(a) A woman looking out from the window of a building at a height of 30m, observed that the angle of depression of the top of a flag pole was 44°. If the foot of the pole is 25m from the foot of the building and on the same horizontal ground, find, correct to the nearest whole number, the (i) angle of depression of the foot of the pole from the woman ; (ii) height of the flag pole.
(b) In the diagram, O is the centre of the circle, < OQR = 32° and < TPQ = 15°. Calculate, (i) < QPR ; (ii) < TQo.
(b) In the diagram, O is the centre of the circle, < OQR = 32° and < TPQ = 15°. Calculate, (i) < QPR ; (ii) < TQo.
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Correct Answer: Option n
Explanation:
(a)
(i) From \(\Delta\) TFO,
\(\tan x = \tan \theta\) (alternate angles)
\(\tan x = \frac{25}{30} = 0.8333\)
\(x = \tan^{-1} (0.8333)\)
= \(39.80° \approxeq 40°\)
Angle of depression of the pole from the woman
= 90° - 40°
= 50°.
(ii) From \(\Delta\) TPQ,
\(\tan 46 = \frac{25}{TQ}\)
\(TQ = \frac{25}{\tan 46}\)
\(TQ = \frac{25}{1.0355}\)
= \(24.1429 m\)
Height of the pole : \(TF - TQ\)
= \(30 m - 24.1429 m\)
= \(5.8571 m\)
\(\approxeq 6m\)
(a)
(i) From \(\Delta\) TFO,
\(\tan x = \tan \theta\) (alternate angles)
\(\tan x = \frac{25}{30} = 0.8333\)
\(x = \tan^{-1} (0.8333)\)
= \(39.80° \approxeq 40°\)
Angle of depression of the pole from the woman
= 90° - 40°
= 50°.
(ii) From \(\Delta\) TPQ,
\(\tan 46 = \frac{25}{TQ}\)
\(TQ = \frac{25}{\tan 46}\)
\(TQ = \frac{25}{1.0355}\)
= \(24.1429 m\)
Height of the pole : \(TF - TQ\)
= \(30 m - 24.1429 m\)
= \(5.8571 m\)
\(\approxeq 6m\)