Question

Question 12
If the height of a tower and the distance of the point of observation from its foot, both are increased by $10\%$, then the angle of elevation of its top remains unchanged.

Answer 1

The statement is true.

Case I:
Let the height of tower be h and the distance of the point of observation from its foot is x.

In $\Delta ABC$,

$tan{\theta }_1=\frac{AC}{BC}=\frac{h}{x}$

$\Rightarrow {\theta }_1=ta{n}^{-1}\left(\frac{h}{x}\right)\dots \dots \left(i\right)$


Case II
Now, the height of a tower increased by $10\%=h+10\%ofh=h+h\times \frac{10}{100}=\frac{11h}{10}$

And the distance of the point of observation from its foot $=x+10\%ofx$

$=x+x\times \frac{10}{100}=\frac{11x}{10}$
Let the new angle be ${\theta }_2$.


$tan{\theta }_2=\frac{\left(\frac{11h}{10}\right)}{\left(\frac{11x}{10}\right)}$

$\Rightarrow tan{\theta }_2=\frac{h}{x}$

$\Rightarrow {\theta }_2=ta{n}^{-1}\left(\frac{h}{x}\right)\dots \dots \left(ii\right)$

Therefore,

${\theta }_1={\theta }_2$

Hence, the required angle of elevation of its top remains unchanged.