Trigonometrical Ratios And Identities Ques 9
- $P Q R$ is a triangular park with $P Q=P R=200 \mathrm{m}$. A TV tower stands at the mid-point of $Q R$. If the angles of elevation of the top of the tower at $P, Q$ and $R$ are respectively $45^{\circ}, 30^{\circ}$ and $30^{\circ}$, then the height of the tower (in $\mathrm{m}$ ) is
(2018 Main)
(a) $100$
(b) $50$
(c) $100 \sqrt{3}$
(d) $50 \sqrt{2}$
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Answer:
Correct Answer: 9.(a)
Solution: (a) Let height of tower $T M$ be $h$.
In $\triangle P M T, $
$\tan 45^{\circ}=\frac{T M}{P M}$
$\Rightarrow \quad 1=\frac{h}{P M}$
$\Rightarrow \quad P M=h$
In $\triangle T Q M, \tan 30^{\circ}=\frac{h}{Q M} ; Q M=\sqrt{3} h$
$ \begin{aligned} & \text { In } \triangle P M Q, P M^2+Q M^2=P Q^2 \\ & h^2+(\sqrt{3} h)^2=(200)^2 \\ & \Rightarrow \quad 4 h^2=(200)^2 \\ & \Rightarrow \quad h=100 \mathrm{m} \end{aligned} $
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