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 $59^\circ$, and the angle of depression from ship b to the submarine is $47^\circ$.
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 triangle's top angle

Let the submarine be point C. The angle at ship A (∠CAB) is equal to the angle of depression, so $\angle CAB = 59^\circ$. The angle at ship B (∠CBA) is equal to its angle of depression, so $\angle CBA = 47^\circ$. The sum of angles in a triangle is $180^\circ$, so:
$\angle ACB = 180^\circ - 59^\circ - 47^\circ = 74^\circ$

Step2: Apply Law of Sines for ship A to sub

Law of Sines: $\frac{BC}{\sin(\angle CAB)} = \frac{AB}{\sin(\angle ACB)}$. Solve for $BC$ (distance from ship A to submarine):
$BC = \frac{AB \cdot \sin(59^\circ)}{\sin(74^\circ)}$
Substitute $AB=1425$, $\sin(59^\circ)\approx0.8572$, $\sin(74^\circ)\approx0.9613$:
$BC = \frac{1425 \cdot 0.8572}{0.9613} \approx \frac{1221.51}{0.9613} \approx 1270.69$

Step3: Apply Law of Sines for ship B to sub

Law of Sines: $\frac{AC}{\sin(\angle CBA)} = \frac{AB}{\sin(\angle ACB)}$. Solve for $AC$ (distance from ship B to submarine):
$AC = \frac{AB \cdot \sin(47^\circ)}{\sin(74^\circ)}$
Substitute $AB=1425$, $\sin(47^\circ)\approx0.7314$, $\sin(74^\circ)\approx0.9613$:
$AC = \frac{1425 \cdot 0.7314}{0.9613} \approx \frac{1042.245}{0.9613} \approx 1084.20$

Answer:

The distance from ship A to the submarine is about 1270.69 feet.
The distance from ship B to the submarine is about 1084.20 feet.