Theory question 103
Trigonometry(a) Given that sin x = 5/13, 0° ≤ x ≤ 90°, find (cos x − 2 sin x)/(2 tan x)
(b) The diagram represents the vertical cross-section of a mountain with height NQ standing on a horizontal ground PRN. If the angles of elevation of the top Q of the mountain from P and R are 30° and 70° respectively and PR = 500m, calculate, correct to 3 significant figures: (i) |QP|; (ii) the height of the mountain.
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(a) Opposite = 5, hypotenuse = 13, so adjacent = 12: cos x = 12/13, tan x = 5/12.
(cos x − 2 sin x)/(2 tan x) = (12/13 − 10/13) ÷ (10/12) = 2/13 × 12/10 = 12/65.
(b) In triangle PQR, ∠PRQ = 110°, so ∠PQR = 180 − 30 − 110 = 40°.
(i) Sine rule: QP/sin 110° = 500/sin 40°, QP = 500 × 0.9397/0.6428 ≈ 731 m.
(ii) Height NQ = QP sin 30° = 730.9 × 0.5 ≈ 365 m.
Use the exterior angle at R to find the angle at Q, then the sine rule.