(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| = 5 cm, and |BC| = 8 cm. 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)].
Explanation
(a) (frac{h}{100} = sin 24)
(h = 100 sin 24)
= (100 times 0.4067)
= (40.67m)
(b) (sin theta = frac{4}{5} = 0.8)
(theta = sin^{-1} (0.8) = 53.13°)
(< BAC = 2 theta = 2(53.13°))
= (106.26°)
(approxeq 106°) (to the nearest degree).
(c)
In (Delta PSQ),
(frac{h}{PQ} = tan 71.6)
(PQ = frac{h}{tan 71.6})
In (Delta QSR),
(frac{h}{QR} = tan 45)
(QR = frac{h}{tan 45})
(PR = PQ + QR)
(frac{h}{3} + h = 70 implies frac{4}{3}h = 70)
(h = frac{70 times 3}{4} = 52.5m)