Bearings Problem set A One question at a time

Write down the bearing of the first point from the second point when the bearing of the second from the first is:

1. (a) 064°;

   • \(064^\circ + 180^\circ = 244^\circ\)
   • Bearing: 244°

2. (b) 281°;

   • \(281^\circ - 180^\circ = 101^\circ\)
   • Bearing: 101°

3. (c) 150°.

   • \(150^\circ + 180^\circ = 330^\circ\)
   • Bearing: 330°

P, Q and R are three buoys. Q is 8 km from P on a bearing of 063°. R is 11 km from Q on a bearing of 150°. Find the distance PR.

• The back-bearing of PQ is 243°, so the angle at Q is \(|243^\circ - 150^\circ| = 93^\circ\)
• \(PR^2 = 8^2 + 11^2 - 2(8)(11)\cos 93^\circ = 194.2\)
• \(PR = 13.9\) km

Three villages P, Q and R: Q is 9 km from P on a bearing of 025°, and R is 7 km from Q on a bearing of 110°. Find the bearing of P from R, to the nearest degree.

• Angle at Q is 95° and PR = 11.87 km (cosine rule).
• \(\dfrac{\sin P}{7} = \dfrac{\sin 95^\circ}{11.87}\), so \(\angle QPR = 35.97^\circ\)
• The route turns right at Q, so R is clockwise from PQ. Bearing of R from P: \(025^\circ + 35.97^\circ = 60.97^\circ\)
• Bearing of P from R: \(60.97^\circ + 180^\circ = 241.0^\circ\), which is 241°

A ship leaves harbour P and sails 18 nautical miles on a bearing of 075° to a lighthouse Q. It then sails 25 nautical miles on a bearing of 160° to an island R.

1. (a) Find the distance PR.

   • The back-bearing of PQ is 255°, so the angle at Q is \(|255^\circ - 160^\circ| = 95^\circ\)
   • \(PR^2 = 18^2 + 25^2 - 2(18)(25)\cos 95^\circ = 1027\)
   • \(PR = 32.1\) nautical miles

2. (b) Find the bearing on which the ship must sail from R to return directly to P, to the nearest degree.

   • \(\dfrac{\sin P}{25} = \dfrac{\sin 95^\circ}{32.05}\), so \(\angle QPR = 50.98^\circ\)
   • The route turns right at Q, so R is clockwise from PQ. Bearing of R from P: \(075^\circ + 50.98^\circ = 126.0^\circ\)
   • Bearing of P from R: \(126.0^\circ + 180^\circ = 306.0^\circ\), which is 306°

A drone flies 35 km from its base P on a bearing of 048°, then 28 km on a bearing of 300°. Find how far it is from P.

• The back-bearing of PQ is 228°, so the angle at Q is \(|228^\circ - 300^\circ| = 72^\circ\)
• \(PR^2 = 35^2 + 28^2 - 2(35)(28)\cos 72^\circ = 1403\)
• \(PR = 37.5\) km

Two coastguard stations X and Y are 15 km apart, with Y due east of X. A boat B is seen from X on a bearing of 035° and from Y on a bearing of 300°.

1. (a) Find the distance of the boat from X.

   • Angle at X: \(90^\circ-35^\circ=55^\circ\). Angle at Y: \(300^\circ-270^\circ=30^\circ\)
   • Angle at B: \(180^\circ-55^\circ-30^\circ=95^\circ\)
   • \(\frac{XB}{\sin30^\circ}=\frac{15}{\sin95^\circ}\), so \(XB=7.529\), which is 7.53 km

2. (b) Find the distance of the boat from Y.

   • \(\frac{YB}{\sin55^\circ}=\frac{15}{\sin95^\circ}\)
   • \(YB=12.334\), which is 12.3 km

A yacht leaves a marina P and sails 16 km on a bearing of 250° to Q. It then sails 12 km on a bearing of 170° to R.

1. (a) Find the distance PR.

   • The back-bearing of PQ is 070°, so the angle at Q is \(|070^\circ - 170^\circ| = 100^\circ\)
   • \(PR^2 = 16^2 + 12^2 - 2(16)(12)\cos 100^\circ = 466.7\)
   • \(PR = 21.6\) km

2. (b) Find the bearing of P from R, to the nearest degree.

   • \(\dfrac{\sin P}{12} = \dfrac{\sin 100^\circ}{21.60}\), so \(\angle QPR = 33.16^\circ\)
   • The route turns left at Q, so R is anticlockwise from PQ. Bearing of R from P: \(250^\circ - 33.16^\circ = 216.8^\circ\)
   • Bearing of P from R: \(216.8^\circ + 180^\circ = 396.8^\circ\)\(\;\text{ so }\; 396.8^\circ - 360^\circ = 36.84^\circ\), which is 037°