Trigonometry question with bearings

Question

The "good ship" is at A, 1km due north of the "enemy ship" at noon. After this, the good ship moves on a bearing of 045 degrees while the enemy remains stationary .

When the good ship has travelled 1.5km from A, how far is it from the enemy ship?

Answer 1

This forms an obtuse triangle with sides of 1 mile and 1.5 miles with the angle between these two sides being 135°.

Using the law of cosines, x² = 1² + 1.5² - 2(1)(1.5)COS(135°) = 5.3713

x = √(5.3713) = 2.32 miles