(B) Let $PQ$ represent the pillar with height $h$.In $\Delta PAQ$,$\tan(30^o) = \frac{PQ}{AQ}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AQ}$ $\Right

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(B) Let $PQ$ represent the pillar with height $h$.In $\Delta PAQ$,$\tan(30^o) = \frac{PQ}{AQ}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AQ}$ $\Rightarrow AQ = h\sqrt{3}$.In $\Delta PQB$,$\tan(60^o) = \frac{PQ}{BQ}$ $\Rightarrow \sqrt{3} = \frac{h}{BQ}$ $\Rightarrow BQ = \frac{h}{\sqrt{3}}$.The distance covered in $10 \text{ minutes}$ is $AB = AQ - BQ = h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$.Since the speed is uniform,the time taken is proportional to the distance.Time taken to cover $AB = 10 \text{ minutes}$.Time taken to cover $BQ = \frac{BQ}{AB} \times 10 = \frac{h/\sqrt{3}}{2h/\sqrt{3}} \times 10 = \frac{1}{2} \times 10 = 5 \text{ minutes}$.

Class 11 Mathematics Trigonometrical Equations Height and Distance

Difficult

$A$ man is walking towards a vertical pillar in a straight path,at a uniform speed. At a certain point $A$ on the path,he observes that the angle of elevation of the top of the pillar is $30^o$. After walking for $10 \text{ minutes}$ from $A$ in the same direction,at a point $B$,he observes that the angle of elevation of the top of the pillar is $60^o$. Then the time taken (in minutes) by him,from $B$ to reach the pillar,is:

JEE MAIN 2016 All JEE MAIN

A
$20$
B
$5$
C
$6$
D
$10$

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Trigonometrical Equations

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$A$ balloon is descending at a rate of $4 \, m/min$. Its angle of elevation from a point on the ground is $45^\circ$,which reduces to $30^\circ$ after $10 \, minutes$. At what distance from the observer will the balloon touch the ground?

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