyle \Rightarrow \sqrt{3}h-h=6
\displaystyle \Rightarrow (\sqrt{3}-1)h=6
\displaystyle \Rightarrow h=\frac{6}{\sqrt{3}-1}
\displaystyle =\frac{6(\sqrt{3}+1)}{3-1}=3(\sqrt{3}+1)
\displaystyle \Rightarrow h=3(1.73+1)
\displaystyle \Rightarrow h=3\times2.73=8.19\text{ m}
\displaystyle \therefore \text{Height of tower }=8.19\text{ m.}
\\

\displaystyle \textbf{Question 15. }\text{A man in a boat rowing away from a light house }100\text{ m high takes }2\text{ minutes}
\displaystyle \text{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 \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle AB\text{ is light house of height }100\text{ m}
\displaystyle \text{The boat is at }C\text{ making angle of elevation }60^\circ\text{ with the top of light house.}
\displaystyle \text{After }2\text{ minutes, boat is at }D,\text{ making angle }30^\circ\text{ with the top of light house.}
\displaystyle \text{Let }BC=y\text{ m and }CD=x\text{ m.}
\displaystyle \text{In }\triangle ABC,
\displaystyle \tan60^\circ=\frac{100}{y}
\displaystyle \Rightarrow \sqrt{3}=\frac{100}{y}
\displaystyle \Rightarrow y=\frac{100}{\sqrt{3}}\qquad (i)
\displaystyle \text{In }\triangle ABD,
\displaystyle \tan30^\circ=\frac{100}{x+y}
\displaystyle \Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{x+y}
\displaystyle \Rightarrow x+y=100\sqrt{3}\qquad (ii)
\displaystyle \text{From }(i),
\displaystyle x=100\sqrt{3}-y
\displaystyle =100\sqrt{3}-\frac{100}{\sqrt{3}}
\displaystyle =\frac{100(3-1)}{\sqrt{3}}=\frac{200}{\sqrt{3}}\text{ m}
\displaystyle \text{Speed of boat}=\frac{\text{distance}}{\text{time}}
\displaystyle =\frac{\frac{200}{\sqrt{3}}}{2}
\displaystyle =\frac{100}{\sqrt{3}}\text{ m/minute}
\displaystyle =\frac{100}{1.732}=57.73\text{ m/minute}
\\

\displaystyle \textbf{Question 16. }\text{Two poles of equal heights are standing opposite to each other on either side}
\displaystyle \text{of the road, which is }80\text{ m wide. From a point between them on the road, the angles}
\displaystyle \text{of elevation of the top of the poles are }60^{\circ}\text{ and }30^{\circ}\text{ respectively. Find the}
\displaystyle \text{height of the poles and the distances of the point from the poles.}
\displaystyle \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AB\text{ and }CD\text{ are two poles of equal heights }h\text{ m.}
\displaystyle \text{Let }AB=CD=h\text{ m}\qquad [\text{Height of the poles}]
\displaystyle \text{Given:}
\displaystyle BC=80\text{ m}\qquad [\text{Width of the road}]
\displaystyle \text{Let }CE=x\text{ m}
\displaystyle \therefore BE=(80-x)\text{ m}
\displaystyle \text{In }\triangle CDE,
\displaystyle \frac{CD}{CE}=\tan30^\circ
\displaystyle \Rightarrow \frac{h}{x}=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow x=\sqrt{3}h\qquad (i)
\displaystyle \text{In }\triangle ABE,
\displaystyle \frac{AB}{BE}=\tan60^\circ
\displaystyle \Rightarrow \frac{h}{80-x}=\sqrt{3}
\displaystyle \Rightarrow h=80\sqrt{3}-\sqrt{3}x
\displaystyle \Rightarrow x=\frac{80\sqrt{3}-h}{\sqrt{3}}\qquad (ii)
\displaystyle \text{From equation }(i)\text{ and }(ii),\text{ we get}
\displaystyle \sqrt{3}h=\frac{80\sqrt{3}-h}{\sqrt{3}}
\displaystyle \Rightarrow 3h=80\sqrt{3}-h
\displaystyle \Rightarrow 4h=80\sqrt{3}
\displaystyle \Rightarrow h=20\sqrt{3}\text{ m}
\displaystyle \text{Hence, height of poles is }20\sqrt{3}\text{ m}
\displaystyle \text{Substituting }h\text{ in equation }(i),
\displaystyle x=h\sqrt{3}=20\sqrt{3}\times\sqrt{3}=60\text{ m}
\displaystyle \text{Hence, position of the point is at a distance of }60\text{ m}
\displaystyle \text{from pole }CD\text{ and }20\text{ m from pole }AB.
\\

