(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 fo...
Question 1 Report
(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.
Answer Details
(a) The woman is at a window \(30\text{ m}\) above the ground. The foot of the flag pole is \(25\text{ m}\) horizontally from the foot of the building, on the same level ground.
(i) Angle of depression of the foot of the pole. From the window, the horizontal distance to the pole is \(25\text{ m}\) and the vertical drop to the foot of the pole is the full window height \(30\text{ m}\):
\[\tan\theta=\frac{30}{25}=1.2\]\[\theta=\tan^{-1}(1.2)\approx 50.19^\circ\approx\mathbf{50^\circ}.\](ii) Height of the flag pole. The angle of depression of the top of the pole is \(44^\circ\). Over the horizontal distance \(25\text{ m}\), the vertical drop from window level down to the top of the pole is
\[h_1=25\tan 44^\circ=25\times0.9657=24.14\text{ m}.\]So the top of the pole is \(30-24.14=5.86\text{ m}\) above the ground. Since the pole stands on the same ground, its height is
\[|\text{pole}|=30-24.14\approx\mathbf{6\text{ m}}.\](b) \(O\) is the centre of the circle, \(\angle OQR=32^\circ\) and \(\angle TPQ=15^\circ\).
(i) \(\angle QPR\): \(OQ\) and \(OR\) are radii, so triangle \(OQR\) is isosceles and
\[\angle ORQ=\angle OQR=32^\circ.\]\[\angle QOR=180^\circ-32^\circ-32^\circ=116^\circ.\]\(\angle QOR\) is the angle subtended by chord \(QR\) at the centre, and \(\angle QPR\) is subtended by the same chord at the circumference, so it is half:
\[\angle QPR=\tfrac{1}{2}\times116^\circ=\mathbf{58^\circ}.\](ii) \(\angle TQO\): \(\angle TPQ=15^\circ\) stands on chord \(TQ\), so the central angle on the same chord is
\[\angle TOQ=2\times15^\circ=30^\circ.\]\(OT\) and \(OQ\) are radii, so triangle \(TOQ\) is isosceles:
\[\angle TQO=\angle OTQ=\frac{180^\circ-30^\circ}{2}=\mathbf{75^\circ}.\]Download The App On Google Playstore
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