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 \textbf{larger angle corresponds to the nearer observer}. Set up distances accordingly and use \(\tan\theta\).

Answer 1

The Correct Option is B

Step 1: Model the situation
Let the observers be \(A\) and \(B\) with \(AB=10\) km, and let the balloon be vertically above point \(P\) on the same line. Suppose \(\angle APB=30^\circ\) at the farther observer and \(\angle BP A=60^\circ\) at the nearer observer (larger angle \(\Rightarrow\) nearer).
Let \(AP=d\) so \(BP=d-10\), and let the height be \(h\).
Step 2: Use \(\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}\)
From \(A\): \(\tan 30^\circ=\dfrac{h}{d}\Rightarrow h=\dfrac{d}{\sqrt{3}}\).
From \(B\): \(\tan 60^\circ=\dfrac{h}{d-10}\Rightarrow h=(d-10)\sqrt{3}\).
Step 3: Equate the two expressions for \(h\)
\(\dfrac{d}{\sqrt{3}}=(d-10)\sqrt{3}\ \Rightarrow\ d=3(d-10)\ \Rightarrow\ 2d=30\ \Rightarrow\ d=15\).
Hence \(h=\dfrac{15}{\sqrt{3}}=5\sqrt{3}\ \text{km}\).
\[ \boxed{h=5\sqrt{3}\ \text{km}} \]