\displaystyle \textbf{Question 17. }\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 depression from the top of tower to the foot of hill is }30^{\circ}.
\displaystyle \text{If tower is }50\text{ metre high, find the height of the hill.} \hspace{0.2cm}\text{[CBSE 2018 (C)]}
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,
\displaystyle \tan60^\circ=\frac{AB}{BC}
\displaystyle \Rightarrow \sqrt{3}=\frac{AB}{BC}
\displaystyle \Rightarrow BC=\frac{AB}{\sqrt{3}}\qquad (i)
\displaystyle \text{In }\triangle DBC,
\displaystyle \tan30^\circ=\frac{DC}{BC}
\displaystyle \Rightarrow \frac{1}{\sqrt{3}}=\frac{DC}{BC}
\displaystyle \Rightarrow BC=\sqrt{3}\,DC\qquad (ii)
\displaystyle \text{Equating }(i)\text{ and }(ii),
\displaystyle \frac{AB}{\sqrt{3}}=\sqrt{3}\,DC
\displaystyle \Rightarrow AB=3(DC)
\displaystyle \text{Put }DC=50\text{ m}
\displaystyle \Rightarrow AB=3\times50=150\text{ m}
\displaystyle \therefore \text{Height of hill }=150\text{ m}
\\

\displaystyle \textbf{Question 18. }\text{A ladder }15\text{ m long makes an angle of }60^{\circ}\text{ with the wall. Find the}
\displaystyle \text{height of the point where the ladder touches the wall.} \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let ladder }AB=15\text{ m and wall } \\ AC=h\text{ m}
\displaystyle \text{In right }\triangle ABC,
\displaystyle \cos60^\circ=\frac{h}{15}
\displaystyle \frac{1}{2}=\frac{h}{15}
\displaystyle \Rightarrow h=\frac{15}{2}\text{ m}
\displaystyle \Rightarrow h=7.5\text{ m}
\\

\displaystyle \textbf{Question 19. }\text{An observer, }1.5\text{ m tall, is }28.5\text{ m away from a }30\text{ m high tower.}
\displaystyle \text{Determine the angle of elevation of the top of tower from the eye of the observer.}
\displaystyle \hspace{0.2cm}\text{[CBSE 2017(C)]}
\displaystyle \text{Answer:}
\displaystyle \text{Draw }AM\parallel BD.
\displaystyle \text{So, }AB=DM=1.5\text{ m}
\displaystyle \text{So, }CM=CD-DM=30-1.5 \\  =28.5\text{ m}
\displaystyle \text{Let }\theta\text{ be the angle of elevation} \\ \text{of the top of the tower from} \\ \text{the eye of the observer.}
\displaystyle \text{In }\triangle ACM,
\displaystyle \tan\theta=\frac{CM}{AM}=\frac{28.5}{28.5}
\displaystyle \tan\theta=1
\displaystyle \Rightarrow \theta=45^\circ
\\

\displaystyle \textbf{Question 20. }\text{The shadow of a tower at a time is three times as long as its shadow}
\displaystyle \text{when the angle of elevation of the sun is }60^{\circ}. \text{Find the angle of elevation of}
\displaystyle \text{the sun at the time of the longer shadow.} \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }h\text{ be the height of tower.}
\displaystyle \text{In }\triangle ABC,
\displaystyle \tan60^\circ=\frac{h}{x}
\displaystyle \Rightarrow \sqrt{3}=\frac{h}{x}\qquad (i)
\displaystyle \text{In }\triangle ABD,
\displaystyle \tan\theta=\frac{h}{3x}\qquad (ii)
\displaystyle \text{Dividing }(ii)\text{ by }(i),\text{ we get}
\displaystyle \frac{\tan\theta}{\sqrt{3}}=\frac{h}{3x}\times\frac{x}{h}
\displaystyle \Rightarrow \tan\theta=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow \theta=30^\circ
\displaystyle \text{Hence, }\theta=30^\circ\text{ is the angle of elevation at the time of the longer shadow.}
\\

