\displaystyle \textbf{Question 1: }\text{Find }AD. \displaystyle \textbf{Answer:}
\displaystyle \text{In right-angled triangle }ADE,\ DE=CB=20\text{ m and }\angle ADE=32^\circ.
\displaystyle \cos32^\circ=\frac{DE}{AD}
\displaystyle \cos32^\circ=\frac{20}{AD}
\displaystyle \therefore AD=\frac{20}{\cos32^\circ}
\displaystyle =\frac{20}{0.8480}=23.58\text{ m (approx.).}
\displaystyle \therefore \text{The length of }AD\text{ is }23.58\text{ m (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In the following diagram, }AB\text{ is a floor-board. }PQRS\text{ is a cubical box}
\displaystyle \text{with each edge }1\text{ m and }\angle B=60^\circ.\text{ Calculate the length of the board }AB. \displaystyle \textbf{Answer:}
\displaystyle \text{Since each edge of the cubical box is }1\text{ m, }PS=PQ=1\text{ m.}
\displaystyle \text{In right-angled triangle }BPS,
\displaystyle \sin60^\circ=\frac{PS}{PB}=\frac{1}{PB}
\displaystyle \therefore PB=\frac{1}{\sin60^\circ}=\frac{1}{\frac{\sqrt3}{2}}=\frac{2}{\sqrt3}\text{ m}
\displaystyle \text{Since }PQ\parallel BS,\ \angle APQ=\angle PBS=60^\circ.
\displaystyle \text{In right-angled triangle }APQ,
\displaystyle \cos60^\circ=\frac{PQ}{AP}=\frac{1}{AP}
\displaystyle \therefore AP=\frac{1}{\cos60^\circ}=\frac{1}{\frac12}=2\text{ m}
\displaystyle AB=PB+PA=\frac{2}{\sqrt3}+2
\displaystyle =1.1547+2=3.1547\text{ m}
\displaystyle \therefore \text{The length of the board }AB\text{ is }3.15\text{ m (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Calculate }BC. \displaystyle \textbf{Answer:}
\displaystyle \text{In right-angled triangle }ACD,\ AD=20\text{ m and }\angle CAD=42^\circ.
\displaystyle \tan42^\circ=\frac{CD}{AD}=\frac{CD}{20}
\displaystyle \therefore CD=20\tan42^\circ
\displaystyle \text{In right-angled triangle }ABD,\ \angle ABD=35^\circ.
\displaystyle \tan35^\circ=\frac{AD}{BD}=\frac{20}{BC+CD}
\displaystyle BC+CD=\frac{20}{\tan35^\circ}
\displaystyle \therefore BC=\frac{20}{\tan35^\circ}-CD
\displaystyle =\frac{20}{\tan35^\circ}-20\tan42^\circ
\displaystyle =28.563-18.008
\displaystyle =10.555\text{ m}
\displaystyle \therefore \text{The length of }BC\text{ is }10.56\text{ m (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Calculate }AB. \displaystyle \textbf{Answer:}
\displaystyle \text{In right-angled triangle }ADE,\ DE=6\text{ m and }\angle AED=30^\circ.
\displaystyle \cos30^\circ=\frac{AE}{DE}=\frac{AE}{6}
\displaystyle \therefore AE=6\cos30^\circ
\displaystyle =6\times\frac{\sqrt3}{2}=3\sqrt3=5.196\text{ m}
\displaystyle \text{Since }EB\perp BC\text{ and }\angle ECB=47^\circ,
\displaystyle \angle CEB=90^\circ-47^\circ=43^\circ.
\displaystyle \text{In right-angled triangle }EBC,\ EC=5\text{ m.}
\displaystyle \cos43^\circ=\frac{EB}{EC}=\frac{EB}{5}
\displaystyle \therefore EB=5\cos43^\circ=5\times0.7314=3.657\text{ m}
\displaystyle AB=AE+EB
\displaystyle =5.196+3.657=8.853\text{ m}
\displaystyle \therefore \text{The length of }AB\text{ is }8.85\text{ m (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The radius of a circle is }15\text{ cm and chord }AB\text{ subtends an angle of}
\displaystyle 131^\circ\text{ at the centre }C\text{ of the circle. Using trigonometry, calculate:}
\displaystyle \text{(i) the length of }AB;\qquad\text{(ii) the distance of }AB\text{ from the centre }C.
\displaystyle \textbf{Answer:}
\displaystyle \text{Draw }CX\perp AB,\text{ meeting }AB\text{ at }X.
