Trigonometrical Ratios And Identities Ques 4
- $A B C$ is a triangular park with $A B=A C=100 \mathrm{m}$. A vertical tower is situated at the mid-point of $B C$. If the angles of elevation of the top of the tower at $A$ and $B$ are $\cot ^{-1}(3 \sqrt{2})$ and $\operatorname{cosec}^{-1}(2 \sqrt{2})$ respectively, then the height of the tower (in $\mathrm{m}$ ) is
(2019 Main, 10 April I)
(a) $25$
(b) $20$
(c) $10 \sqrt{5}$
(d) $\frac{100}{3 \sqrt{3}}$
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Answer:
Correct Answer: 4.(b)
Solution: (b) Given $A B C$ is a triangular park with $A B=A C=100 \mathrm{m}$.
A vertical tower is situated at the mid-point of $B C$.
Let the height of the tower is $h \mathrm{m}$.
Now, according to given information, we have the following figure.
From the figure and given information, we have
$ \begin{aligned} & \beta=\cot ^{-1}(3 \sqrt{2}) \\ & \alpha=\operatorname{cosec}^{-1}(2 \sqrt{2}) \end{aligned} $
Now, in $\triangle Q P A$,
$ \begin{aligned} & \cot \beta=\frac{l}{h} \\ & \Rightarrow \quad l=(3 \sqrt{2}) h \quad ……..(i) \\ \end{aligned} $
and in $\triangle B P Q, \tan \alpha=\frac{h}{B P}$
$\Rightarrow \quad \cot \alpha=\frac{B P}{h}=\frac{\sqrt{(100)^2-l^2}}{h}$
$[\because \quad p$ is mid-point of isosceles $\triangle A B C, A P \perp B C]$
$\Rightarrow \quad h^2 \cot ^2 \alpha=(100)^2-l^2$
$\Rightarrow \quad h^2\left(\operatorname{cosec}^2 \alpha-1\right)=(100)^2-(3 \sqrt{2} h)^2 \quad $ [from Eq. (i)]
$\Rightarrow \quad h^2(8-1)=(100)^2-18 h^2$
$\Rightarrow \quad 25 h^2=(100)^2$
$\Rightarrow \quad h^2=\left(\frac{100}{5}\right)^2 \Rightarrow h=20 \mathrm{m}$
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