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Two poles of equal heights are standing opposite each other on either side of the road, which is $80 \text{ m}$ 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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Solution:
correct figure (1 Mark)
Let $\text{AB}$ and $\text{CD}$ be the poles of height $h \text{ m}$ and $\text{O}$ is the observation point.
In $\triangle \text{DCO}$, $\tan 30^\circ = \frac{h}{80-x}$ (1 Mark)
$\Rightarrow 80 - x = h\sqrt{3}$ (i) (
frac{1}{2} Mark)
In $\triangle \text{BAO}$, $\tan 60^\circ = \frac{h}{x}$ (1 Mark)
$\Rightarrow h = x\sqrt{3}$ (ii)
Using (i) and (ii) we get $x = 20$, i.e., $\text{AO = 20 m}$ and $h = 20\sqrt{3} \text{ m}$. (
frac{1}{2} +
frac{1}{2} Mark)
$\therefore \text{CO = 60 m}$. (
frac{1}{2} Mark)
$\therefore$ The height of the pole is $20\sqrt{3} \text{ m}$ and the distance of the point from the poles is $20 \text{ m}$ and $60 \text{ m}$. (
frac{1}{2} Mark)
correct figure (1 Mark)
Let $\text{AB}$ and $\text{CD}$ be the poles of height $h \text{ m}$ and $\text{O}$ is the observation point.
In $\triangle \text{DCO}$, $\tan 30^\circ = \frac{h}{80-x}$ (1 Mark)
$\Rightarrow 80 - x = h\sqrt{3}$ (i) (
frac{1}{2} Mark)
In $\triangle \text{BAO}$, $\tan 60^\circ = \frac{h}{x}$ (1 Mark)
$\Rightarrow h = x\sqrt{3}$ (ii)
Using (i) and (ii) we get $x = 20$, i.e., $\text{AO = 20 m}$ and $h = 20\sqrt{3} \text{ m}$. (
frac{1}{2} +
frac{1}{2} Mark)
$\therefore \text{CO = 60 m}$. (
frac{1}{2} Mark)
$\therefore$ The height of the pole is $20\sqrt{3} \text{ m}$ and the distance of the point from the poles is $20 \text{ m}$ and $60 \text{ m}$. (
frac{1}{2} Mark)