The angle of elevation of the top of a tower from the top of a building whose height is $680 ext{ m}$ is $45^\circ$ and the angle of elevation of the top of the same tower from the foot of the same building is $60^\circ$. What is the height (in m) of the tower?

Question

The angle of elevation of the top of a tower from the top of a building whose height is $680 	ext{ m}$ is $45^\circ$ and the angle of elevation of the top of the same tower from the foot of the same building is $60^\circ$. What is the height (in m) of the tower?

Accepted Answer

**Solution:**

1. Let the height of the building be $h = 680 	ext{ m}$.
2. Let the height of the tower be $H$ and the horizontal distance between the building and the tower be $x$.
3. From the top of the building, the angle of elevation to the top of the tower is $45^\circ$.
   - Using the tangent function:
   $$	an 45^\circ = \frac{H - h}{x} = 1 \implies H - h = x$$

4. From the foot of the building, the angle of elevation to the top of the tower is $60^\circ$.
   - Using the tangent function:
   $$	an 60^\circ = \frac{H}{x} = \sqrt{3} \implies H = x \sqrt{3}$$

5. From step 3, $x = H - h$.
   Substitute into step 4:
   $$H = (H - h) \sqrt{3}$$

6. Rearranging:
   $$H = H \sqrt{3} - h \sqrt{3}$$
   $$H - H \sqrt{3} = - h \sqrt{3}$$
   $$H (1 - \sqrt{3}) = - h \sqrt{3}$$

7. Multiply both sides by $-1$:
   $$H (\sqrt{3} - 1) = h \sqrt{3}$$

8. Solve for $H$:
   $$H = \frac{h \sqrt{3}}{\sqrt{3} - 1}$$

9. Rationalize the denominator:
   $$H = \frac{h \sqrt{3}}{\sqrt{3} - 1} 	imes \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{h \sqrt{3} (\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} = \frac{h \sqrt{3} (\sqrt{3} + 1)}{3 - 1} = \frac{h \sqrt{3} (\sqrt{3} + 1)}{2}$$

10. Simplify numerator:
    $$\sqrt{3} (\sqrt{3} + 1) = 3 + \sqrt{3}$$

11. Substitute $h = 680$:
    $$H = \frac{680 (3 + \sqrt{3})}{2} = 340 (3 + \sqrt{3})$$

**Answer:** $340(3 + \sqrt{3})$ meters

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