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Two poles of equal heights are standing opposite to each other on either side of the road which is $90$ m wide. From a point between them on the road, the angles of elevation of the top of the poles are $30^{\circ}$ and $60^{\circ}$ respectively. Find the height of the poles and the distances of the point from the poles.
[Use $\sqrt{3} = 1.732$]
[Use $\sqrt{3} = 1.732$]
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Let CB be the road and AB, DC are the poles of equal heights ($h$).
P be the point on the road.
$\angle APB = 60^{\circ}$ and $\angle DPC = 30^{\circ}$.
Let BP be $x$, then PC = $(90 - x)$ ($\frac{1}{2}$ Mark)
In $\triangle ABP$
$\tan 60^{\circ} = \frac{h}{x} = \sqrt{3}$ (1 Mark)
$h = \sqrt{3}x$ ($\frac{1}{2}$ Mark)
In $\triangle DCP$
$\tan 30^{\circ} = \frac{h}{90 - x} = \frac{1}{\sqrt{3}}$ (1 Mark)
$\Rightarrow \sqrt{3}h = 90 - x$ ($\frac{1}{2}$ Mark)
On solving $x = 22.5$ ($\frac{1}{2}$ Mark)
AB = $\sqrt{3}x = 22.5 \times 1.732 = 38.97$ m ($\frac{1}{2}$ Mark)
Height of the poles = $38.97$ m
Distances of the point P from the poles are $22.5$ m and $67.5$ m ($\frac{1}{2}$ Mark)
P be the point on the road.
$\angle APB = 60^{\circ}$ and $\angle DPC = 30^{\circ}$.
Let BP be $x$, then PC = $(90 - x)$ ($\frac{1}{2}$ Mark)
In $\triangle ABP$
$\tan 60^{\circ} = \frac{h}{x} = \sqrt{3}$ (1 Mark)
$h = \sqrt{3}x$ ($\frac{1}{2}$ Mark)
In $\triangle DCP$
$\tan 30^{\circ} = \frac{h}{90 - x} = \frac{1}{\sqrt{3}}$ (1 Mark)
$\Rightarrow \sqrt{3}h = 90 - x$ ($\frac{1}{2}$ Mark)
On solving $x = 22.5$ ($\frac{1}{2}$ Mark)
AB = $\sqrt{3}x = 22.5 \times 1.732 = 38.97$ m ($\frac{1}{2}$ Mark)
Height of the poles = $38.97$ m
Distances of the point P from the poles are $22.5$ m and $67.5$ m ($\frac{1}{2}$ Mark)