急!F5 Solving Triangles 10exq1

3.
(a)
In ΔABC :
cosA = (AB² + AC² - BC²) / (2 x AB x AC) (cosine law)
cosA = (65² + 45² - 30²) / (2 x 65 x 45)
A = 23.9°

cosC = (AC² + BC² - AB²) / (2 x AC x BC) (cosine law)
cosC = (45² + 30² - 65²) / (2 x 45 x 30)
C = 118.8°

The true bearing of B from A
= (118.8° + 130° - 180°) + 23.9°
= 092.7°


(b)
Draw the altitude CH of ΔABC.

In ΔACH :
sinA = CH/AC
sin23.9° = CH/(45 km)
CH = 18.2 km

The shortest distance of the ship to island C = 18.2 km