Class 10: Heights and Distances

$\displaystyle \textbf{Question 1: }\text{The height of a tree is }\sqrt{3}\text{ times the length of its shadow. Find the} \\ \text{angle of elevation of the sun.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the shadow }=x.$
$\displaystyle \therefore \text{Tree height }=\sqrt{3}x.$
$\displaystyle \text{Let the angle of elevation be }\theta.$
$\displaystyle \tan\theta=\frac{\sqrt{3}x}{x}=\sqrt{3}.$
$\displaystyle \therefore \theta=60^{\circ}.$
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$\displaystyle \textbf{Question 2: }\text{The angle of elevation of the top of a tower from a point on the ground, }160\text{ m}$
$\displaystyle \text{from its foot, is }60^\circ.\text{ Find the height of the tower.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the height of the tower }=x\text{ m.}$
$\displaystyle \tan60^\circ=\frac{x}{160}.$
$\displaystyle \therefore x=160\tan60^\circ=160\sqrt{3}\text{ m}.$
$\displaystyle =160\times1.732=277.13\text{ m (approx.).}$
$\displaystyle {\therefore \text{Height of the tower }=277.13\text{ m (approx.).}}$
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$\displaystyle \textbf{Question 3: }\text{A ladder is placed along a wall such that its upper end is resting against a}$
$\displaystyle \text{vertical wall. The foot of the ladder is }2.4\text{ m from the wall and the ladder is}$
$\displaystyle \text{making an angle of }68^\circ\text{ with the ground. Find the height, up to which the}$
$\displaystyle \text{ladder reaches.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the height reached by the ladder }=h\text{ m.}$
$\displaystyle \text{Distance of the foot of the ladder from the wall }=2.4\text{ m.}$
$\displaystyle \tan68^\circ=\frac{h}{2.4}.$
$\displaystyle \therefore h=2.4\tan68^\circ=2.4\times2.475=5.94\text{ m (approx.).}$
$\displaystyle {\therefore \text{The ladder reaches a height of }5.94\text{ m (approx.).}}$
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$\displaystyle \textbf{Question 4: }\text{Two persons are standing on the opposite sides of a tower. They observe the}$
$\displaystyle \text{angles of elevation of the top of the tower to be }30^\circ\text{ and }38^\circ\text{ respectively.}$
$\displaystyle \text{Find the distance between them, if the height of the tower is }50\text{ m.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the distance of the first person from the tower }=x_1\text{ m.}$
$\displaystyle \text{Let the distance of the second person from the tower }=x_2\text{ m.}$
$\displaystyle \tan30^\circ=\frac{50}{x_1}.$
$\displaystyle \therefore x_1=\frac{50}{\tan30^\circ}=50\sqrt{3}=86.60\text{ m (approx.).}$
$\displaystyle \tan38^\circ=\frac{50}{x_2}.$
$\displaystyle \therefore x_2=\frac{50}{\tan38^\circ}=64.00\text{ m (approx.).}$
$\displaystyle \therefore \text{Distance between the two persons}=x_1+x_2=86.60+64.00=150.60\text{ m (approx.).}$
$\displaystyle {\therefore \text{The distance between the two persons is }150.60\text{ m (approx.).}}$
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$\displaystyle \textbf{Question 5: }\text{A kite is attached to a string. Find the length of the string, when the height}$
$\displaystyle \text{of the kite is }60\text{ m and the string makes an angle of }30^\circ\text{ with the ground.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Height of the kite }=60\text{ m.}$
$\displaystyle \text{Let the length of the string }=x\text{ m.}$
$\displaystyle \sin30^\circ=\frac{60}{x}.$
$\displaystyle \therefore x=\frac{60}{\sin30^\circ}=60\times2=120\text{ m.}$
$\displaystyle {\therefore \text{The length of the string is }120\text{ m.}}$
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$\displaystyle \textbf{Question 6: }\text{A boy, }1.6\text{ m tall, is }20\text{ m away from a tower and observes the angle}$
$\displaystyle \text{of elevation of the top of the tower to be (i) }45^\circ\text{ (ii) }60^\circ\text{. Find the}$
$\displaystyle \text{height of the tower in each case.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \textbf{Case (i): }\text{Angle of elevation }=45^\circ.$
$\displaystyle \text{Distance of the boy from the tower }=20\text{ m.}$
