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Two poles of equal heights are standing opposite each other on either side of the road, which is $80$ m wide. From a point between them on the road, the angles of elevation of the top of the poles are $60^{\circ}$ and $30^{\circ}$ respectively. Find the height of the poles and the distance of the point from the poles.
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Correct figure (1 Mark)
Let AB and CD be the poles of height $h$ m and O is the observation point.
In $\triangle DCO$, $\tan 30^{\circ} = \frac{h}{80-x}$ (1 Mark)
$\Rightarrow 80-x = h\sqrt{3}$ (i) (1/2 Mark)
In $\triangle ABO$, $\tan 60^{\circ} = \frac{h}{x}$ (1 Mark)
$\Rightarrow h = x\sqrt{3}$ (ii) (1 Mark)
Using (i) and (ii) we get $x = 20$, i.e., AO = $20$ m and $h = 20\sqrt{3}$ m. (1/2 Mark + 1/2 Mark)
$\therefore$ CO = $60$ m.
$\therefore$ The height of the pole is $20\sqrt{3}$ m and the distance of the point from the poles is $20$ m and $60$ m
Let AB and CD be the poles of height $h$ m and O is the observation point.
In $\triangle DCO$, $\tan 30^{\circ} = \frac{h}{80-x}$ (1 Mark)
$\Rightarrow 80-x = h\sqrt{3}$ (i) (1/2 Mark)
In $\triangle ABO$, $\tan 60^{\circ} = \frac{h}{x}$ (1 Mark)
$\Rightarrow h = x\sqrt{3}$ (ii) (1 Mark)
Using (i) and (ii) we get $x = 20$, i.e., AO = $20$ m and $h = 20\sqrt{3}$ m. (1/2 Mark + 1/2 Mark)
$\therefore$ CO = $60$ m.
$\therefore$ The height of the pole is $20\sqrt{3}$ m and the distance of the point from the poles is $20$ m and $60$ m