\displaystyle \text{The perpendicular from the centre to a chord bisects the chord and the central angle.}
\displaystyle \therefore AX=XB\text{ and }\angle ACX=\frac{131^\circ}{2}=65.5^\circ
\displaystyle \angle CAX=90^\circ-65.5^\circ=24.5^\circ
\displaystyle \text{In right-angled triangle }ACX,\ AC=15\text{ cm.}
\displaystyle \cos24.5^\circ=\frac{AX}{AC}=\frac{AX}{15}
\displaystyle \therefore AX=15\cos24.5^\circ=13.64\text{ cm}
\displaystyle AB=2AX=2\times13.64=27.28\text{ cm}
\displaystyle \therefore AB=27.3\text{ cm (approx.).}
\displaystyle \sin24.5^\circ=\frac{CX}{AC}=\frac{CX}{15}
\displaystyle \therefore CX=15\sin24.5^\circ=6.22\text{ cm}
\displaystyle \therefore \text{The distance of chord }AB\text{ from the centre }C\text{ is }6.22\text{ cm (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{At a point on level ground, the tangent of the angle of elevation of a}
\displaystyle \text{vertical tower is }\frac{5}{12}.\text{ After walking }192\text{ m towards the tower, the tangent of}
\displaystyle \text{the angle of elevation is }\frac{3}{4}.\text{ Find the height of the tower.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the height of the tower }AB=h\text{ m and }DB=x\text{ m.}
\displaystyle \text{Since }CD=192\text{ m, }CB=CD+DB=192+x.
\displaystyle \text{From point }D,
\displaystyle \tan\angle ADB=\frac{AB}{DB}=\frac{h}{x}=\frac{3}{4}
\displaystyle \therefore h=\frac{3x}{4}
\displaystyle \text{From point }C,
\displaystyle \tan\angle ACB=\frac{AB}{CB}=\frac{h}{192+x}=\frac{5}{12}
\displaystyle \therefore h=\frac{5}{12}(192+x)
\displaystyle \frac{5}{12}(192+x)=\frac{3x}{4}
\displaystyle 5(192+x)=9x
\displaystyle 960+5x=9x
\displaystyle 4x=960
\displaystyle x=240\text{ m}
\displaystyle h=\frac{3}{4}\times240=180\text{ m}
\displaystyle \therefore \text{The height of the tower is }180\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A vertical tower stands on a horizontal plane and is surmounted by a}
\displaystyle \text{vertical flagstaff of height }h\text{ m. At a point on the plane, the angle of elevation}
\displaystyle \text{of the bottom of the flagstaff is }\alpha\text{ and that of the top of the flagstaff is }\beta.
\displaystyle \text{Prove that the height of the tower is }\frac{h\tan\alpha}{\tan\beta-\tan\alpha}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }DB\text{ be the height of the tower and }CB\text{ be the horizontal distance.}
\displaystyle \text{In right-angled triangle }CDB,
\displaystyle \tan\alpha=\frac{DB}{CB}
\displaystyle \therefore CB=\frac{DB}{\tan\alpha}
\displaystyle \text{In right-angled triangle }CAB,
\displaystyle \tan\beta=\frac{AB}{CB}=\frac{DB+h}{CB}
\displaystyle \therefore \tan\beta=\frac{DB+h}{\frac{DB}{\tan\alpha}}
\displaystyle \therefore DB+h=\frac{DB\tan\beta}{\tan\alpha}
\displaystyle \therefore h=DB\left(\frac{\tan\beta}{\tan\alpha}-1\right)
\displaystyle =DB\left(\frac{\tan\beta-\tan\alpha}{\tan\alpha}\right)
\displaystyle \therefore DB=\frac{h\tan\alpha}{\tan\beta-\tan\alpha}
\displaystyle \therefore \text{The height of the tower is }\frac{h\tan\alpha}{\tan\beta-\tan\alpha}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{With reference to the given figure, a man stands on the ground at point }A,
\displaystyle \text{which is on the same horizontal plane as }B,\text{ the foot of the vertical pole }BC.
\displaystyle \text{The height of the pole is }10\text{ m and the man's eye is }2\text{ m above the ground.}
\displaystyle \text{He observes the angle of elevation of the top of the pole as }x^\circ,\text{ where }\tan x=\frac{2}{5}.
