Question

The angles of elevation of the top of a tower from two points A and B lying on the horizontal plane through the foot of the tower are respectively ${45}^0$ and ${30}^0$. If A and B are on the same side of the tower and $AB=48metre$, then the height of the tower is:

• A. $24(\sqrt{3}+1)$ metre
• B. 24 metre
• C. $14\sqrt{2}$ metre
• D. 96 metre

Answer 1

The correct option is A $24(\sqrt{3}+1)$ metre

Let $QA=xmQB=AB+QA=(48+x)m$

In $\Delta PQA,\angle PQA=90°,\angle PAQ=45°\tan 45°=\frac{PQ}{QA}=\frac{PQ}{x}\therefore PQ=xm\to \left(1\right)$

In$\Delta PQB,\angle PQB=90°,\angle PBQ=30°\tan 30°=\frac{PQ}{QB}=\frac{PQ}{(48+x)}\Rightarrow PQ=\frac{(48+x)}{\sqrt{3}}m\to \left(2\right)$

From $\left(1\right)\&\left(2\right)$

$x=\frac{(48+x)}{\sqrt{3}}\Rightarrow x=\frac{48}{\sqrt{3}-1}m\therefore heightoftower\left(PQ\right)=x=\frac{48}{\sqrt{3}-1}=24(\sqrt{3}+1)m$

Option A is correct