If the angle of elevation of a cloud from a point $P$ which is $25\, m$ above a lake be $30^\circ$ and the angle of depression of reflection of the cloud in the lake from $P$ be $60^\circ$, then the height of the cloud (in meters) from the surface of the lake is :

Question

If \[ \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}, \] where $ \alpha, \beta \in \mathbb{N} $, then the value of $ \alpha + \beta $ is ___________.

Answer 1

To solve this problem, we need to understand the scenario and apply trigonometric concepts. The point $ P $ is 25 meters above the lake. The angle of elevation to the cloud is $ 30^\circ $, and the angle of depression to the reflection of the cloud in the lake is $ 60^\circ $.

Let's denote:

$ AB $ as the distance from the lake surface to the cloud, which is what we want to find.
$ CD $ as the distance from the point $ P $ to the cloud.
$ BD $ as the horizontal distance from the point directly below $ P $ to the base of the cloud.
$ C $ as the reflection of the cloud in the lake, making $ CD = CB $.

Using trigonometry in $\triangle PBD$:

The angle of elevation of the cloud is $ 30^\circ $.

Using the tangent of $ 30^\circ $:

\[ \tan 30^\circ = \frac{AB}{BD} \]

\[ \frac{1}{\sqrt{3}} = \frac{AB}{BD} \]

\[ BD = AB \cdot \sqrt{3} \]

For the reflection, the angle of depression is $ 60^\circ $.

Using the tangent of $ 60^\circ $:

\[ \tan 60^\circ = \frac{2 \times AB + 50}{BD} \]

\[ \sqrt{3} = \frac{2 \times AB + 50}{AB \cdot \sqrt{3}} \]

Cross multiplying, we get:

\[ 3AB = 2 \times AB + 50 \]

\[ AB = 50 \]

Therefore, the height of the cloud from the surface of the lake is 50 meters.

Answer 2