\displaystyle \textbf{Question 21. }\text{From a point on the ground, the angles of elevation of the bottom and top}
\displaystyle \text{of a transmission tower fixed on the top of a }20\text{ m high building are }45^{\circ}
\displaystyle \text{and }60^{\circ}\text{ respectively. Find the height of the tower.} \hspace{0.2cm}\text{[CBSE 2017(C)]}
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle BCD,
\displaystyle \tan45^\circ=\frac{BC}{x}
\displaystyle 1=\frac{20}{x}
\displaystyle \Rightarrow x=20\text{ m}
\displaystyle \text{In }\triangle ACD,
\displaystyle \tan60^\circ=\frac{h+20}{x}
\displaystyle \sqrt{3}=\frac{h+20}{20}
\displaystyle \Rightarrow 20\sqrt{3}-20=h
\displaystyle \Rightarrow 20(\sqrt{3}-1)=h
\displaystyle \Rightarrow 20(1.73-1)=h
\displaystyle \Rightarrow 20\times0.73=h
\displaystyle \Rightarrow h=14.6\text{ m}
\\

\displaystyle \textbf{Question 22. }\text{The angle of elevation of a cloud from a point }60\text{ m above the surface}
\displaystyle \text{of the water of a lake is }30^{\circ}\text{ and the angle of depression of its shadow in water}
\displaystyle \text{of lake is }60^{\circ}. \text{Find the height of the cloud from the surface of water.}
\displaystyle \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let height of cloud }C\text{ from the lake be }h\text{ m.}
\displaystyle \text{'P' is the position of the point }60\text{ m above the lake.}
\displaystyle \text{'T' is the reflection of the cloud in the lake.}
\displaystyle \text{Let }PR=x\text{ m, }CS=h\text{ m, }CR=(h-60)\text{ m}
\displaystyle \text{Now, }TR=TS+SR=CS+SR=(h+60)\text{ m}
\displaystyle \qquad [\because\ CS=TS,\text{ by law of reflection in plane mirror}]
\displaystyle \text{In right-angled triangle }PRC,
\displaystyle \frac{PR}{CR}=\cot30^\circ
\displaystyle \Rightarrow PR=CR\cot30^\circ
\displaystyle \Rightarrow PR=(h-60)\sqrt{3}\qquad (i)
\displaystyle \text{In right-angled triangle }PRT,
\displaystyle \frac{PR}{TR}=\cot60^\circ
\displaystyle \Rightarrow PR=TR\cot60^\circ
\displaystyle \Rightarrow PR=\frac{h+60}{\sqrt{3}}\qquad (ii)
\displaystyle \text{From }(i)\text{ and }(ii),\text{ we get}
\displaystyle (h-60)\sqrt{3}=\frac{h+60}{\sqrt{3}}
\displaystyle \Rightarrow 3h-180=h+60
\displaystyle \Rightarrow 2h=240
\displaystyle \Rightarrow h=120\text{ m}
\displaystyle \therefore \text{Height of the cloud above the lake is }120\text{ m}
\\

\displaystyle \textbf{Question 23. }\text{An observer finds the angle of elevation of the top of the tower from a}
\displaystyle \text{certain point on the ground as }30^{\circ}. \text{If the observer moves }20\text{ m towards the}
\displaystyle \text{base of the tower, the angle of elevation to the top increases by }15^{\circ}.
\displaystyle \text{Find the height of tower.} \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AB\text{ be the tower and let}
\displaystyle D\text{ be the point from where the observer observes the top of the tower.}
\displaystyle \text{Let }AB=h\text{ m, }CD=20\text{ m, }BC=x\text{ m}
\displaystyle \text{In right-angled triangle }ABC,
\displaystyle \frac{AB}{BC}=\tan45^\circ
\displaystyle \Rightarrow \frac{h}{x}=1
\displaystyle \Rightarrow h=x\qquad (i)
\displaystyle \text{In right-angled triangle }ABD,
\displaystyle \frac{AB}{BD}=\tan30^\circ
\displaystyle \Rightarrow \frac{h}{x+20}=\frac{1}{\sqrt{3}}\qquad (ii)
\displaystyle \text{From equation }(i)\text{ and }(ii),\text{ we get}
\displaystyle \frac{h}{h+20}=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow \sqrt{3}h=h+20
\displaystyle \Rightarrow h(\sqrt{3}-1)=20
\displaystyle \Rightarrow h=\frac{20}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}
\displaystyle \Rightarrow h=\frac{20(\sqrt{3}+1)}{2}
\displaystyle \Rightarrow h=10(\sqrt{3}+1)\text{ m}
\displaystyle \therefore \text{Height of the tower is }10(\sqrt{3}+1)\text{ m}
\\

