Question

The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. From point Y, 40 m vertically above X, the angle of elevation of the top Q of the tower is 45°. Find the height of the tower PQ and the distance PX.

Hint:

To solve problems involving angles of elevation, use the tangent function, and solve the system of equations using the relationships between the height of the object and the horizontal distance from the point of observation.

Answer 1

Let the height of the tower be \( h \) and the distance from the point X to the base of the tower be \( x \).

Step 1: Using the information that the angle of elevation from point X is 60°, we can use the tangent of the angle to write the equation: \[ \tan 60^\circ = \frac{h}{x}. \] Since \( \tan 60^\circ = \sqrt{3} \), the equation becomes: \[ \sqrt{3} = \frac{h}{x} \quad \Rightarrow \quad h = \sqrt{3}x \quad \text{(Equation 1)}.
\]
Step 2: Next, from point Y, which is 40 m vertically above X, the angle of elevation of Q is 45°. Using the tangent of 45°, we write: \[ \tan 45^\circ = \frac{h - 40}{x}. \] Since \( \tan 45^\circ = 1 \), the equation becomes: \[ 1 = \frac{h - 40}{x} \quad \Rightarrow \quad h - 40 = x \quad \Rightarrow \quad h = x + 40 \quad \text{(Equation 2)}.
\]
Step 3: Now, solve the system of equations (Equation 1 and Equation 2). From Equation 1, we have: \[ h = \sqrt{3}x. \] Substitute this into Equation 2: \[ \sqrt{3}x = x + 40. \] Solve for \( x \): \[ \sqrt{3}x - x = 40 \quad \Rightarrow \quad x(\sqrt{3} - 1) = 40 \quad \Rightarrow \quad x = \frac{40}{\sqrt{3} - 1}. \] To simplify, multiply the numerator and denominator by \( \sqrt{3} + 1 \): \[ x = \frac{40(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{40(\sqrt{3} + 1)}{3 - 1} = 20(\sqrt{3} + 1).
\]
Step 4: Now substitute this value of \( x \) back into Equation 1 to find \( h \): \[ h = \sqrt{3} \times 20(\sqrt{3} + 1) = 20\sqrt{3}(\sqrt{3} + 1). \] Simplify: \[ h = 20 \times (3 + \sqrt{3}) = 60 + 20\sqrt{3}.
\] Thus, the height of the tower \( h \) is \( 60 + 20\sqrt{3} \) meters, and the distance \( PX = 20(\sqrt{3} + 1) \) meters.

Conclusion: The height of the tower PQ is \( 60 + 20\sqrt{3} \) meters, and the distance \( PX \) is \( 20(\sqrt{3} + 1) \) meters.