Theory question 112
Trigonometry(a) Given that 5 cos (x + 8.5)° − 1 = 0, 0° ≤ x ≤ 90°, calculate, correct to the nearest degree, the value of x.
(b) The bearing of Q from P is 150° and the bearing of P from R is 015°. If Q and R are 24 km and 32 km respectively from P: (i) represent this information in a diagram; (ii) calculate the distance between Q and R, correct to two decimal places; (iii) find the bearing of R from Q, correct to the nearest degree.
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(a) cos(x + 8.5)° = 0.2 → x + 8.5 = 78.46 → x ≈ 70°.
(b)(i) Sketch with north lines at P, Q and R: PQ = 24 km on bearing 150° from P; R is 32 km from P on bearing 195° (back bearing of 015°).
(ii) ∠QPR = 195° − 150° = 45°. QR2 = 322 + 242 − 2(32)(24) cos 45° = 1600 − 1086.1 = 513.9, so QR ≈ 22.67 km.
(iii) sin ∠PQR = 32 sin 45°/22.67 ≈ 0.998, so ∠PQR ≈ 86.5° (the angles 45° + 86.5° + 48.5° add to 180°).
The bearing of P from Q is 330°, so the bearing of R from Q = 330° − 86.5° ≈ 243.5°, i.e. about 243° (244° accepted).
Find the angle at P from the two bearings, then use the cosine rule.