0580/41 May/Jun 2022, Question 7: The Cosine Rule

Part (a) · 4 marks

Work out the size of angle ACD using triangle ACD with sides AD = 9 km, DC = 12 km and AC = 14 km.

1. 19² = 14² + 12² − 2 × 14 × 12 × cos ACD
2. 2cos ACD = (196 + 144 − 81)/(2 × 14 × 12) = 259/336 = 0.7708…
3. 3Angle ACD = 39.57… ≈ 39.6°

Answer39.6 or 39.57....

Part (b) · 3 marks

Use triangle ABC with angle BAC = 25°, angle BCA = 32° and AC = 14 km to demonstrate that BC equals 7.05 km to 2 decimal places.

1. 1BC/sin 25 = 14/sin 123
2. 2BC = 14 sin 25 ÷ sin 123
3. 3= 7.054… = 7.05 km (2 d.p.)

Answer(14 sin 25)/(sin 123) = 7.054…

Part (c) · 3 marks

Find the perpendicular distance from B to the line AC.

1. 1The shortest distance is the perpendicular from B to AC
2. 2Distance = 7.05 × sin 32
3. 3= 3.735… ≈ 3.74 km

Answer3.74 or 3.735 to 3.739

Part (d) · 4 marks

Determine the straight-line distance BD.

1. 1Angle BCD = 32 + 39.57… = 71.57…°
2. 2BD² = 12² + 7.05² − 2 × 12 × 7.05 × cos 71.57…
3. 3BD² = 140.2…
4. 4BD = 11.84… ≈ 11.8 km

Answer11.8 or 11.83 to 11.85

Part (e) · 2 marks

Given that C lies due east of A, work out the bearing of D measured from C.

1. 1C is due east of A, so the bearing of A from C is 270°.
2. 2From part (a), angle ACD = 39.57…°.
3. 3D is further clockwise than A, so add: 270 + 39.57… = 309.57…
4. 4Bearing of D from C ≈ 309.6°

Answer309.6 or 309.57...

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