Trigonometrical Ratios And Identities Ques 2
- 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 $30^{\circ}$. After walking for $10 \mathrm{min}$ from $A$ in the same direction, at a point $B$, he observes that the angle of elevation of the top of the pillar is $60^{\circ}$. Then, the time taken (in minutes) by him, from $B$ to reach the pillar, is
(2016 Main)
(a) $6$
(b) $10$
(c) $20$
(d) $5$
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Answer:
Correct Answer: 2.(d)
Solution: (d) According to given information, we have the following figure
Now, from $\triangle A C D$ and $\triangle B C D$, we have
$ \tan 30^{\circ}=\frac{h}{x+y} $
and
$ \tan 60^{\circ}=\frac{h}{y} $
$ \begin{aligned} & h=\frac{x+y}{\sqrt{3}} \quad ……..(i) \\ & h=\sqrt{3} y \quad ……..(ii) \end{aligned} $
From Eqs. (i) and (ii), $\frac{x+y}{\sqrt{3}}=\sqrt{3} y$
$\Rightarrow \quad x+y=3 y$
$\Rightarrow \quad x-2 y=0$
$\Rightarrow \quad y=\frac{x}{2}$
$\because \quad $ Speed is uniform and distance $x$ covered in $10 $ $\mathrm{min}$.
$\therefore \quad$ Distance $\frac{x}{2}$ will be cover in $5 $ $\mathrm{min}$.
$\therefore \quad$ Distance $y$ will be cover in $5 $ $\mathrm{min}$.
BETA
