(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.

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(a) Without using Mathematical tables or calculators, simplify : \(\frac{2\tan 60° ...

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(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.

Explanation

(a)
From

Δ

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

Δ

POT,

\tan 36 ° = \frac{| T O |}{1050}

| O T | = 1050 \tan 36 ° = 1050 \times 0.7265

= 762.87m
In

Δ

POB,

\tan 41 ° = \frac{| O B |}{1050}

| O B | = 1050 \tan 41 ° = 1050 \times 0.8693

= 912.751m

| T B | = 912.751 m - 762.87 m = 149.88 m

Hence, the height of the control tower = 150 m (to the nearest meter).
(ii) In

Δ

POB,

\cos 41 ° = \frac{1050}{| P B |}

| P B | = \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).

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