Question

A tower $PQ$ stands on a horizontal ground with base $Q$ on the ground The point $R$ divides the tower in two parts such that $QR =15 \,m$ If from a point $A$ on the ground the angle of elevation of $R$ is $60^{\circ}$ and the part $PR$ of the tower subtends an angle of $15^{\circ}$ at $A$, then the height of the tower is :

• A. $5(2 \sqrt{3}+3) m$
• B. $5(\sqrt{3}+3) m$
• C. $10(\sqrt{3}+1) m$
• D. $10(2 \sqrt{3}+1) m$

Answer 1

The Correct Option is A

To determine the height of the tower \(PQ\), we need to analyze the given information and use trigonometric principles.

1. Let the entire height of the tower be \(PQ = h\), and the height of part \(QR = 15 \, m\) as given.
2. So, the height of part \(PR\) would be \(h - 15 \, m\).
3. The angle of elevation from point \(A\) to point \(R\) is \(60^\circ\), and the part \(PR\) subtends an angle of \(15^\circ\) at point \(A\).
4. Let's denote the distance \(AQ = x\).
5. Using the tangent formula for the angle of elevation to \(R\):
   • \(\tan 60^\circ = \frac{QR}{AQ} = \frac{15}{x}\)
   • Given \(\tan 60^\circ = \sqrt{3}\), so \(\sqrt{3} = \frac{15}{x}\)
   • Therefore, \(x = \frac{15}{\sqrt{3}} = 5\sqrt{3}\)
6. For the part \(PR\) which subtends an angle of \(15^\circ\) at \(A\):
   • \(\tan 15^\circ = \frac{PR}{AQ} = \frac{h - 15}{x}\)
   • Using \(\tan 15^\circ = 2 - \sqrt{3}\) (standard trigonometric value):
   • \(2 - \sqrt{3} = \frac{h - 15}{5\sqrt{3}}\)
   • So, \(h - 15 = (2 - \sqrt{3}) \times 5\sqrt{3}\)
   • \(h - 15 = 10\sqrt{3} - 15\)
   • Therefore, \(h = 10\sqrt{3} - 15 + 15\)
   • Finally, \(h = 10\sqrt{3} + 15\)
7. Now, simplifying \(h\):
   • As \(h = 5(2\sqrt{3} + 3)\), this matches the correct answer from the options.

Thus, the height of the tower \(PQ\) is \(5(2\sqrt{3} + 3) \, m\).