\displaystyle \textbf{Question 24. }\text{An aeroplane is flying at a height of }300\text{ m above the ground. Flying}
\displaystyle \text{at this height, the angles of depression from the aeroplane of two points on both}
\displaystyle \text{banks of a river in opposite direction are }45^{\circ}\text{ and }60^{\circ}\text{ respectively.}
\displaystyle \text{Find the width of the river. [Use }\sqrt{3}=1.732\text{]} \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A\text{ be the position of the aeroplane and let }C\text{ and}
\displaystyle D\text{ be two points on the two banks of the river such that}
\displaystyle \text{the angles of depression at }C\text{ and }D\text{ are }60^\circ\text{ and }45^\circ
\displaystyle \text{respectively.}
\displaystyle \text{Let}
\displaystyle BC=x\text{ metres}
\displaystyle \text{and}
\displaystyle BD=y\text{ metres}
\displaystyle \text{We have to find }CD\text{ i.e. }(x+y)\text{ m}
\displaystyle \text{In right triangle }ABC,
\displaystyle \frac{AB}{BC}=\tan60^\circ
\displaystyle \Rightarrow \frac{AB}{BC}=\sqrt{3}
\displaystyle \Rightarrow \frac{300}{x}=\sqrt{3}
\displaystyle \Rightarrow x=\frac{300}{\sqrt{3}}
\displaystyle \Rightarrow x=\frac{300}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=100\sqrt{3}\text{ m}\qquad (i)
\displaystyle \text{In right triangle }ABD,
\displaystyle \frac{AB}{BD}=\tan45^\circ
\displaystyle \Rightarrow \frac{300}{y}=1
\displaystyle \Rightarrow y=300\text{ m}\qquad (ii)
\displaystyle \text{From equation }(i)\text{ and }(ii),\text{ we get}
\displaystyle CD=x+y=(100\sqrt{3}+300)\text{ m}
\displaystyle =100(\sqrt{3}+3)\text{ m}
\displaystyle =100(1.732+3)\text{ m}
\displaystyle =473.2\text{ m}
\displaystyle \therefore \text{The width of the river is }473.2\text{ metres.}
\\

\displaystyle \textbf{Question 25. }\text{A man observes a car from the top of a tower, which is moving towards the}
\displaystyle \text{tower with a uniform speed. If the angle of depression of the car changes from }30^{\circ}\text{ to}
\displaystyle 45^{\circ}\text{ in }12\text{ minutes, find the time taken by the car now to reach the tower.} \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AB\text{ be the tower of height }h\text{ metres.}
\displaystyle \text{Let }C\text{ be the original position of the car and after }12
\displaystyle \text{minutes let the car reaches at position }D.
\displaystyle \text{It is given that the angles of depression at }C\text{ and }D
\displaystyle \text{are }30^\circ\text{ and }45^\circ\text{ respectively.}
\displaystyle \text{Let the speed of the car be }x\text{ m/minute.}
\displaystyle \text{Now, distance covered by the car in }12\text{ minutes is }CD
\displaystyle CD=12x\text{ metres}
\displaystyle \text{Let car takes }t\text{ minutes to reach the tower }AB\text{ from the}
\displaystyle \text{position }D.
\displaystyle \text{Then }AD=xt\text{ metres.}
\displaystyle \text{In right triangle }DAB,
\displaystyle \frac{AB}{AD}=\tan45^\circ
\displaystyle \Rightarrow \frac{h}{xt}=1
\displaystyle \Rightarrow h=xt\qquad (i)
\displaystyle \text{In right triangle }CAB,
\displaystyle \frac{AB}{AC}=\tan30^\circ
\displaystyle \Rightarrow \frac{h}{xt+12x}=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow \sqrt{3}h=xt+12x\qquad (ii)
\displaystyle \text{Substituting the value of }h\text{ from equation }(i)\text{ in equation }(ii),
\displaystyle \sqrt{3}\,xt=xt+12x
\displaystyle \Rightarrow \sqrt{3}\,t=t+12
\displaystyle \Rightarrow (\sqrt{3}-1)t=12
\displaystyle \Rightarrow t=\frac{12}{\sqrt{3}-1}=\frac{12}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}
\displaystyle \Rightarrow t=6(\sqrt{3}+1)\text{ minutes}
\displaystyle =(6\times2.732)\text{ minutes}
\displaystyle =16.392\text{ minutes}\qquad (\because \sqrt{3}=1.732)
\displaystyle \text{Now }0.392\text{ minutes}=0.392\times60=23.52\text{ seconds}
\displaystyle \therefore t\approx16\text{ minutes }24\text{ seconds}
\displaystyle \text{Thus the car will reach the tower from }D\text{ in }16\text{ minutes}
\displaystyle \text{and }24\text{ seconds.}
\\

