(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 \(\tan 71.6° = 3\)].
(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 \(\Delta QSR\),
\(\frac{h}{QR} = \tan 45\)
\(QR = \frac{h}{\tan 45}\)
\(PR = PQ + QR\)
\(\frac{h}{3} + h = 70 \implies \frac{4}{3}h = 70\)
\(h = \frac{70 \times 3}{4} = 52.5m\)
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}