Answer to Question #151693 in Trigonometry for Mahaba Kamal

Question

A submarine sailed 30 nautical miles from X to Y on a course of 040°. It then sailed
55 nautical miles due east to Z. Find the distance and bearing of Z from X.

Answer 1

Using cosine rule:

a² = b² + c² -2bccosA

So,

(xz)² = (xy)² + (yz)² - 2(xy)(yz)cos(XZ)

(xz)² = 30² + 55² - (2*30*55cos(130^o))

(xz)² = 30² + 55² - (-2121.199)

(xz)² = 6046.199

xz = 77.76 nautical miles

Using sine rule,

(sin A/a) = (sin B/b)

So,

(sin YZ/yz) = (sin XZ/xz)

(sin thetha/55) = (sin 130/77.76)

sin thetha = (55 * sin 130)/77.76

sin thetha = 0.542

thetha = sin^-1(0.542) = 32.82^o

Therefore, the bearing of Z from X is 40+32.82 = 72.82^o.