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$\mathrm{ABC}$ is a triangular park with $\mathrm{AB}=\mathrm{AC}=100 \mathrm{~m} . \mathrm{A}$ TV tower stands at the mid-point of $\mathrm{BC}$. 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
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The correct answer is:
$50 \sqrt{3} \mathrm{~m}$
$\tan 45^{\circ}=\frac{\mathrm{PQ}}{\mathrm{AQ}}=\frac{\mathrm{h}}{\mathrm{AQ}} \Rightarrow \mathrm{h}=\mathrm{AQ}$
Where $\mathrm{PQ}$ is tower and $\mathrm{ABC}$ is the park, with $\mathrm{Q}$ being mid point of the side $\mathrm{BC}$ and $\mathrm{PQ}=\mathrm{h}$ Also, $A Q^{2}+B Q^{2}=100^{2}$
$\Rightarrow \mathrm{h}^{2}+\mathrm{h}^{2} \cot ^{2} 60^{\circ}=100^{2}$
$\Rightarrow \mathrm{h}^{2}\left[1+\frac{1}{3}\right]=100^{2}$
$\Rightarrow \mathrm{h}^{2}=\frac{3 \times 100^{2}}{4} \Rightarrow \mathrm{h}=50 \sqrt{3}$
Where $\mathrm{PQ}$ is tower and $\mathrm{ABC}$ is the park, with $\mathrm{Q}$ being mid point of the side $\mathrm{BC}$ and $\mathrm{PQ}=\mathrm{h}$ Also, $A Q^{2}+B Q^{2}=100^{2}$
$\Rightarrow \mathrm{h}^{2}+\mathrm{h}^{2} \cot ^{2} 60^{\circ}=100^{2}$
$\Rightarrow \mathrm{h}^{2}\left[1+\frac{1}{3}\right]=100^{2}$
$\Rightarrow \mathrm{h}^{2}=\frac{3 \times 100^{2}}{4} \Rightarrow \mathrm{h}=50 \sqrt{3}$
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