From the top of a vertical tower, the angles of depression of two cars, in the same strait line with the base of the tower, at an instant are found to be $ 45^{o}$ and$\ 60^{o}$. If the cars are 100 in apart and are on the same side of the tower. Find the height of the tower. $ \left( use\ \sqrt{3} =1.73\right)$

Given: A vertical tower and angle of the depressions off two cars from the top of the tower $\angle ABC=45^{o}$ and $\angle ADC=60^{o}$ and distance BD between the two cars$=100$
.
To do: To find the height of the tower.

Solution:
Let us say AC is the tower. And B, d are the points where the cars are located at the time.
As given distance between the cars, $BD=100\ $m 

$\angle ABC=45^{o} \ and\ \angle ADC=60^{o}$

In $\vartriangle ADC,\ tan60^{o} =\frac{AC}{DC} =\sqrt{3}$                        $\left( \because \ tan60^{o} =\sqrt{3}\right)$

$\Rightarrow DC=\frac{AC}{\sqrt{3}}$

In $\vartriangle ABC$, $tan45^{o} =\frac{AC}{BC} =1$                                    $( \because \ tan45^{o} =1)$

$\Rightarrow AC=BC$

and $BC=BD+DC$

$\Rightarrow AC=BD+\frac{AC}{\sqrt{3}}$ 

$\Rightarrow AC-\frac{AC}{\sqrt{3}} =BD$
 
$\Rightarrow AC\left( 1-\frac{1}{\sqrt{3}}\right) =100$

$\Rightarrow AC=\frac{100}{\left( 1-\frac{1}{\sqrt{3}}\right)}$

$=\frac{\left( 100\sqrt{3}\right)}{\left(\sqrt{3} -1\right)}$

$=236.98$ m
 

Thus, the height of the tower is 236.98 m.

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Updated on: 10-Oct-2022

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