(a) Without using Mathematical tables or calculators, simplify : \(\frac{2\tan 60° + \cos 30°}{\sin 60°}\)
(b) From an aeroplane in the air and at a horizontal distance of 1050m, the angles of depression of the top and base of a control tower at an instance are 36° and 41° respectively. Calculate, correct to the nearest meter, the :
(i) height of the control tower ; (ii) shortest distance between the aeroplane and the base of the control tower.
(b) From an aeroplane in the air and at a horizontal distance of 1050m, the angles of depression of the top and base of a control tower at an instance are 36° and 41° respectively. Calculate, correct to the nearest meter, the :
(i) height of the control tower ; (ii) shortest distance between the aeroplane and the base of the control tower.
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Correct Answer: Option n
Explanation:
(a)
From \(\Delta\) AMC, \(\sin 60° = \frac{\sqrt{3}}{2}\)
\(\cos 30° = \frac{\sqrt{3}}{2}\)
\(\tan 60° = \frac{\sqrt{3}}{1} = \sqrt{3}\)
Hence, \(\frac{2 \tan 60° + \cos 30°}{\sin 60} = \frac{2 \times \sqrt{3} + \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}\)
= \(\frac{4\sqrt{3} + \sqrt{3}}{2} \div \frac{\sqrt{3}}{2}\)
= \(\frac{5\sqrt{3}}{2} \times \frac{2}{\sqrt{3}} = 5\)
(b)
In \(\Delta\) POT, \(\tan 36° = \frac{|TO|}{1050}\)
\(|OT| = 1050 \tan 36° = 1050 \times 0.7265\)
= 762.87m
In \(\Delta\) POB, \(\tan 41° = \frac{|OB|}{1050}\)
\(|OB| = 1050 \tan 41° = 1050 \times 0.8693\)
= 912.751m
\(|TB| = 912.751m - 762.87m = 149.88m\)
Hence, the height of the control tower = 150 m (to the nearest meter).
(ii) In \(\Delta\) POB, \(\cos 41° = \frac{1050}{|PB|}\)
\(|PB| = \frac{1050}{\cos 41°} = \frac{1050}{0.7547}\)
= 1,391.28m
The shortest distance between the aeroplane and the base of the control tower = 1,391m (to the nearest metre).
(a)
From \(\Delta\) AMC, \(\sin 60° = \frac{\sqrt{3}}{2}\)
\(\cos 30° = \frac{\sqrt{3}}{2}\)
\(\tan 60° = \frac{\sqrt{3}}{1} = \sqrt{3}\)
Hence, \(\frac{2 \tan 60° + \cos 30°}{\sin 60} = \frac{2 \times \sqrt{3} + \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}\)
= \(\frac{4\sqrt{3} + \sqrt{3}}{2} \div \frac{\sqrt{3}}{2}\)
= \(\frac{5\sqrt{3}}{2} \times \frac{2}{\sqrt{3}} = 5\)
(b)
In \(\Delta\) POT, \(\tan 36° = \frac{|TO|}{1050}\)
\(|OT| = 1050 \tan 36° = 1050 \times 0.7265\)
= 762.87m
In \(\Delta\) POB, \(\tan 41° = \frac{|OB|}{1050}\)
\(|OB| = 1050 \tan 41° = 1050 \times 0.8693\)
= 912.751m
\(|TB| = 912.751m - 762.87m = 149.88m\)
Hence, the height of the control tower = 150 m (to the nearest meter).
(ii) In \(\Delta\) POB, \(\cos 41° = \frac{1050}{|PB|}\)
\(|PB| = \frac{1050}{\cos 41°} = \frac{1050}{0.7547}\)
= 1,391.28m
The shortest distance between the aeroplane and the base of the control tower = 1,391m (to the nearest metre).