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



(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 \(\tan 71.6° = 3\)].
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    Correct Answer: Option n
    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 \(\Delta QSR\),
    \(\frac{h}{QR} = \tan 45\)
    \(QR = \frac{h}{\tan 45}\)
    \(PR = PQ + QR\)
    \(\frac{h}{3} + h = 70 \implies \frac{4}{3}h = 70\)
    \(h = \frac{70 \times 3}{4} = 52.5m\)

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