\displaystyle \textbf{Question 26. }\text{The angles of depression of two ships from an aeroplane flying at the height}
\displaystyle \text{of }7500\text{ m are }30^{\circ}\text{ and }45^{\circ}. \text{If both the ships are in the same line and}
\displaystyle \text{on the same side of the aeroplane such that one ship is exactly behind the other,}
\displaystyle \text{find the distance between the ships. [Use }\sqrt{3}=1.73\text{]} \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,
\displaystyle \tan45^\circ=\frac{7500}{y}
\displaystyle \Rightarrow 1=\frac{7500}{y}
\displaystyle \Rightarrow y=7500
\displaystyle \text{In }\triangle ABD,
\displaystyle \tan30^\circ=\frac{7500}{x+y}
\displaystyle \Rightarrow \frac{1}{\sqrt{3}}=\frac{7500}{x+y}
\displaystyle \Rightarrow x+y=7500\sqrt{3}
\displaystyle \Rightarrow x+7500=7500\sqrt{3}\qquad [\because y=7500]
\displaystyle \Rightarrow x=7500(\sqrt{3}-1)
\displaystyle \Rightarrow x=7500(1.73-1)
\displaystyle \Rightarrow x=7500\times0.73=5475\text{ m}
\\

\displaystyle \textbf{Question 27. }\text{If Figure, }AB\text{ is a }6\text{ m high pole and }CD\text{ is a ladder inclined at}
\displaystyle \text{an angle of }60^{\circ}\text{ to the horizontal and reaches up to a point }D\text{ of pole. If }AD=2.54\text{ m,}
\displaystyle \text{find the length of the ladder. (use }\sqrt{3}=1.73\text{)} \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle BD=AB-AD
\displaystyle =6\text{ m}-2.54\text{ m}=3.46\text{ m}
\displaystyle \text{In }\triangle DBC,
\displaystyle \frac{BD}{CD}=\sin60^\circ
\displaystyle \frac{3.46}{CD}=\frac{\sqrt{3}}{2}
\displaystyle \Rightarrow CD=\frac{2\times3.46}{\sqrt{3}}
\displaystyle =\frac{2\times3.46}{1.73}
\displaystyle =2\times2=4\text{ m}
\displaystyle \therefore \text{Length of the ladder is }4\text{ m}
\\

\displaystyle \textbf{Question 28. }\text{An observer, }1.7\text{ m tall, is }20\sqrt{3}\text{ m away from a tower.}
\displaystyle \text{The angle of elevation from the eye of observer to the top of tower is }30^{\circ}.
\displaystyle \text{Find the height of tower.} \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }CD\text{ be the tower of height }h.
\displaystyle \text{In }\triangle DEA,
\displaystyle \frac{DE}{AE}=\tan30^\circ
\displaystyle \frac{h-1.7}{20\sqrt{3}}=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow h-1.7=20
\displaystyle \Rightarrow h=20+1.7=21.7\text{ m}
\displaystyle \therefore \text{Height of the tower is }21.7\text{ m}
\\

