Question

consult the figure. to find the length of the span of a proposed ski lift from a to b, a surveyor measures the angle dab to be 25° and then walks off a distance of l = 1000 feet to c and measures the angle acb to be 15°. what is the distance from a to b? the distance from a to b is approximately □ feet. (do not round until the final answer. then round to two decimal places as needed.)

Explanation:

Step1: Find angle \( \angle BAC \)

Since \( \angle DAB = 25^{\circ} \), then \( \angle BAC=180^{\circ}- 25^{\circ}=155^{\circ} \)

Step2: Find angle \( \angle ABC \)

Using the angle - sum property of a triangle (\( \angle BAC+\angle ABC+\angle ACB = 180^{\circ} \)). Given \( \angle BAC = 155^{\circ} \) and \( \angle ACB=15^{\circ} \), then \( \angle ABC=180^{\circ}-(155^{\circ}+15^{\circ}) = 10^{\circ} \)

Step3: Apply the Law of Sines

The Law of Sines states that \( \frac{AB}{\sin\angle ACB}=\frac{AC}{\sin\angle ABC} \). We know \( AC = L=1000 \) feet, \( \angle ACB = 15^{\circ} \), and \( \angle ABC=10^{\circ} \)
So \( AB=\frac{AC\times\sin\angle ACB}{\sin\angle ABC} \)
Substitute the values: \( AB=\frac{1000\times\sin15^{\circ}}{\sin10^{\circ}} \)
We know that \( \sin15^{\circ}=\sin(45^{\circ}-30^{\circ})=\sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.2588 \) and \( \sin10^{\circ}\approx0.1736 \)
\( AB=\frac{1000\times0.2588}{0.1736}\approx1490.79 \)

Answer:

\( 1490.79 \)