A man on the top of a vertical tower observes a car moving at uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how long will the car take to reach the observation tower from this point?                                                   (OR)Prove that tanΘ+sinΘtanΘ−sinθ=secΘ+1secΘ−1

# A man on the top of a vertical tower observes a car moving at uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how long will the car take to reach the observation tower from this point?                                                   (OR)Prove that $\frac{\mathrm{tan\Theta }+\mathrm{sin\Theta }}{\mathrm{tan\Theta }-\mathrm{sin}\theta }=\frac{\mathrm{sec\Theta }+1}{\mathrm{sec\Theta }-1}$

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### Solution:

let the speed of the car be ‘𝑥’ 𝑚/𝑚𝑖𝑛

⇒ 𝐵𝐶 = 𝑥 ✕12

In △ AOC

$\begin{array}{r}\mathrm{tan}{30}^{\circ }=\frac{AO}{OC}\\ ⇒\frac{AO}{OC}=\frac{1}{\sqrt{3}}\\ ⇒OC=\sqrt{3}AO\\ ⇒OC=\sqrt{3}OB\\ ⇒\frac{AO}{OB}=1\\ ⇒\sqrt{3}AO=\sqrt{3}OB\\ ⇒OB+BC=\sqrt{3}OB\\ ⇒OB\left(\sqrt{3}-1\right)=BC\\ ⇒\frac{OB}{BC}=\frac{1}{\sqrt{3}-1}\end{array}$

Let 𝑡 be the time taken for a car to reach the observation tower from the point Now, $t=\frac{OB}{x}$

$\begin{array}{r}t=\frac{BC}{x\left(\sqrt{3-1\right)}}\\ t=\frac{12}{\sqrt{3}-1}×\frac{\sqrt{3}+1}{\sqrt{3}+1}\\ t=\frac{12\left(\sqrt{3}+1\right)}{3-1}\\ t=6\left(\sqrt{3+1}\right)\mathrm{min}\end{array}$

Hence, the car takes 6 𝑚𝑖𝑛 to reach the observation tower from this point.

(OR)

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