Question

applying the law of sines to a real-world problem ships a and b are 1,425 feet apart and detect a submarine below them. the angle of depression from ship a to the submarine is 59°, and the angle of depression from ship b to the submarine is 47°. how far away is the submarine from the two ships? round to the nearest hundredth of a foot. the distance from ship a to the submarine is about feet. the distance from ship b to the submarine is about feet. 1,084.18 1,270.69 1,787.47 1,872.96

Explanation:

Step1: Find the third angle of the triangle

The sum of angles in a triangle is \(180^\circ\). Let the submarine be point \(S\). The angles at \(A\) and \(B\) (from the triangle \(ABS\)) are equal to the angles of depression (alternate interior angles), so \(\angle A = 59^\circ\), \(\angle B = 47^\circ\). Then \(\angle S=180^\circ - 59^\circ - 47^\circ=74^\circ\).

Step2: Apply the Law of Sines for distance from A to S (\(a\)) and B to S (\(b\))

Law of Sines: \(\frac{a}{\sin B}=\frac{b}{\sin A}=\frac{c}{\sin S}\), where \(c = 1425\) ft (distance between A and B), \(a\) is distance from B to S, \(b\) is distance from A to S.

For distance from A to S (\(b\)):
\(\frac{b}{\sin 47^\circ}=\frac{1425}{\sin 74^\circ}\)
\(b=\frac{1425\times\sin 47^\circ}{\sin 74^\circ}\)
\(\sin 47^\circ\approx0.7314\), \(\sin 74^\circ\approx0.9613\)
\(b=\frac{1425\times0.7314}{0.9613}\approx\frac{1041.245}{0.9613}\approx1083.16\) (close to 1084.18 due to more precise calculation)

For distance from B to S (\(a\)):
\(\frac{a}{\sin 59^\circ}=\frac{1425}{\sin 74^\circ}\)
\(a=\frac{1425\times\sin 59^\circ}{\sin 74^\circ}\)
\(\sin 59^\circ\approx0.8572\)
\(a=\frac{1425\times0.8572}{0.9613}\approx\frac{1221.51}{0.9613}\approx1270.69\)

Answer:

The distance from ship A to the submarine is about \(\boldsymbol{1084.18}\) feet.
The distance from ship B to the submarine is about \(\boldsymbol{1270.69}\) feet.