Question

Ram and Shyam are 10 km apart. They both see a hot-air balloon making angles of elevation \(60^\circ\) and \(30^\circ\) respectively. What is the height at which the balloon could be flying?

• A. \(4\sqrt{3}\)
• B. \(5\sqrt{3}\)
• C. \(3\sqrt{3}\)
• D. \(2\sqrt{3}\)

Hint:

In height–distance problems with two angles from points on a line, the larger angle corresponds to the nearer observer}. Set up distances accordingly and use \(\tan\theta\).

Answer 1

The Correct Option is B

Step 1: Define the Problem
Two observers, \(A\) and \(B\), are 10 km apart. A balloon is directly above point \(P\). \(\angle APB=30^\circ\) for the observer farthest from the balloon, and \(\angle BPA=60^\circ\) for the nearer observer (larger angle means closer).
Let \(AP=d\), so \(BP=d-10\), and let the height of the balloon be \(h\).
Step 2: Apply the Tangent Function
From observer \(A\): \(\tan 30^\circ=\dfrac{h}{d}\), so \(h=\dfrac{d}{\sqrt{3}}\).
From observer \(B\): \(\tan 60^\circ=\dfrac{h}{d-10}\), so \(h=(d-10)\sqrt{3}\).
Step 3: Solve for Height
Equate the two expressions for \(h\): \(\dfrac{d}{\sqrt{3}}=(d-10)\sqrt{3}\). This simplifies to \(d=3(d-10)\), then \(2d=30\), and finally \(d=15\).
Therefore, \(h=\dfrac{15}{\sqrt{3}}=5\sqrt{3}\ \text{km}\).