(a) Without using Mathematical tables or calculators, simplify : (frac{2tan 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.
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{4sqrt{3} + sqrt{3}}{2} div frac{sqrt{3}}{2})
= (frac{5sqrt{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).