$\displaystyle \text{Let }h_1\text{ be the height of the tower above the boy's eye level.}$
$\displaystyle \tan45^\circ=\frac{h_1}{20}.$
$\displaystyle \therefore h_1=20\text{ m.}$
$\displaystyle \therefore \text{Height of the tower}=20+1.6=21.6\text{ m.}$
$\displaystyle \textbf{Case (ii): }\text{Angle of elevation }=60^\circ.$
$\displaystyle \text{Distance of the boy from the tower }=20\text{ m.}$
$\displaystyle \text{Let }h_2\text{ be the height of the tower above the boy's eye level.}$
$\displaystyle \tan60^\circ=\frac{h_2}{20}.$
$\displaystyle \therefore h_2=20\sqrt{3}=34.64\text{ m (approx.).}$
$\displaystyle \therefore \text{Height of the tower}=34.64+1.6=36.24\text{ m (approx.).}$
$\displaystyle {\therefore \text{The height of the tower is (i) }21.6\text{ m \quad (ii) }36.24\text{ m (approx.).}}$
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$\displaystyle \textbf{Question 7: }\text{The upper part of a tree, broken over by the wind, makes an angle of }45^\circ\text{ with}$
$\displaystyle \text{the ground; and the distance from the root to the point where the top of the tree}$
$\displaystyle \text{touches the ground is }15\text{ m. What was the height of the tree before it was}$
$\displaystyle \text{broken?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let }h_1\text{ m be the height of the standing part of the tree and }h_2\text{ m be the length of the broken part.}$
$\displaystyle \tan45^\circ=\frac{h_1}{15}.$
$\displaystyle \therefore h_1=15\text{ m.}$
$\displaystyle \sin45^\circ=\frac{15}{h_2}.$
$\displaystyle \therefore h_2=\frac{15}{\sin45^\circ}=15\sqrt{2}=21.21\text{ m (approx.).}$
$\displaystyle \therefore \text{Height of the tree before it was broken}=15+21.21=36.21\text{ m (approx.).}$
$\displaystyle {\therefore \text{The height of the tree before it was broken is }36.21\text{ m (approx.).}}$
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$\displaystyle \textbf{Question 8: }\text{The angle of elevation of the top of an unfinished tower at a point }80\text{ m}$
$\displaystyle \text{from its base is }30^\circ\text{. How much higher must the tower be raised so that its}$
$\displaystyle \text{angle of elevation at the same point may be }60^\circ\text{?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the height of the unfinished tower }=h\text{ m.}$
$\displaystyle \tan30^\circ=\frac{h}{80}.$
$\displaystyle \therefore h=80\tan30^\circ=\frac{80}{\sqrt{3}}=46.19\text{ m (approx.).}$
$\displaystyle \text{Let the tower be raised by }x\text{ m.}$
$\displaystyle \tan60^\circ=\frac{x+46.19}{80}.$
$\displaystyle \therefore x=80\tan60^\circ-46.19=92.37\text{ m (approx.).}$
$\displaystyle {\therefore \text{The tower must be raised by }92.37\text{ m (approx.).}}$
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$\displaystyle \textbf{Question 9: }\text{At a particular time when the sun's altitude is }30^\circ\text{, the length of the}$
$\displaystyle \text{shadow of a vertical tower is }45\text{ m. Calculate:}$
$\displaystyle \text{(i) the height of the tower,}$
$\displaystyle \text{(ii) the length of the shadow of the tower, when the sun's altitude is (a) }45^\circ\text{ (b) }60^\circ.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the height of the tower }=h\text{ m.}$
$\displaystyle \textbf{(i)}\ \tan30^\circ=\frac{h}{45}.$
$\displaystyle \therefore h=45\tan30^\circ=\frac{45}{\sqrt3}=15\sqrt3=25.98\text{ m (approx.).}$
$\displaystyle \textbf{(ii) Let the length of the shadow be }x_1\text{ m when the sun's altitude is }45^\circ\text{ and }x_2\text{ m when the sun's altitude is }60^\circ.$
$\displaystyle \tan45^\circ=\frac{25.98}{x_1}.$
$\displaystyle \therefore x_1=25.98\text{ m (approx.).}$
$\displaystyle \tan60^\circ=\frac{25.98}{x_2}.$
$\displaystyle \therefore x_2=\frac{25.98}{\tan60^\circ}=15\text{ m.}$
$\displaystyle {\therefore \text{(i) Height of the tower }=25.98\text{ m (approx.)}}$
$\displaystyle {\therefore \text{(ii) Length of the shadow: (a) }25.98\text{ m (approx.) \quad (b) }15\text{ m}}$
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$\displaystyle \textbf{Question 10: }\text{Two vertical poles are on either side of a road. A }30\text{ m long ladder is}$
