Question

a pilot sees two birds. the angle between them from the pilots location is 58°. the distance to bird b is 6.2 mi, and the angle at bird b between the pilot and bird a is 81°. how far is the pilot from bird a? distance from bird a = ? mi round your answer to the nearest hundredth.

Explanation:

Step1: Find the third angle of the triangle

In a triangle, the sum of interior angles is \(180^\circ\). Let the pilot's location be \(P\), Bird \(A\) be \(A\), and Bird \(B\) be \(B\). We know \(\angle P = 58^\circ\) and \(\angle B = 81^\circ\). So \(\angle A=180^\circ - 58^\circ - 81^\circ=41^\circ\).

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\), where \(a, b, c\) are the lengths of the sides opposite angles \(A, B, C\) respectively. Let the distance from the pilot to Bird \(A\) be \(a\) (opposite \(\angle B\)), the distance from the pilot to Bird \(B\) be \(b = 6.2\) mi (opposite \(\angle A\)), and the distance between Bird \(A\) and Bird \(B\) be \(c\) (opposite \(\angle P\)). Wait, actually, let's correct: Let \(PB = 6.2\) mi (side opposite \(\angle A\)), \(PA=x\) (side opposite \(\angle B\)). So by Law of Sines, \(\frac{PA}{\sin B}=\frac{PB}{\sin A}\). Substituting values: \(\frac{x}{\sin 81^\circ}=\frac{6.2}{\sin 41^\circ}\).

Step3: Solve for \(x\)

\(x=\frac{6.2\times\sin 81^\circ}{\sin 41^\circ}\). Calculate \(\sin 81^\circ\approx0.9877\), \(\sin 41^\circ\approx0.6561\). Then \(x=\frac{6.2\times0.9877}{0.6561}\approx\frac{6.12374}{0.6561}\approx9.33\).

Answer:

\(9.33\)