\displaystyle \textbf{Question 1: }\text{With reference to the given figure, a man stands on the ground}
\displaystyle \text{at point }A,\text{ which is on the same horizontal plane as }B,\text{ the foot of the}
\displaystyle \text{vertical pole }BC.\text{ The height of the pole is }10\text{ m. The man's eye is }2\text{ m}
\displaystyle \text{above the ground. He observes the angle of elevation of }C,\text{ the top of the pole,}
\displaystyle \text{as }x^\circ,\text{ where }\tan x=\frac{2}{5}.\text{ Calculate: (i) the distance }AB\text{ in meters;}
\displaystyle \text{(ii) the angle of elevation of the top of the pole when he is standing}
\displaystyle 15\text{ meters from the pole. Give your answer to the nearest degree.}\hfill \text{[ICSE 1999]}
\displaystyle \text{Answer:}
\displaystyle \tan x=\frac{2}{5}
\displaystyle CE=10-2=8\text{ m}
\displaystyle \text{In }\triangle CDE,\quad \tan x=\frac{CE}{DE}
\displaystyle \frac{2}{5}=\frac{8}{DE}
\displaystyle DE=\frac{5}{2}\times8=20\text{ m}
\displaystyle \text{(i) }AB=DE=20\text{ m}
\displaystyle \text{(ii) When the man is }15\text{ m from the pole,}
\displaystyle \tan y=\frac{10-2}{15}=\frac{8}{15}=0.5333
\displaystyle y=28.07^\circ\approx28^\circ
\displaystyle \therefore \text{(i) }AB=20\text{ m and (ii) the angle of elevation is }28^\circ.
\\

\displaystyle \textbf{Question 2: }\text{From a window }A,\ 10\text{ m above the ground,}
\displaystyle \text{the angle of elevation of the top }C\text{ of a tower is }x^\circ,\text{ where }
\displaystyle \tan x=\frac{5}{2},\text{ and the angle of depression of the foot }D\text{ of the tower}
\displaystyle \text{is }y^\circ,\text{ where }\tan y=\frac{1}{4}.\text{ Calculate the height }CD\text{ of the tower.}
\displaystyle \hfill \text{[ICSE 2000]}
\displaystyle \text{Answer:}
\displaystyle ED=10\text{ m}
\displaystyle \tan y=\frac{ED}{AE}=\frac{1}{4}
\displaystyle \frac{10}{AE}=\frac{1}{4}
\displaystyle AE=40\text{ m}
\displaystyle \tan x=\frac{EC}{AE}=\frac{5}{2}
\displaystyle \frac{EC}{40}=\frac{5}{2}
\displaystyle EC=40\times\frac{5}{2}=100\text{ m}
\displaystyle CD=CE+ED=100+10=110\text{ m}
\displaystyle \therefore \text{The height of the tower is }110\text{ m}.
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\displaystyle \textbf{Question 3: }\text{A man standing on the bank of a river observes that}
\displaystyle \text{the angle of elevation of the top of the tower is }60^\circ.\text{ When he moves }
\displaystyle 50\text{ m away from the bank, he finds the angle of elevation to be }30^\circ.
\displaystyle \text{Calculate: (i) the width of the river and (ii) the height of the tower.}
\displaystyle \hfill \text{[ICSE 2003]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the width of the river be }w\text{ m}
\displaystyle \text{and the height of the tower be }h\text{ m}.
\displaystyle \text{From the first position, }\tan60^\circ=\frac{h}{w}
\displaystyle \sqrt{3}=\frac{h}{w}
\displaystyle h=\sqrt{3}w
\displaystyle \text{From the second position, }\tan30^\circ=\frac{h}{w+50}
\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{w+50}
\displaystyle w+50=\sqrt{3}h
\displaystyle w+50=\sqrt{3}(\sqrt{3}w)
\displaystyle w+50=3w
\displaystyle 2w=50
\displaystyle w=25\text{ m}
\displaystyle h=\sqrt{3}w=25\sqrt{3}
\displaystyle =25\times1.732=43.3\text{ m}
\displaystyle \therefore \text{Width of the river}=25\text{ m and height of the tower}=43.3\text{ m}.
\\

