Question

When viewed from a point P which is 56 metres above a lake, the angle of elevation is 30 and from the same point, the angle of depression of its reflection in the lake 60. What is the height of the cloud?

• A. 112m
• B. 110m
• C. 108m
• D. 106m

Hint:

For problems involving angles of elevation and depression with reflections in water, remember that the object and its reflection are equidistant from the surface of the water. Draw a horizontal line from the observation point to create right-angled triangles and use tangent ratios.

Answer 1

The Correct Option is A

Step 1: Set up the problem with a diagram and variables.
Let P be the point of observation, and its height above the lake surface is \( PE = 56 \) m.
Let the cloud be at point C, and its height above the lake surface be \( CE' = H \) m.
Let C' be the reflection of the cloud in the lake. The depth of the reflection below the lake surface will also be \( CE' = H \) m.
Draw a horizontal line from P parallel to the lake surface, intersecting the vertical line from C (and C') at point D.
So, \( PD = EE' \). Also, \( DE' = PE = 56 \) m.
Step 2: Use the angle of elevation to form an equation.
The angle of elevation from P to the cloud C is \( \angle CPD = 30^\circ \). In the right-angled triangle \( \triangle PDC \): \[ \tan(30^\circ) = \frac{CD}{PD} \] We know that \( CD = CE' - DE' = H - 56 \). So, \[ \tan(30^\circ) = \frac{H - 56}{PD} \] \[ \frac{1}{\sqrt{3}} = \frac{H - 56}{PD} \quad \Rightarrow \quad PD = \sqrt{3}(H - 56) \quad \cdots (1) \] Step 3: Use the angle of depression to form an equation.
The angle of depression from P to the reflection C' is \( \angle DPC' = 60^\circ \).
In the right-angled triangle \( \triangle PDC' \): \[ \tan(60^\circ) = \frac{DC'}{PD} \] We know that \( DC' = DE' + E'C' = 56 + H \). So, \[ \tan(60^\circ) = \frac{56 + H}{PD} \] \[ \sqrt{3} = \frac{56 + H}{PD} \quad \Rightarrow \quad PD = \frac{56 + H}{\sqrt{3}} \quad \cdots (2) \] Step 4: Solve the system of equations to find the height H.
Equate the expressions for PD from (1) and (2):
\[ \sqrt{3}(H - 56) = \frac{56 + H}{\sqrt{3}} \] Multiply both sides by \( \sqrt{3} \): \[ 3(H - 56) = 56 + H \] Distribute 3 on the left side: \[ 3H - 168 = 56 + H \] Bring all terms with H to one side and constants to the other: \[ 3H - H = 56 + 168 \] \[ 2H = 224 \] Divide by 2: \[ H = \frac{224}{2} \] \[ H = 112 \text{ m} \] Thus, the height of the cloud from the surface of the lake is 112 meters.