Two Ships From Port

Part of the Triangles Without Right Angles section of Coddy's Trigonometry journey. Lesson 29 of 37.

Two ships leave a port at the same time on different bearings. The angle between their courses is the difference of the bearings (or 360° minus it, when the courses lie either side of north and the difference is over 180°), and after a while the distance between them is the third side of a triangle.

$Ships from P on bearings 030° and 100°, 20 km and 26 km out$

One ship sails 20 km on bearing 030°, the other 26 km on bearing 100°. The angle between the courses is 100° − 30° = 70° (adjacent angles at the port), so ${d}^{2} = {20}^{2}+{26}^{2}-2\cdot 20\cdot 26\cdot \cos\left(70^{\circ}\right)$ and the ships are about 26.84 km apart.

Challenge

Two ships leave port P at noon: one on bearing 040° at 18 km/h, the other on bearing 115° at 24 km/h. At 2 pm they are 36 km and 48 km from P. Find the distance x between them, to two decimal places, then press Check.

Try it yourself

$Use the figure to find x$