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From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. $(Use \sqrt{3} = 1.73 )$

Answers (1)

Given :

$BC=20\; m$

$\angle B\!DC=45^{\circ}$

$\angle A\!DC=60^{\circ}$

To find : $AB=h=?$

In $\bigtriangleup BCD \;\;\;\; (\angle C =90^{\circ})$

$\tan 45=\frac{BC}{CD}$

$1=\frac{20}{CD}\;\;\;\; (\because \tan 45 =1)$

$CD=20\; m \;\;\;\;\;-(1)$

In $\bigtriangleup AC\!D$ ,

$\tan 60=\frac{AC}{CD}$

$\sqrt{3}=\frac{AC}{20}\;\;\;\;\;\;\;(CD=20\;m(\text{from (1)}))$

$\Rightarrow AC=20\sqrt{3}\; m$

Now,

$AC=AB+BC=20\sqrt{3}$

$\Rightarrow h+20=20\sqrt{3}$

$\Rightarrow h=20(\sqrt{3}-1)$

$=20(1.732-1)$

$=20(0.732)$

$h=14.64\; m$

Posted by

Safeer PP

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