\displaystyle \text{Calculate: (i) the distance }AB\text{ in metres; (ii) the angle of elevation of the top of}
\displaystyle \text{the pole when he is standing }15\text{ m from the pole. Give your answer to the nearest degree.}
\displaystyle \hfill\text{[ICSE 1999]} \displaystyle \textbf{Answer:}
\displaystyle \text{The height of the pole above the man's eye level is}
\displaystyle CE=BC-BE=10-2=8\text{ m}
\displaystyle \text{(i) In right-angled triangle }DCE,
\displaystyle \tan x=\frac{CE}{DE}=\frac{2}{5}
\displaystyle \frac{8}{DE}=\frac{2}{5}
\displaystyle 2DE=40
\displaystyle DE=20\text{ m}
\displaystyle \text{Since }ABED\text{ is a rectangle, }AB=DE.
\displaystyle \therefore AB=20\text{ m}
\displaystyle \text{(ii) Let the required angle of elevation be }y^\circ.
\displaystyle \tan y=\frac{8}{15}
\displaystyle y=\tan^{-1}\left(\frac{8}{15}\right)
\displaystyle y=28.07^\circ
\displaystyle \therefore \text{The required angle of elevation is }28^\circ\text{, to the nearest degree.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The angles of elevation of the top of a tower from two points on the}
\displaystyle \text{ground at distances }a\text{ and }b\text{ m from the base of the tower and in the same}
\displaystyle \text{straight line with it are complementary. Prove that the height of the tower is}
\displaystyle \sqrt{ab}\text{ m}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the height of the tower be }h\text{ m.}
\displaystyle \tan\alpha=\frac{h}{a}
\displaystyle \tan\beta=\frac{h}{b}
\displaystyle \text{Since }\alpha+\beta=90^\circ,
\displaystyle \tan(\alpha+\beta)\text{ is undefined.}
\displaystyle \therefore 1-\tan\alpha\tan\beta=0
\displaystyle \therefore \tan\alpha\tan\beta=1
\displaystyle \therefore \frac{h}{a}\times\frac{h}{b}=1
\displaystyle \therefore \frac{h^2}{ab}=1
\displaystyle \therefore h^2=ab
\displaystyle \therefore h=\sqrt{ab}\text{ m}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{From a window }A,\ 10\text{ m above the ground, the angle of elevation of the}
\displaystyle \text{top }C\text{ of a tower is }x^\circ,\text{ where }\tan x^\circ=\frac{5}{2},\text{ and the angle of depression}
\displaystyle \text{of the foot }D\text{ of the tower is }y^\circ,\text{ where }\tan y^\circ=\frac{1}{4}.\text{ Calculate the}
\displaystyle \text{height }CD\text{ of the tower in metres.}\hfill\text{[ICSE 2000]}
\displaystyle \textbf{Answer:}
\displaystyle \text{Since }A\text{ is }10\text{ m above the ground, }ED=AB=10\text{ m.}
\displaystyle \text{In right-angled triangle }AED,
\displaystyle \tan y^\circ=\frac{ED}{AE}=\frac{1}{4}
\displaystyle \frac{10}{AE}=\frac{1}{4}
\displaystyle \therefore AE=40\text{ m}
\displaystyle \text{In right-angled triangle }AEC,
\displaystyle \tan x^\circ=\frac{EC}{AE}=\frac{5}{2}
\displaystyle \frac{EC}{40}=\frac{5}{2}
\displaystyle \therefore EC=40\times\frac{5}{2}=100\text{ m}
\displaystyle CD=CE+ED
\displaystyle =100+10=110\text{ m}
\displaystyle \therefore \text{The height of the tower is }110\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A vertical tower is }20\text{ m high. A man standing at some distance}
\displaystyle \text{from the tower knows that the cosine of the angle of elevation of the top}
\displaystyle \text{of the tower is }0.53.\text{ How far is he standing from the foot of the tower?}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the distance of the man from the foot of the tower }AB=x\text{ m.}
\displaystyle \text{Then }CB=\sqrt{x^2+20^2}=\sqrt{x^2+400}.