\displaystyle \textbf{Question 29. }\text{The angles of depression of the top and bottom of a }50\text{ m high building}
\displaystyle \text{from the top of a tower are }45^{\circ}\text{ and }60^{\circ}\text{ respectively. Find the height of the tower}
\displaystyle \text{and the horizontal distance between the tower and the building. (use }\sqrt{3}=1.73\text{)} \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }CD\text{ is the building of height }50\text{ m and }AB\text{ be the tower.}
\displaystyle \text{Let horizontal distance between the tower and building is }BC=x\text{ metre.}
\displaystyle \therefore BCDE\text{ is a rectangle}
\displaystyle \text{So, }ED=BC\text{ and }BE=CD
\displaystyle \text{Also, }ED=x\text{ and }BE=50\text{ m}
\displaystyle \text{Let }AE=y.
\displaystyle \text{Now, in }\triangle AED,
\displaystyle \frac{y}{x}=\tan45^\circ
\displaystyle \Rightarrow \frac{y}{x}=1
\displaystyle \Rightarrow y=x\qquad (i)
\displaystyle \text{Now, in }\triangle ABC,
\displaystyle \frac{AB}{BC}=\tan60^\circ
\displaystyle \Rightarrow \frac{AE+BE}{BC}=\sqrt{3}
\displaystyle \Rightarrow \frac{y+50}{x}=\sqrt{3}
\displaystyle \Rightarrow x+50=\sqrt{3}x\qquad [\because y=x,\text{ using }(i)]
\displaystyle \Rightarrow \sqrt{3}x-x=50
\displaystyle \Rightarrow (\sqrt{3}-1)x=50
\displaystyle \Rightarrow x=\frac{50}{\sqrt{3}-1}
\displaystyle \Rightarrow x=\frac{50(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}
\displaystyle \Rightarrow x=\frac{50(\sqrt{3}+1)}{2}
\displaystyle \Rightarrow x=25(\sqrt{3}+1)=25\times2.73
\displaystyle \Rightarrow x=68.25\text{ m}
\displaystyle \therefore \text{Height of the tower }=50+y=50+68.25\quad [\because x=y]
\displaystyle =118.25\text{ m}
\displaystyle \text{Horizontal distance between the tower and the building}
\displaystyle =x=68.25\text{ m}
\\

\displaystyle \textbf{Question 30. }\text{A man standing on the deck of a ship, which is }10\text{ m above water level,}
\displaystyle \text{observes the angle of elevation of the top of a hill as }60^{\circ}\text{ and angle of depression}
\displaystyle \text{of the base of the hill as }30^{\circ}. \text{Find the distance of the hill from the ship and}
\displaystyle \text{the height of the hill.} \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AB\text{ be the water level, } \\ DA\text{ be the height of ship }=10\text{ m.}
\displaystyle \text{Let }BC\text{ be the hill of height }h \\ \text{from water level.}
\displaystyle \text{Let }AB=x\text{ m.}
\displaystyle \text{In }\triangle DEB,
\displaystyle \frac{BE}{DE}=\tan30^\circ
\displaystyle \Rightarrow \frac{10}{x}=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow x=10\sqrt{3}\text{ m}\qquad (i)
\displaystyle \text{Now, in }\triangle CED,
\displaystyle \frac{CE}{DE}=\tan60^\circ
\displaystyle \Rightarrow \frac{h-10}{10\sqrt{3}}=\sqrt{3}\qquad [\text{From }(i)]
\displaystyle \Rightarrow h-10=30
\displaystyle \Rightarrow h=40\text{ m}
\displaystyle \text{So, distance of hill from ship }=10\sqrt{3}\text{ m and the height of the hill }=40\text{ m.}
\\

