Class 10: Heights and Distances – Very Short Answer Type Questions

$\displaystyle \textbf{Question 1: }\text{The height of a tower is }10\text{ m. What is the length of its shadow when}$
$\displaystyle \text{Sun's altitude is }45^\circ\text{?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the length of the shadow be }x\text{ m.}$
$\displaystyle \tan45^\circ=\frac{10}{x}$
$\displaystyle 1=\frac{10}{x}\Rightarrow x=10\text{ m}$
$\displaystyle \therefore \text{The length of the shadow is }10\text{ m.}$
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$\displaystyle \textbf{Question 2: }\text{If the ratio of the height of a tower and the length of its shadow is }$
$\displaystyle \sqrt{3}:1, \ \text{what is the angle of elevation of the Sun?}\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the angle of elevation of the Sun be }\theta.$
$\displaystyle \tan\theta=\frac{\text{Height of the tower}}{\text{Length of the shadow}}=\frac{\sqrt{3}}{1}=\sqrt{3}$
$\displaystyle \tan\theta=\tan60^\circ$
$\displaystyle \therefore \theta=60^\circ.$
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$\displaystyle \textbf{Question 3: }\text{What is the angle of elevation of the Sun when the length of the shadow of a}$
$\displaystyle \text{vertical pole is equal to its height?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the pole and the length of its shadow be }h\text{ and the angle be }\theta.$
$\displaystyle \tan\theta=\frac{\text{Height of the pole}}{\text{Length of the shadow}}=\frac{h}{h}=1$
$\displaystyle \tan\theta=\tan45^\circ$
$\displaystyle \therefore \theta=45^\circ.$
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$\displaystyle \textbf{Question 4: }\text{From a point on the ground, }20\text{ m away from the foot of a vertical tower,}$
$\displaystyle \text{the angle of elevation of the top of the tower is }60^\circ.\text{ What is the height of the tower?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower be }h\text{ m.}$
$\displaystyle \tan60^\circ=\frac{h}{20}$
$\displaystyle \sqrt{3}=\frac{h}{20}\Rightarrow h=20\sqrt{3}\text{ m}$
$\displaystyle \therefore \text{The height of the tower is }20\sqrt{3}\text{ m.}$
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$\displaystyle \textbf{Question 5: }\text{If the angles of elevation of the top of a tower from two points at a}$
$\displaystyle \text{distance of }4\text{ m and }9\text{ m from the base of the tower and in the same straight line}$
$\displaystyle \text{with it are complementary, find the height of the tower.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower be }h\text{ m and the angles be }\alpha\text{ and }\beta.$
$\displaystyle \tan\alpha=\frac{h}{4},\qquad \tan\beta=\frac{h}{9}$
$\displaystyle \text{Since }\alpha+\beta=90^\circ,\text{ we have }\tan\alpha\tan\beta=1.$
$\displaystyle \frac{h}{4}\times\frac{h}{9}=1$
$\displaystyle \frac{h^2}{36}=1\Rightarrow h^2=36$
$\displaystyle h=6\text{ m}\qquad [\text{Since height is positive}]$
$\displaystyle \therefore \text{The height of the tower is }6\text{ m.}$
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$\displaystyle \textbf{Question 6: }\text{In the adjoining figure, what are the angles of depression from the observing}$
$\displaystyle \text{positions }O_1\text{ and }O_2\text{ of the object at }A\text{?}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{For }O_1,\text{ the angle between the line of sight }AO_1\text{ and the vertical is }60^\circ.$
$\displaystyle \therefore \text{Angle of elevation of }O_1\text{ from }A=90^\circ-60^\circ=30^\circ.$
$\displaystyle \therefore \text{Angle of depression of }A\text{ from }O_1=30^\circ.$
$\displaystyle \text{The angle of elevation of }O_2\text{ from }A=45^\circ.$
$\displaystyle \therefore \text{Angle of depression of }A\text{ from }O_2=45^\circ.$
$\displaystyle \therefore \text{The required angles of depression from }O_1\text{ and }O_2\text{ are }30^\circ\text{ and }45^\circ\text{ respectively.}$
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$\displaystyle \textbf{Question 7: }\text{The tops of two towers of height }x\text{ and }y,\text{ standing on level ground,}$
$\displaystyle \text{subtend angles of }30^\circ\text{ and }60^\circ\text{ respectively at the centre of the line joining their feet.}$
$\displaystyle \text{Find }x:y.\hfill\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the equal horizontal distance from the centre to each tower be }d.$
$\displaystyle \tan30^\circ=\frac{x}{d},\qquad \tan60^\circ=\frac{y}{d}$
$\displaystyle \therefore \frac{x}{y}=\frac{\tan30^\circ}{\tan60^\circ}=\frac{1/\sqrt{3}}{\sqrt{3}}=\frac{1}{3}$
$\displaystyle \therefore x:y=1:3.$
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$\displaystyle \textbf{Question 8: }\text{The angle of elevation of the top of a tower at a point on the ground is }30^\circ.$
