Question

At a height of 25,000 ft. ,a pilot finds the angle of depression of the top of the tower as 25 degrees 30'. How far is he from the top of the tower? The tower is 500 ft. high.

Answer 1

At a height of 25000 ft, a pilot finds the angle of depression of the top of the tower as $25{}^{\circ}30'$. How far is he from the top of the tower? The tower is 500 ft high.

Solution.

The angle of depression of the top of the tower is $\alpha$. We must find a distance from the pilot to the top of the tower $AC$.

Use the sine function to find it:

$$\sin \alpha = \frac{AB}{AC}$$ $$AC = \frac{AB}{\sin \alpha}$$ $$AB = AD - BD = AD - CE = 25000 - 500 = 24500 \text{ (ft)}$$ $$\alpha = 25{}^{\circ}30' = 25{}^{\circ} + 30' \cdot \frac{1{}^{\circ}}{1{}^{\circ}} = |1{}^{\circ} = 60'| = 25{}^{\circ} + 30' \cdot \frac{1{}^{\circ}}{60'} = 25{}^{\circ} + \frac{30'}{60'} \cdot 1{}^{\circ} = 25.5{}^{\circ}$$

So

$$AC = \frac{24500}{\sin 25.5{}^{\circ}} \approx \frac{24500}{0.43} = 56976.7 \text{ (ft)}$$

Answer: 56976.7 ft