Bearings Problem set B One question at a time

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

1. (a) 118°;

   • \(118^\circ + 180^\circ = 298^\circ\)
   • Bearing: 298°

2. (b) 305°;

   • \(305^\circ - 180^\circ = 125^\circ\)
   • Bearing: 125°

3. (c) 037°.

   • \(037^\circ + 180^\circ = 217^\circ\)
   • Bearing: 217°

P, Q and R are three camps in a forest. Q is 7 km from P on a bearing of 048°. R is 12 km from Q on a bearing of 137°. Find the distance PR.

• The back-bearing of PQ is 228°, so the angle at Q is \(|228^\circ - 137^\circ| = 91^\circ\)
• \(PR^2 = 7^2 + 12^2 - 2(7)(12)\cos 91^\circ = 195.9\)
• \(PR = 14.0\) km

A cyclist rides from town P to town Q, 10 km on a bearing of 070°, and then 8 km on a bearing of 152° to town R. Find the bearing of P from R, to the nearest degree.

• Angle at Q is 98° and PR = 13.65 km (cosine rule).
• \(\dfrac{\sin P}{8} = \dfrac{\sin 98^\circ}{13.65}\), so \(\angle QPR = 35.48^\circ\)
• The route turns right at Q, so R is clockwise from PQ. Bearing of R from P: \(070^\circ + 35.48^\circ = 105.5^\circ\)
• Bearing of P from R: \(105.5^\circ + 180^\circ = 285.5^\circ\), which is 285°

A walker leaves a car park P and walks 14 km on a bearing of 330° to a hut Q. She then walks 20 km on a bearing of 060° to a lake R.

1. (a) Find the distance PR.

   • The back-bearing of PQ is 150°, so the angle at Q is \(|150^\circ - 060^\circ| = 90^\circ\)
   • \(PR^2 = 14^2 + 20^2 - 2(14)(20)\cos 90^\circ = 596.0\)
   • \(PR = 24.4\) km

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

   • \(\dfrac{\sin P}{20} = \dfrac{\sin 90^\circ}{24.41}\), so \(\angle QPR = 55.01^\circ\)
   • The route turns right at Q, so R is clockwise from PQ. Bearing of R from P: \(330^\circ + 55.01^\circ = 385.0^\circ\)\(\;\text{ so }\; 385.0^\circ - 360^\circ = 25.01^\circ\)
   • Bearing of P from R: \(25.01^\circ + 180^\circ = 205.0^\circ\), which is 205°

A rescue helicopter flies 40 km from its base P on a bearing of 200°, then 30 km on a bearing of 110°. Find how far it is from P.

• The back-bearing of PQ is 020°, so the angle at Q is \(|020^\circ - 110^\circ| = 90^\circ\)
• \(PR^2 = 40^2 + 30^2 - 2(40)(30)\cos 90^\circ = 2500\)
• \(PR = 50.0\) km

Two fire lookout towers P and Q are 20 km apart, with Q due east of P. Smoke F is seen from P on a bearing of 062° and from Q on a bearing of 318°.

1. (a) Find the distance of the smoke from P.

   • Angle at P: \(90^\circ-62^\circ=28^\circ\). Angle at Q: \(318^\circ-270^\circ=48^\circ\)
   • Angle at F: \(180^\circ-28^\circ-48^\circ=104^\circ\)
   • \(\frac{PF}{\sin48^\circ}=\frac{20}{\sin104^\circ}\), so \(PF=15.318\), which is 15.3 km

2. (b) Find the distance of the smoke from Q.

   • \(\frac{QF}{\sin28^\circ}=\frac{20}{\sin104^\circ}\)
   • \(QF=9.677\), which is 9.68 km

A walker leaves a station P and walks 15 km on a bearing of 290° to Q. She then walks 11 km on a bearing of 205° to R.

1. (a) Find the distance PR.

   • The back-bearing of PQ is 110°, so the angle at Q is \(|110^\circ - 205^\circ| = 95^\circ\)
   • \(PR^2 = 15^2 + 11^2 - 2(15)(11)\cos 95^\circ = 374.8\)
   • \(PR = 19.4\) km

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

   • \(\dfrac{\sin P}{11} = \dfrac{\sin 95^\circ}{19.36}\), so \(\angle QPR = 34.48^\circ\)
   • The route turns left at Q, so R is anticlockwise from PQ. Bearing of R from P: \(290^\circ - 34.48^\circ = 255.5^\circ\)
   • Bearing of P from R: \(255.5^\circ + 180^\circ = 435.5^\circ\)\(\;\text{ so }\; 435.5^\circ - 360^\circ = 75.52^\circ\), which is 076°