$\displaystyle \text{What will be the angle of elevation, if the height of the tower is tripled?}\hfill\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original height be }h\text{ and the horizontal distance be }x.$
$\displaystyle \tan30^\circ=\frac{h}{x}=\frac{1}{\sqrt{3}}$
$\displaystyle \text{If the height is tripled, let the new angle of elevation be }\theta.$
$\displaystyle \tan\theta=\frac{3h}{x}=3\tan30^\circ=\sqrt{3}=\tan60^\circ$
$\displaystyle \therefore \theta=60^\circ.$
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$\displaystyle \textbf{Question 9: }\text{AB is a pole of height }6\text{ m standing at a point }B\text{ and }CD\text{ is a ladder}$
$\displaystyle \text{inclined at an angle of }60^\circ\text{ to the horizontal and reaches up to a point }D\text{ of the pole.}$
$\displaystyle \text{If }AD=2.54\text{ m, find the length of the ladder. (Use }\sqrt{3}=1.73\text{).}\hfill\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle AB=6\text{ m and }AD=2.54\text{ m}$
$\displaystyle \therefore BD=AB-AD=6-2.54=3.46\text{ m}$
$\displaystyle \text{Let the length of the ladder }CD=l\text{ m.}$
$\displaystyle \sin60^\circ=\frac{BD}{CD}$
$\displaystyle \frac{\sqrt{3}}{2}=\frac{3.46}{l}$
$\displaystyle \frac{1.73}{2}=\frac{3.46}{l}$
$\displaystyle l=\frac{3.46\times2}{1.73}=4\text{ m}$
$\displaystyle \therefore \text{The length of the ladder is }4\text{ m.}$
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$\displaystyle \textbf{Question 10: }\text{An observer, }1.7\text{ m tall, is }20\sqrt{3}\text{ m away from a tower. The angle of}$
$\displaystyle \text{elevation from the eye of an observer to the top of tower is }30^\circ.\text{ Find the height of the}$
$\displaystyle \text{tower.}\hfill\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower above the observer's eye level be }h\text{ m.}$
$\displaystyle \tan30^\circ=\frac{h}{20\sqrt{3}}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{20\sqrt{3}}$
$\displaystyle h=20\text{ m}$
$\displaystyle \text{Height of the tower}=20+1.7=21.7\text{ m}$
$\displaystyle \therefore \text{The height of the tower is }21.7\text{ m.}$
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$\displaystyle \textbf{Question 11: }\text{An observer, }1.5\text{ m tall, is }28.5\text{ m away from a }30\text{ m high tower.}$
$\displaystyle \text{Determine the angle of elevation of the top of the tower from the eye of the observer.}$
$\displaystyle \hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Height of the tower above the observer's eye}=30-1.5=28.5\text{ m}$
$\displaystyle \text{Let the angle of elevation be }\theta.$
$\displaystyle \tan\theta=\frac{28.5}{28.5}=1$
$\displaystyle \tan\theta=\tan45^\circ$
$\displaystyle \therefore \theta=45^\circ.$
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$\displaystyle \textbf{Question 12: }\text{The ratio of the length of a vertical rod and the length of its shadow is}$
$\displaystyle 1:\sqrt{3}.\text{ What is the angle of elevation of the sun at that moment?}\hfill\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the angle of elevation of the sun be }\theta.$
$\displaystyle \tan\theta=\frac{\text{Height of the rod}}{\text{Length of the shadow}}=\frac{1}{\sqrt{3}}$
$\displaystyle \tan\theta=\tan30^\circ$
$\displaystyle \therefore \theta=30^\circ.$
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$\displaystyle \textbf{Question 13: }\text{The length of the shadow of a tower on the plane ground is }\sqrt{3}\text{ times}$
$\displaystyle \text{the height of the tower. Find the angle of elevation of the sun.}\hfill\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower be }h\text{ and the angle of elevation of the sun be }\theta.$
$\displaystyle \therefore \text{Length of the shadow}=\sqrt{3}h.$
$\displaystyle \tan\theta=\frac{\text{Height of the tower}}{\text{Length of the shadow}}=\frac{h}{\sqrt{3}h}=\frac{1}{\sqrt{3}}$
$\displaystyle \tan\theta=\tan30^\circ$
$\displaystyle \therefore \theta=30^\circ.$
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$\displaystyle \textbf{Question 14: }\text{Find the length of the shadow on the ground of a pole of height }18\text{ m when}$
$\displaystyle \text{angle of elevation }\theta\text{ of the sun is such that }\tan\theta=\frac{6}{7}.\hfill\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the length of the shadow be }x\text{ m.}$
$\displaystyle \tan\theta=\frac{\text{Height of the pole}}{\text{Length of the shadow}}$
$\displaystyle \frac{6}{7}=\frac{18}{x}$
$\displaystyle 6x=18\times7$
$\displaystyle x=21\text{ m}$
$\displaystyle \therefore \text{The length of the shadow is }21\text{ m.}$
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