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.

Explanation:

Step1: Find the third - angle of the triangle

The sum of angles in a triangle is $180^{\circ}$. The two angles of depression give us two angles of the triangle formed by the two ships and the submarine. Let the third angle $\theta=180^{\circ}-(59^{\circ}+47^{\circ}) = 74^{\circ}$.

Step2: Apply the Law of Sines to find the distance from ship A to the submarine

Let the distance from ship A to the submarine be $a$ and from ship B to the submarine be $b$. The Law of Sines states that $\frac{a}{\sin47^{\circ}}=\frac{b}{\sin59^{\circ}}=\frac{1425}{\sin74^{\circ}}$.
To find $a$, we have $a=\frac{1425\times\sin47^{\circ}}{\sin74^{\circ}}$.
$a=\frac{1425\times0.7314}{0.9613}\approx1084.92$ feet.

Step3: Apply the Law of Sines to find the distance from ship B to the submarine

To find $b$, we have $b = \frac{1425\times\sin59^{\circ}}{\sin74^{\circ}}$.
$b=\frac{1425\times0.8572}{0.9613}\approx1261.57$ feet.

Answer:

The distance from ship A to the submarine is about $1084.92$ feet.
The distance from ship B to the submarine is about $1261.57$ feet.