\displaystyle \textbf{Question 4: }\text{As observed from the top of an }80\text{ m tall lighthouse,}
\displaystyle \text{the angles of depression of two ships on the same side of the lighthouse}
\displaystyle \text{in a horizontal line with its base are }30^\circ\text{ and }40^\circ\text{ respectively.}
\displaystyle \text{Find the distance between the two ships. Give your answer correct}
\displaystyle \text{to the nearest meter.}\hfill \text{[ICSE 2012]}
\displaystyle \text{Answer:} \displaystyle \text{Let }AB\text{ be the lighthouse and }AB=80\text{ m}.
\displaystyle \text{The angles of elevation from the ships are }40^\circ\text{ and }30^\circ.
\displaystyle \text{For the nearer ship at }C,\quad \tan40^\circ=\frac{AB}{BC}
\displaystyle \tan40^\circ=\frac{80}{BC}
\displaystyle BC=\frac{80}{\tan40^\circ}=\frac{80}{0.839}=95.34\text{ m}
\displaystyle \text{For the farther ship at }D,\quad \tan30^\circ=\frac{AB}{BD}
\displaystyle \tan30^\circ=\frac{80}{BD}
\displaystyle BD=\frac{80}{\tan30^\circ}=80\sqrt{3}=138.56\text{ m}
\displaystyle CD=BD-BC=138.56-95.34=43.22\text{ m}
\displaystyle CD\approx43\text{ m}
\displaystyle \therefore \text{The distance between the two ships is }43\text{ m}.
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\displaystyle \textbf{Question 5: }\text{An airplane, at an altitude of }250\text{ m, observes}
\displaystyle \text{the angles of depression of two boats on the opposite banks of a river}
\displaystyle \text{to be }45^\circ\text{ and }60^\circ\text{ respectively. Find the width of the river.}
\displaystyle \text{Write the answer correct to the nearest whole number.}\hfill \text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle AD=250\text{ m}
\displaystyle \text{In }\triangle ABD,\quad \tan45^\circ=\frac{AD}{BD}
\displaystyle 1=\frac{250}{BD}
\displaystyle BD=250\text{ m}
\displaystyle \text{In }\triangle ACD,\quad \tan60^\circ=\frac{AD}{CD}
\displaystyle \sqrt{3}=\frac{250}{CD}
\displaystyle CD=\frac{250}{\sqrt{3}}=144.34\text{ m}
\displaystyle \text{Width of the river}=BC=BD+CD
\displaystyle =250+144.34=394.34\text{ m}
\displaystyle \approx394\text{ m}
\displaystyle \therefore \text{The width of the river is }394\text{ m}.
\\

\displaystyle \textbf{Question 6: }\text{From the top of a lighthouse }100\text{ m high,}
\displaystyle \text{the angles of depression of two ships are observed as }48^\circ\text{ and }36^\circ
\displaystyle \text{respectively. Find the distance between the two ships, to the nearest meter, if:}
\displaystyle \text{(i) the ships are on the same side of the lighthouse,}
\displaystyle \text{(ii) the ships are on the opposite sides of the lighthouse.}\hfill \text{[ICSE 2010]}
\displaystyle \text{Answer:}
\displaystyle AB=100\text{ m}
\displaystyle \text{From }\triangle ABC,\quad \tan48^\circ=\frac{AB}{BC}
\displaystyle BC=\frac{100}{\tan48^\circ}=\frac{100}{1.1106}=90.04\text{ m}
\displaystyle \text{From }\triangle ADB,\quad \tan36^\circ=\frac{AB}{BD}
\displaystyle BD=\frac{100}{\tan36^\circ}=\frac{100}{0.7265}=137.64\text{ m}
\displaystyle \text{(i) If the ships are on the same side of the lighthouse,}
\displaystyle \text{distance between the ships}=BD-BC
\displaystyle =137.64-90.04=47.60\text{ m}\approx48\text{ m}
\displaystyle \text{(ii) If the ships are on opposite sides of the lighthouse,}
\displaystyle \text{distance between the ships}=BD+BC
\displaystyle =137.64+90.04=227.68\text{ m}\approx228\text{ m}
\displaystyle \therefore \text{The required distances are }48\text{ m and }228\text{ m respectively.}
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\displaystyle \textbf{Question 7: }\text{From the top of a hill, the angles of depression}
\displaystyle \text{of two consecutive kilometer stones, due east, are found to be }30^\circ
\displaystyle \text{and }45^\circ\text{ respectively. Find the distances of the two stones}
\displaystyle \text{from the foot of the hill.}\hfill \text{[ICSE 2007]}
\displaystyle \text{Answer:} \displaystyle \text{Let }AB=h\text{ km and }BC=x\text{ km}.
\displaystyle \text{Since the stones are consecutive kilometer stones, }CD=1\text{ km}
\displaystyle \therefore BD=x+1\text{ km}
\displaystyle \text{From }\triangle ABC,\quad \tan45^\circ=\frac{AB}{BC}
\displaystyle 1=\frac{h}{x}
\displaystyle h=x
\displaystyle \text{From }\triangle ABD,\quad \tan30^\circ=\frac{AB}{BD}
\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{x+1}
\displaystyle \frac{1}{\sqrt{3}}=\frac{x}{x+1}
\displaystyle \sqrt{3}x=x+1
\displaystyle x(\sqrt{3}-1)=1
\displaystyle x=\frac{1}{\sqrt{3}-1}=1.366\text{ km}
\displaystyle \therefore BC=1.366\text{ km}
\displaystyle BD=BC+CD=1.366+1=2.366\text{ km}
\displaystyle \therefore \text{The distances of the two stones from the foot of the hill are}
\displaystyle 1.366\text{ km and }2.366\text{ km.}
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