\displaystyle \cos\theta=\frac{AB}{CB}=0.53
\displaystyle \therefore \frac{x}{\sqrt{x^2+400}}=0.53
\displaystyle \therefore x=0.53\sqrt{x^2+400}
\displaystyle \text{Squaring both sides,}
\displaystyle x^2=0.2809(x^2+400)
\displaystyle x^2-0.2809x^2=112.36
\displaystyle 0.7191x^2=112.36
\displaystyle x^2=156.25
\displaystyle x=\sqrt{156.25}=12.5\text{ m}
\displaystyle \therefore \text{The man is standing }12.5\text{ m from the foot of the tower.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{A man standing on the bank of a river observes that the angle of elevation}
\displaystyle \text{of the top of a tree is }60^\circ.\text{ When he moves }50\text{ m away from the bank, he finds}
\displaystyle \text{the angle of elevation to be }30^\circ.\text{ Calculate: (i) the width of the river and}
\displaystyle \text{(ii) the height of the tree.}\hfill\text{[ICSE 2003]}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the width of the river }AB=w\text{ m and the height of the tree }BC=h\text{ m.}
\displaystyle \text{In right-angled triangle }ABC,
\displaystyle \tan60^\circ=\frac{BC}{AB}=\frac{h}{w}
\displaystyle \therefore \sqrt3=\frac{h}{w}
\displaystyle \therefore h=\sqrt3\,w
\displaystyle \text{After moving }50\text{ m away from the bank, }DB=w+50.
\displaystyle \text{In right-angled triangle }DBC,
\displaystyle \tan30^\circ=\frac{BC}{DB}=\frac{h}{w+50}
\displaystyle \frac{1}{\sqrt3}=\frac{h}{w+50}
\displaystyle \therefore \sqrt3\,h=w+50
\displaystyle \therefore \sqrt3(\sqrt3\,w)=w+50
\displaystyle 3w=w+50
\displaystyle 2w=50
\displaystyle w=25\text{ m}
\displaystyle h=\sqrt3\,w=25\sqrt3\text{ m}
\displaystyle =43.3\text{ m (approx.).}
\displaystyle \therefore \text{(i) The width of the river is }25\text{ m.}
\displaystyle \therefore \text{(ii) The height of the tree is }25\sqrt3\text{ m, i.e. }43.3\text{ m (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{A }20\text{ m high vertical pole and a vertical tower stand on the same}
\displaystyle \text{level ground. The angle of elevation of the top of the tower, as seen from the}
\displaystyle \text{foot of the pole, is }60^\circ,\text{ and the angle of elevation of the top of the pole,}
\displaystyle \text{as seen from the foot of the tower, is }30^\circ.\text{ Find: (i) the height of the tower;}
\displaystyle \text{(ii) the horizontal distance between the pole and the tower.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }CD=20\text{ m be the height of the pole, }AB\text{ the height of the tower}
\displaystyle \text{and }CB\text{ the horizontal distance between the pole and the tower.}
\displaystyle \text{In right-angled triangle }DCB,
\displaystyle \tan30^\circ=\frac{CD}{CB}=\frac{20}{CB}
\displaystyle \frac{1}{\sqrt3}=\frac{20}{CB}
\displaystyle \therefore CB=20\sqrt3\text{ m}
\displaystyle =34.64\text{ m (approx.).}
\displaystyle \text{In right-angled triangle }ACB,
\displaystyle \tan60^\circ=\frac{AB}{CB}
\displaystyle \sqrt3=\frac{AB}{20\sqrt3}
\displaystyle \therefore AB=20\sqrt3\times\sqrt3=60\text{ m}
\displaystyle \therefore \text{(i) The height of the tower is }60\text{ m.}
\displaystyle \therefore \text{(ii) The horizontal distance is }20\sqrt3\text{ m, i.e. }34.64\text{ m (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{A vertical pole and a vertical tower stand on the same level ground. From}
\displaystyle \text{the top of the pole, the angle of elevation of the top of the tower is }60^\circ,\text{ and}
\displaystyle \text{the angle of depression of the foot of the tower is }30^\circ.\text{ Find: (i) the height of}
\displaystyle \text{the tower, if the pole is }20\text{ m high; (ii) the height of the pole, if the tower is}
\displaystyle 75\text{ m high}.
\displaystyle \textbf{Answer:}
\displaystyle \text{(i) Given }AB=20\text{ m. Since }AE\parallel BC,\ CE=AB=20\text{ m.}
\displaystyle \text{In right-angled triangle }AEC,
\displaystyle \tan30^\circ=\frac{EC}{AE}=\frac{20}{AE}
\displaystyle \frac{1}{\sqrt3}=\frac{20}{AE}
\displaystyle \therefore AE=20\sqrt3\text{ m}
\displaystyle \text{In right-angled triangle }AED,
\displaystyle \tan60^\circ=\frac{DE}{AE}
\displaystyle \sqrt3=\frac{DE}{20\sqrt3}
\displaystyle \therefore DE=20\sqrt3\times\sqrt3=60\text{ m}
\displaystyle CD=CE+DE=20+60=80\text{ m}
\displaystyle \therefore \text{The height of the tower is }80\text{ m.}
\displaystyle \text{(ii) Let the height of the pole }AB=h\text{ m.}
\displaystyle \text{Since the height of the tower is }75\text{ m, }CE=h\text{ and }DE=75-h.