\displaystyle \textbf{Question 31. }\text{An aeroplane, when flying at a height of }4000\text{ m from the ground passes}
\displaystyle \text{vertically above another aeroplane at an instant when the angles of elevation of the two}
\displaystyle \text{planes from the same point on the ground are }60^{\circ}\text{ and }45^{\circ}\text{ respectively. Find the}
\displaystyle \text{vertical distance between the aeroplanes at that instant. (Take }\sqrt{3}=1.73\text{)} \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P\text{ and }Q\text{ be the positions of the two aeroplanes.}
\displaystyle Q\text{ is vertically below }P\text{ and }OP=4000\text{ m}
\displaystyle \text{Let the distance between the two aeroplanes be }y\text{ m.}
\displaystyle \text{i.e. }PQ=y\text{ m and }AO=x\text{ m.}
\displaystyle \text{In }\triangle AOP,
\displaystyle \frac{OP}{AO}=\tan60^\circ
\displaystyle \Rightarrow \frac{4000}{x}=\sqrt{3}
\displaystyle \Rightarrow x=\frac{4000}{\sqrt{3}}
\displaystyle \text{In }\triangle AOQ,
\displaystyle \frac{OQ}{AO}=\tan45^\circ=\frac{OQ}{AO}=1
\displaystyle \Rightarrow OQ=AO
\displaystyle \therefore OQ=\frac{4000}{\sqrt{3}}\text{ m}
\displaystyle \text{Vertical distance between the two planes is given by}
\displaystyle y=OP-OQ
\displaystyle y=4000-\frac{4000}{\sqrt{3}}
\displaystyle y=4000\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)\text{ m}
\displaystyle y=1687.86\text{ m}
\displaystyle \therefore \text{Distance between the aeroplanes is }1687.86\text{ m.}
\\

\displaystyle \textbf{Question 32. }\text{A bird is sitting on the top of a }80\text{ m high tree. From a point on}
\displaystyle \text{the ground, the angle of elevation of the bird is }45^{\circ}. \text{The bird flies away horizontally}
\displaystyle \text{in such a way that it remained at a constant height from the ground. After }2\text{ seconds,}
\displaystyle \text{the angle of elevation of the bird from the same point is }30^{\circ}. \text{Find the speed of flying}
\displaystyle \text{of the bird. (Take }\sqrt{3}=1.732\text{)} \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let 'C' be the position of the bird and }BC\text{ is the tree,}
\displaystyle 80\text{ m high.}
\displaystyle \text{After }2\text{ seconds, position of bird is }E.
\displaystyle \text{Let }CE=x
\displaystyle \text{In }\triangle CBA,
\displaystyle \frac{BC}{AB}=\tan45^\circ
\displaystyle \Rightarrow \frac{80}{AB}=1
\displaystyle \Rightarrow AB=80\text{ m}
\displaystyle \text{In }\triangle EDA,
\displaystyle \frac{ED}{AD}=\tan30^\circ
\displaystyle \Rightarrow \frac{80}{AB+BD}=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow \frac{80}{80+x}=\frac{1}{\sqrt{3}}\qquad [\because AB=80\text{ m}]
\displaystyle \Rightarrow 80\sqrt{3}=80+x
\displaystyle \Rightarrow x=80\sqrt{3}-80
\displaystyle \Rightarrow x=80(\sqrt{3}-1)
\displaystyle \Rightarrow x=80(1.732-1)
\displaystyle \Rightarrow x=80\times0.732
\displaystyle \Rightarrow x=58.56\text{ m}
\displaystyle \therefore BD=x=58.56\text{ m}
\displaystyle \text{So, the speed of flying of the bird}
\displaystyle =\frac{\text{Distance }(BD)}{\text{Time}}
\displaystyle =\frac{58.56}{2}
\displaystyle =29.28\text{ m/s}
\\

\displaystyle \textbf{Question 33. }\text{The angles of elevation of the top of a tower from two points at a distance}
\displaystyle \text{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 }60^{\circ}\text{ and }30^{\circ}\text{ respectively. Find the height of the tower.} \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AB\text{ be the tower of height }h.
\displaystyle \text{Let }C\text{ and }D\text{ be the two points at distance }4\text{ m and }9\text{ m}
\displaystyle \text{from the base }B\text{ of the tower.}
\displaystyle \text{In }\triangle ABC,
\displaystyle \frac{AB}{BC}=\tan60^\circ
\displaystyle \Rightarrow \frac{h}{4}=\sqrt{3}
\displaystyle \Rightarrow h=4\sqrt{3}\qquad (i)
\displaystyle \text{In }\triangle ABD,
\displaystyle \frac{AB}{BD}=\tan30^\circ
\displaystyle \Rightarrow \frac{h}{9}=\frac{1}{\sqrt{3}}\qquad (ii)
\displaystyle \text{On solving }(i)\text{ and }(ii)\text{ together,}
\displaystyle h^{2}=36
\displaystyle \Rightarrow h=6
\displaystyle \therefore \text{Height of the tower is }6\text{ m.}
\\

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