How to solve this sec 4 qns.?

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# Question

# Answer

Adwin

a) Bearing of C from B: 228°

∠ABC → 360° – 228° – 90° (anti-clockwise turn from B to C)

= **42****°**

b) Bearing of C from A: 113°

∠BAC → 113° – 90° = 23°

Using Law of Sines,

sin 115 ÷ AB = sin 23 ÷ BC

sin 115 ÷ 700 = sin 23 ÷ BC

BC = 700 sin 23 ÷ sin 115

≈ **301.79 m**

Using Law of Cosines,

(QC)^{2} = (BQ)^{2} + (BC)^{2} – 2(BQ)(BC) x cos (∠ABC)

(QC)^{2} = 450^{2} + 301.79^{2} – 2(450)(301.79) x c0s 42

QC = sqrt[450^{2} + 301.79^{2} – 2(450)(301.79) x c0s 42]

≈ **302.87 m**

c) △PQC is a right triangle

sin 12 = 302.87 ÷ PC

PC = 302.78 ÷ sin 12

≈ **1456.29 m**

d) QR is shortest distance from P to a point on BC, ∴ QR must be height of △BQC (with BC as base)

With reference to △BQR,

sin 42 = QR ÷ 450

QR = sin 42 x 450

≈ **301.11 m
**

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