\displaystyle \text{In right-angled triangle }AEC,
\displaystyle \tan30^\circ=\frac{EC}{AE}=\frac{h}{AE}
\displaystyle \frac{1}{\sqrt3}=\frac{h}{AE}
\displaystyle \therefore AE=\sqrt3h
\displaystyle \text{In right-angled triangle }AED,
\displaystyle \tan60^\circ=\frac{DE}{AE}=\frac{75-h}{AE}
\displaystyle \sqrt3=\frac{75-h}{\sqrt3h}
\displaystyle \therefore 75-h=3h
\displaystyle 75=4h
\displaystyle h=\frac{75}{4}=18.75\text{ m}
\displaystyle \therefore \text{The height of the pole is }18.75\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{From a point }36\text{ m above the surface of a lake, the angle of elevation}
\displaystyle \text{of a bird is observed to be }30^\circ,\text{ and the angle of depression of its image in the}
\displaystyle \text{water is observed to be }60^\circ.\text{ Find the actual height of the bird above the}
\displaystyle \text{surface of the lake.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }AC=h\text{ m and }BC=x\text{ m.}
\displaystyle \text{Since the observer is }36\text{ m above the lake, }CE=BD=36\text{ m.}
\displaystyle \text{In right-angled triangle }ABC,
\displaystyle \tan30^\circ=\frac{AC}{BC}=\frac{h}{x}
\displaystyle \frac{1}{\sqrt3}=\frac{h}{x}
\displaystyle \therefore x=\sqrt3h
\displaystyle \text{The actual height of the bird above the lake is}
\displaystyle AE=AC+CE=h+36.
\displaystyle \text{Since the image is as far below the lake as the bird is above it,}
\displaystyle EF=AE=h+36.
\displaystyle \text{Therefore, }BF=BE+EF=36+(h+36)=h+72.
\displaystyle \text{In right-angled triangle }BFC,
\displaystyle \tan60^\circ=\frac{BF}{BC}
\displaystyle \sqrt3=\frac{h+72}{\sqrt3h}
\displaystyle 3h=h+72
\displaystyle 2h=72
\displaystyle h=36\text{ m}
\displaystyle AE=h+36=36+36=72\text{ m}
\displaystyle \therefore \text{The actual height of the bird above the lake is }72\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{A man observes the angle of elevation of the top of a building to be }30^\circ.
\displaystyle \text{He walks towards it in a horizontal line through its base. On covering }60\text{ m, the}
\displaystyle \text{angle of elevation changes to }60^\circ.\text{ Find the height of the building, correct}
\displaystyle \text{to the nearest metre.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the height of the building }AB=h\text{ m.}
\displaystyle \text{In right-angled triangle }ABC,
\displaystyle \tan60^\circ=\frac{AB}{CB}=\frac{h}{CB}
\displaystyle \sqrt3=\frac{h}{CB}
\displaystyle \therefore CB=\frac{h}{\sqrt3}
\displaystyle \text{In right-angled triangle }ABD,
\displaystyle \tan30^\circ=\frac{AB}{DB}=\frac{h}{DB}
\displaystyle \frac{1}{\sqrt3}=\frac{h}{DB}
\displaystyle \therefore DB=\sqrt3h
\displaystyle \text{Since }DC=60\text{ m, }DB-CB=60.
\displaystyle \sqrt3h-\frac{h}{\sqrt3}=60
\displaystyle h\left(\frac{3-1}{\sqrt3}\right)=60
\displaystyle \frac{2h}{\sqrt3}=60
\displaystyle h=30\sqrt3\text{ m}
\displaystyle =51.96\text{ m (approx.).}
\displaystyle \therefore \text{The height of the building is }52\text{ m, correct to the nearest metre.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{From the top of an }80\text{ m tall lighthouse, the angles of depression of}
\displaystyle \text{two ships on the same side of the lighthouse and in a horizontal line with its base}
\displaystyle \text{are }30^\circ\text{ and }40^\circ\text{ respectively. Find the distance between the two ships.}
\displaystyle \text{Give your answer correct to the nearest metre.}\hfill\text{[ICSE 2012]}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }AB=80\text{ m be the height of the lighthouse.}
\displaystyle \text{Let the nearer ship be at }C\text{ and the farther ship be at }D.
