0580/42 Oct/Nov 2023, Question 7: Using Trigonometry and Pythagoras Together

Part (ai) · 2 marks

Pole 2.8 m tall, 7.1 m from Q: angle of elevation from Q.

1. 1tan(angle) = 2.8/7.1
2. 2Angle of elevation = 21.52… ≈ 21.5°

Answer21.5 or 21.52...

Part (aii) · 3 marks

Use the 21° elevation from P to find PQ.

1. 1tan 21° = 2.8/PR, so PR = 2.8 ÷ tan 21° = 7.294…
2. 2PQ² = 7.294…² + 7.1²
3. 3PQ = 10.179… ≈ 10.2 m

Answer10.2 or 10.17 to 10.18

Part (b) · 3 marks

Sine rule in triangle KLM (9.1 cm, 16.7 cm, 32°) for the acute angle at M.

1. 19.1/sin 32° = 16.7/sin M
2. 2sin M = 16.7 × sin 32° ÷ 9.1 = 0.9724…
3. 3Angle KML = 76.52… ≈ 76.5°

Answer76.5 or 76.52 to 76.53

Part (ci) · 2 marks

Use the area 62.89 cm² to show the angle at A is 28.4°.

1. 11/2 × 12.3 × 21.5 × sin BAC = 62.89
2. 2sin BAC = 62.89 ÷ 132.225 = 0.47562…
3. 3Angle BAC = 28.400… = 28.4° to 1 d.p.

Answer1/2 × 12.3 × 21.5 sin(…) = 62.89 leading to 28.40 to 28.41…

Part (cii) · 3 marks

Cosine rule for BC.

1. 1BC² = 12.3² + 21.5² − 2 × 12.3 × 21.5 × cos 28.4°
2. 2BC² = 148.29…
3. 3BC = 12.17… ≈ 12.2 cm

Answer12.2 or 12.17 to 12.18

Part (ciii) · 3 marks

Extend AB to D with a right angle at D and find BD.

1. 1Use the area to find angle A: 1/2 × 21.5 × 12.3 × sin A = 62.89
2. 2sin A = 0.4756…, so angle A = 28.4°
3. 3Triangle ADC is right-angled: AD = 21.5 × cos 28.4° = 18.91…
4. 4BD = AD − AB = 18.91… − 12.3 = 6.61

Answer6.60 to 6.62

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