A B C is a triangular park with A B=A C=100 m . A TV tower stands...

A B C

is a triangular park with

A B=A C=100 \mathrm{ m}

. A TV tower stands at the mid-point of

B C

. The angles of elevation of the top of the tower at

A

B, C

are

45^{{}^{\circ}}, 60^{{}^{\circ}}, 60^{{}^{\circ}}

respectively. The height of the tower is A B C is a triangular park with A B=A C=100 m . A TV tower stands at the mid-point of B C . The angles of elevation of

Options

A50 m
B $50 \sqrt{3} \mathrm{ m}$
C $50 \sqrt{2} \mathrm{ m}$
D $50(3-\sqrt{3}) \mathrm{m}$

Correct answer

B. $50 \sqrt{3} \mathrm{ m}$

Step-by-step solution

According to question, A B C is a triangular park with A B=A C=100 m . A TV tower stands at the mid-point of B C . The angles of elevation of

Let D E=h and C D=D B=x In

\triangle E B D, \quad \tan 60^{{}^{\circ}}=\frac{h}{x} \Rightarrow x=\frac{h}{\sqrt{3}}

Now, in

\triangle A D E

\begin{aligned} & \tan 45^{{}^{\circ}}=\frac{E D}{D A} \\ & \Rightarrow \quad D A=h \end{aligned}

In \triangle A B D

, by applying Pythagoras theorem

\begin{aligned} & \Rightarrow \quad\left(\frac{h}{\sqrt{3}}\right)^2+h^2=100^2 \\ & \Rightarrow \quad \frac{4 h^2}{3}=10000 \quad \Rightarrow h=50 \sqrt{3} \end{aligned}

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