\displaystyle \text{The angles of elevation at }C\text{ and }D\text{ are }40^\circ\text{ and }30^\circ\text{ respectively.}
\displaystyle \text{In right-angled triangle }ABC,
\displaystyle \tan40^\circ=\frac{AB}{BC}=\frac{80}{BC}
\displaystyle \therefore BC=\frac{80}{\tan40^\circ}
\displaystyle =\frac{80}{0.8391}=95.34\text{ m (approx.).}
\displaystyle \text{In right-angled triangle }ABD,
\displaystyle \tan30^\circ=\frac{AB}{BD}=\frac{80}{BD}
\displaystyle \therefore BD=\frac{80}{\tan30^\circ}
\displaystyle =80\sqrt3=138.56\text{ m (approx.).}
\displaystyle CD=BD-BC
\displaystyle =138.56-95.34=43.22\text{ m}
\displaystyle \therefore \text{The distance between the two ships is }43\text{ m, correct to the nearest metre.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In the given figure, from the top of a building }AB=60\text{ m high, the}
\displaystyle \text{angles of depression of the top and bottom of a vertical lamp post }CD\text{ are }30^\circ
\displaystyle \text{and }60^\circ\text{ respectively. Find: (i) the horizontal distance between }AB\text{ and }CD;
\displaystyle \text{(ii) the height of the lamp post.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the height of the lamp post }CD=h\text{ m.}
\displaystyle \text{(i) The angle of elevation of }A\text{ from }C\text{ is }60^\circ.
\displaystyle \text{In right-angled triangle }ABC,
\displaystyle \tan60^\circ=\frac{AB}{BC}=\frac{60}{BC}
\displaystyle \sqrt3=\frac{60}{BC}
\displaystyle \therefore BC=\frac{60}{\sqrt3}=20\sqrt3\text{ m}
\displaystyle =34.64\text{ m (approx.).}
\displaystyle \text{(ii) Draw a horizontal line from }D\text{ to meet }AB\text{ at }E.
\displaystyle \text{Then }AE=AB-EB=60-h\text{ and }ED=BC=20\sqrt3\text{ m.}
\displaystyle \text{The angle of depression of }D\text{ from }A\text{ is }30^\circ.
\displaystyle \tan30^\circ=\frac{AE}{ED}=\frac{60-h}{20\sqrt3}
\displaystyle \frac{1}{\sqrt3}=\frac{60-h}{20\sqrt3}
\displaystyle 60-h=20
\displaystyle h=40\text{ m}
\displaystyle \therefore \text{(i) The horizontal distance is }20\sqrt3\text{ m, i.e. }34.64\text{ m (approx.).}
\displaystyle \therefore \text{(ii) The height of the lamp post is }40\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{An aeroplane at an altitude of }250\text{ m observes the angles of depression}
\displaystyle \text{of two boats on the opposite banks of a river to be }45^\circ\text{ and }60^\circ\text{ respectively.}
\displaystyle \text{Find the width of the river. Give your answer correct to the nearest whole number.}
\displaystyle \hfill\text{[ICSE 2014]}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }AD=250\text{ m be the altitude of the aeroplane.}
\displaystyle \text{The angles of elevation of }A\text{ from }B\text{ and }C\text{ are }45^\circ\text{ and }60^\circ.
\displaystyle \text{In right-angled triangle }ABD,
\displaystyle \tan45^\circ=\frac{AD}{BD}=\frac{250}{BD}
\displaystyle \therefore BD=\frac{250}{\tan45^\circ}=250\text{ m}
\displaystyle \text{In right-angled triangle }ACD,
\displaystyle \tan60^\circ=\frac{AD}{CD}=\frac{250}{CD}
\displaystyle \therefore CD=\frac{250}{\tan60^\circ}=\frac{250}{\sqrt3}
\displaystyle =144.34\text{ m (approx.).}
\displaystyle BC=BD+CD
\displaystyle =250+144.34=394.34\text{ m}
\displaystyle \therefore \text{The width of the river is }394\text{ m, correct to the nearest whole number.}
\displaystyle \\

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