$\displaystyle \text{placed between the two poles. When the ladder rests against one pole, it makes an}$
$\displaystyle \text{angle of }32^\circ24'\text{ with the pole, and when it is turned to rest against the other pole,}$
$\displaystyle \text{it makes an angle of }32^\circ24'\text{ with the road. Calculate the width of the road.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the distances of the foot of the ladder from the two poles be }x_1\text{ m and }x_2\text{ m respectively.}$
$\displaystyle 32^\circ24'=32.4^\circ.$
$\displaystyle \text{In the first position, the angle made by the ladder with the road}=90^\circ-32.4^\circ=57.6^\circ.$
$\displaystyle \cos57.6^\circ=\frac{x_1}{30}.$
$\displaystyle \therefore x_1=30\cos57.6^\circ=16.07\text{ m (approx.).}$
$\displaystyle \text{In the second position, the angle made by the ladder with the road}=32.4^\circ.$
$\displaystyle \cos32.4^\circ=\frac{x_2}{30}.$
$\displaystyle \therefore x_2=30\cos32.4^\circ=25.32\text{ m (approx.).}$
$\displaystyle \therefore \text{Width of the road}=x_1+x_2=16.07+25.32=41.39\text{ m.}$
$\displaystyle {\therefore \text{The width of the road is }41.4\text{ m (approx.).}}$
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$\displaystyle \textbf{Question 11: }\text{Two climbers are at points }A\text{ and }B\text{ on a vertical cliff face. To an}$
$\displaystyle \text{observer }C,\;40\text{ m from the foot of the cliff on the level ground, }A\text{ is at an}$
$\displaystyle \text{elevation of }48^\circ\text{ and }B\text{ at }57^\circ\text{. What is the distance between the climbers?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the heights of climbers }B\text{ and }A\text{ above the ground be }h_B\text{ m and }h_A\text{ m respectively.}$
$\displaystyle \tan57^\circ=\frac{h_B}{40}.$
$\displaystyle \therefore h_B=40\tan57^\circ=61.58\text{ m (approx.).}$
$\displaystyle \tan48^\circ=\frac{h_A}{40}.$
$\displaystyle \therefore h_A=40\tan48^\circ=44.41\text{ m (approx.).}$
$\displaystyle \therefore \text{Distance between the climbers}=h_B-h_A=61.58-44.41=17.17\text{ m (approx.).}$
$\displaystyle {\therefore \text{The distance between the two climbers is }17.17\text{ m (approx.).}}$
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$\displaystyle \textbf{Question 12: }\text{A man stands }9\text{ m away from a flag-pole. He observes that the angle of}$
$\displaystyle \text{elevation of the top of the pole is }28^\circ\text{ and the angle of depression of the bottom}$
$\displaystyle \text{of the pole is }13^\circ\text{. Calculate the height of the pole.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let }h_1\text{ m be the height of the top of the pole above the man's eye level.}$
$\displaystyle \text{Let }h_2\text{ m be the depth of the bottom of the pole below the man's eye level.}$
$\displaystyle \tan28^\circ=\frac{h_1}{9}.$
$\displaystyle \therefore h_1=9\tan28^\circ.$
$\displaystyle \text{Angle of depression}=\text{Angle of elevation}=13^\circ.$
$\displaystyle \tan13^\circ=\frac{h_2}{9}.$
$\displaystyle \therefore h_2=9\tan13^\circ.$
$\displaystyle \therefore \text{Height of the pole}=h_1+h_2=9\tan28^\circ+9\tan13^\circ=6.86\text{ m (approx.).}$
$\displaystyle {\therefore \text{The height of the pole is }6.86\text{ m (approx.).}}$
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$\displaystyle \textbf{Question 13: }\text{From the top of a cliff }92\text{ m high, the angle of depression of a buoy is}$
$\displaystyle \text{ }20^\circ\text{. Calculate, to the nearest metre, the distance of the buoy from the foot of}$
$\displaystyle \text{the cliff.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Angle of depression}=\text{Angle of elevation}=20^\circ.$
$\displaystyle \text{Let the distance of the buoy from the foot of the cliff }=x\text{ m.}$
$\displaystyle \tan20^\circ=\frac{92}{x}.$
$\displaystyle \therefore x=\frac{92}{\tan20^\circ}=252.77\text{ m.}$
$\displaystyle \therefore x\approx253\text{ m.}$
$\displaystyle {\therefore \text{The distance of the buoy from the foot of the cliff is }253\text{ m.}}$
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