(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 $\tan 71.6° = 3$].

Thursday, 03 April 2025

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(a) A surveyor walks 100m up a hill which slopes at an angle of 24° to the ...

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(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

40.67 m

(b)

\sin θ = \frac{4}{5} = 0.8

θ = \sin^{- 1} (0.8) = 53.13 °

< B A C = 2 θ = 2 (53.13 °)

106.26 °

≊ 106 °

(to the nearest degree).
(c)
In

Δ P S Q

\frac{h}{P Q} = \tan 71.6

P Q = \frac{h}{\tan 71.6}

Δ Q S R

\frac{h}{Q R} = \tan 45

Q R = \frac{h}{\tan 45}

P R = P Q + Q R

\frac{h}{3} + h = 70 ⟹ \frac{4}{3} h = 70

h = \frac{70 \times 3}{4} = 52.5 m

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