Class 10: Heights and Distances – Fill in the Blanks

$\displaystyle \textbf{Question 1: }\text{If a pole }6\text{ m high casts a shadow }2\sqrt{3}\text{ m long on the ground, then the Sun's}$
$\displaystyle \text{elevation is }\underline{\hspace{1.5cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the Sun's elevation be }\theta.$
$\displaystyle \tan\theta=\frac{\text{Height of the pole}}{\text{Length of the shadow}}$
$\displaystyle =\frac{6}{2\sqrt{3}}=\sqrt{3}=\tan60^\circ$
$\displaystyle \therefore \theta=60^\circ.$
$\displaystyle \therefore \text{The required answer is }60^\circ.$
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$\displaystyle \textbf{Question 2: }\text{The angle of elevation of the sun when the shadow of a pole }h\text{ metre high is}$
$\displaystyle \sqrt{3}h\text{ metres long is }\underline{\hspace{1.5cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the angle of elevation of the sun be }\theta.$
$\displaystyle \tan\theta=\frac{\text{Height of the pole}}{\text{Length of the shadow}}$
$\displaystyle =\frac{h}{\sqrt{3}h}=\frac{1}{\sqrt{3}}=\tan30^\circ$
$\displaystyle \therefore \theta=30^\circ.$
$\displaystyle \therefore \text{The required answer is }30^\circ.$
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$\displaystyle \textbf{Question 3: }\text{If the height of a tower and the distance of the point of observation from its}$
$\displaystyle \text{foot, both, are increased by }10\%,\text{ then the angle of elevation of its top remains}$
$\displaystyle \underline{\hspace{1.5cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower be }h\text{ and the distance from its foot be }x.$
$\displaystyle \tan\theta=\frac{h}{x}$
$\displaystyle \text{After increasing both by }10\%,\quad \tan\theta'=\frac{1.1h}{1.1x}=\frac{h}{x}$
$\displaystyle \therefore \tan\theta'=\tan\theta\Rightarrow\theta'=\theta.$
$\displaystyle \therefore \text{The angle of elevation remains unchanged.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If the elevation of the sun changes from }30^\circ\text{ to }60^\circ,\text{ then the difference}$
$\displaystyle \text{between the lengths of shadows of a pole }15\text{ m high is }\underline{\hspace{1.5cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the lengths of the shadows at }30^\circ\text{ and }60^\circ\text{ be }x\text{ m and }y\text{ m respectively.}$
$\displaystyle \tan30^\circ=\frac{15}{x}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{15}{x}\Rightarrow x=15\sqrt{3}\text{ m}$
$\displaystyle \tan60^\circ=\frac{15}{y}$
$\displaystyle \sqrt{3}=\frac{15}{y}\Rightarrow y=\frac{15}{\sqrt{3}}=5\sqrt{3}\text{ m}$
$\displaystyle \therefore \text{Difference}=x-y=15\sqrt{3}-5\sqrt{3}=10\sqrt{3}\text{ m.}$
$\displaystyle \therefore \text{The required answer is }10\sqrt{3}\text{ m.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{On the level ground, the angle of elevation of a tower is }30^\circ.\text{ On moving}$
$\displaystyle 20\text{ metres nearer, the angle of elevation is }60^\circ.\text{ The height of the tower is }\underline{\hspace{1.5cm}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height of the tower be }h\text{ m and the nearer distance be }x\text{ m.}$
$\displaystyle \therefore \text{The initial distance from the tower}=(x+20)\text{ m.}$
$\displaystyle \tan60^\circ=\frac{h}{x}$
$\displaystyle \sqrt{3}=\frac{h}{x}\Rightarrow h=\sqrt{3}x \qquad ...(i)$
$\displaystyle \tan30^\circ=\frac{h}{x+20}$
$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{x+20}\Rightarrow x+20=\sqrt{3}h \qquad ...(ii)$
$\displaystyle \text{Substituting }h=\sqrt{3}x\text{ in (ii),}$
$\displaystyle x+20=3x$
$\displaystyle 2x=20\Rightarrow x=10\text{ m}$
$\displaystyle \therefore h=\sqrt{3}\times10=10\sqrt{3}\text{ m.}$
$\displaystyle \therefore \text{The required answer is }10\sqrt{3}\text{ m.}$
$\displaystyle \\$