Question

X and Y are the two points that are 135 m apart on the ground on either side of a pole and in the same line. The angles of elevation of a bird sitting on the top of the pole from X and Y are 30° and 60° respectively. The distance of Y from the foot of the pole (in m) is:

• A. 50.63
• B. 33.75
• C. 67.5
• D. 101.25

Hint:

When solving elevation and distance problems, use the tangent function to relate the height of the object and the distance from the point of observation.

Answer 1

The Correct Option is A

We are given that the distance between points X and Y is 135 m, and the angles of elevation from X and Y are 30° and 60°, respectively. Let the height of the pole be \( h \). Using the tangent formula for both points: \[ \tan 30^\circ = \frac{h}{d_X} \quad \text{and} \quad \tan 60^\circ = \frac{h}{d_Y} \] where \( d_X \) and \( d_Y \) represent the distances from the foot of the pole to points X and Y, respectively. We can solve the above equations for \( h \) and \( d_Y \), where \( d_X = d_Y + 135 \). Substituting these into the formulas: \[ h = d_X \cdot \tan 30^\circ \quad \text{and} \quad h = d_Y \cdot \tan 60^\circ \] Thus, we can equate the two expressions for \( h \): \[ d_X \cdot \tan 30^\circ = d_Y \cdot \tan 60^\circ \] Since \( d_X = d_Y + 135 \), substitute this into the equation: \[ (d_Y + 135) \cdot \tan 30^\circ = d_Y \cdot \tan 60^\circ \] Substitute the values of \( \tan 30^\circ = \frac{1}{\sqrt{3}} \) and \( \tan 60^\circ = \sqrt{3} \): \[ (d_Y + 135) \cdot \frac{1}{\sqrt{3}} = d_Y \cdot \sqrt{3} \] Now, solve for \( d_Y \): \[ \frac{d_Y + 135}{\sqrt{3}} = d_Y \cdot \sqrt{3} \] \[ d_Y + 135 = 3d_Y \] \[ 135 = 2d_Y \] \[ d_Y = \frac{135}{2} = 50.63 \text{ m} \] Thus, the distance of Y from the foot of the pole is 50.63 m.