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 = 1300 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 ABC

In triangle ABC, we know that the sum of interior - angles of a triangle is 180°. Angle BAC = 25° and angle ACB = 15°. So, angle ABC=180°-(25° + 15°)=140°.

Step2: Apply the Law of Sines

The Law of Sines states that in a triangle with sides \(a\), \(b\), \(c\) and opposite angles \(A\), \(B\), \(C\) respectively, \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Let \(AC = L = 1300\) feet, we want to find \(AB\). According to the Law of Sines, \(\frac{AB}{\sin\angle ACB}=\frac{AC}{\sin\angle ABC}\).

Step3: Substitute the values

We know that \(\angle ACB = 15^{\circ}\), \(\angle ABC = 140^{\circ}\), and \(AC = 1300\) feet. So \(AB=\frac{AC\times\sin\angle ACB}{\sin\angle ABC}\). Substituting the values, we have \(\sin15^{\circ}=\sin(45^{\circ}- 30^{\circ})=\sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.2588\), \(\sin140^{\circ}=\sin(180^{\circ}-40^{\circ})=\sin40^{\circ}\approx0.6428\), and \(AC = 1300\) feet. Then \(AB=\frac{1300\times\sin15^{\circ}}{\sin140^{\circ}}\).

Step4: Calculate the value of AB

\(AB=\frac{1300\times0.2588}{0.6428}=\frac{336.44}{0.6428}\approx523.39\) feet.

Answer:

523.39