Exam

BITSAT

Year

2020

Subject

Mathematics

Chapter

Heights and Distances

BITSAT 2020 · Mathematics · Heights and Distances

ABC is a triangular park with AB=AC=100 m. A TV tower stands at the...

$\mathrm{ABC}$ 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

Step-by-step solution

$\tan 45^{{}^{\circ}}=\frac{\mathrm{PQ}}{\mathrm{AQ}}=\frac{\mathrm{h}}{\mathrm{AQ}} \Rightarrow \mathrm{h}=\mathrm{AQ}$ 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, $A Q^{2}+B Q^{2}=100^{2}\Rightarrow \mathrmh^{2}+\mathrmh^{2} \cot ^{2} 60^{{}^{\circ}}=100^{2}$
$\Rightarrow \mathrmh^{2}\left[1+\frac{1}{3}\right]=100^{2}\Rightarrow \mathrmh^{2}=\frac{3 \times 100^{2}}{4} \Rightarrow \mathrm{h}=50 \sqrt{3}$