Theory question 110
Trigonometry(a) A surveyor walks 100m up a hill which slopes at an angle of 24° to the horizontal. Calculate, correct to the nearest metre, the height through which he rises
(b) In the diagram, ABC is an isosceles triangle. |AB| = |AC| = 5cm and |BC| = 8cm. Calculate, correct to the nearest degree, ∠BAC.
(c) Two boats, 70 metres apart and on opposite sides of a light house, are in a straight line with the light-house. The angles of elevation of the top of the light-house from the two boats are 71.6° and 45°. Find the height of the light-house. [Take tan 71.6° = 3]
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(a) Height = 100 sin 24° = 100 × 0.4067 ≈ 41 m.
(b) The perpendicular from A bisects BC: sin(½∠BAC) = 4/5 = 0.8, so ½∠BAC = 53.13° and ∠BAC ≈ 106°.
(c) Let the height be h. Distances from the foot: h/tan 71.6° = h/3 and h/tan 45° = h.
h/3 + h = 70, so 4h/3 = 70 and h = 52.5 m.
In (c) the two horizontal distances add up to the 70 m between the boats.