The angle of elevation of top of tower from certain point is $30^o$. if the observer moves $20\ m$ towards the tower, the angle of elevation of the top increases by $15^o$. Find the height of the tower.

Given: The angle of elevation of top of tower from certain point is $30^o$. if the observer moves $20\ m$ towards the tower, the angle of elevation of the top increases by $15^o$.

To do: To find the height of the tower.

Solution:

Let AB be the tower and C and D be the point of observation 

Given  $CD=20\ m$ And $\angle BCA=30^o$ and $\angle BDA=30+15=45^o$

Let height of tower is $h$

In triangle $BAD$

$tan45^o=\frac{AB}{AD}$

​

$\Rightarrow 1=\frac{h}{AD}$

​

$\Rightarrow AD=h$

In triangle $BAC$

$tan30^o=\frac{AB}{AC}$        $( AC=CD+AD)$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20+h}$

​

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

$\Rightarrow \sqrt{3}h−h=20$

$\Rightarrow h( 1.732−1)=20$

$\Rightarrow h=\frac{20}{0.732}$

​

$=27.3$

Thus the height of the tower is $27.3$.

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Updated on: 10-Oct-2022

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