Two Coast Stations

Part of the Triangles Without Right Angles section of Coddy's Trigonometry journey. Lesson 3 of 37.

Two coastguard stations, S and T, are 15 km apart. They see a ship X at 62° from S and 50° from T. The 15 km side is across from the angle at the ship. Nobody measured that angle, so the law cannot use the 15 km yet.

$Stations S and T 15 km apart, the ship X, angles 62° at S and 50° at T$

Find the third angle first. The angles of a triangle add up to 180°: $180^{\circ}-62^{\circ}-50^{\circ}$ is 68°. Now the 15 km side goes with 68°. The distance d from S is across from the 50° at T. So $\frac{d}{\sin\left(50^{\circ}\right)} = \frac{15}{\sin\left(68^{\circ}\right)}$ , and d is about 12.39 km.

Challenge

Coastguard stations S and T are 12 km apart along a straight coast. A ship X is seen at 47° from S and at 76° from T. Find the distance x from T to the ship. Done means x is in kilometres with 2 decimals.

Then press Check.

Try it yourself

$Use the figure to find x$