Question

The angle of elevation of the top of the tower is ${45}^o$ on walking up a slope inclined at an angle of ${30}^o$ to the horizontal a distance $20mt$, the angle of elevation of top of tower is observed to be ${60}^o$. Then the height of the tower

• A. $10(\sqrt{3}+1)mt$
• B. $20(\sqrt{3}+1)mt$
• C. $100\sqrt{3}mt$
• D. $50(3+\sqrt{3})$

Answer 1

The correct option is A $10(\sqrt{3}+1)mt$

Given,

angle of elevation from top of the tower $={45}^o$

Horizontal distance $=20mt$

Slope of inclination $={30}^o$

Observation angle of elevation from the top of tower $={60}^o$

also given, to find out the height of the tower.

Let us consider fig $\left(a\right)$

In $△ABC$

tangent of $=\frac{Lengthofsides}{Lengthofadjsides}$

(figure) is drawn as per the data given in question

$\iff \tan {60}^o=\frac{AB}{BC}\iff \tan {60}^o=\frac{h+20}{x}[\because \tan {60}^o=\sqrt{3}]$

$\sqrt{3}=\frac{h+20}{x}\iff \sqrt{3}x=h+\mathrm{20......}\left(1\right)[AB=h+20\&BC=xfromfigure]$

Now, let us consider $△ADE,|{|}^{xy}$ we get,

$\tan {45}^o=\frac{AE}{DE}[AE=h,DE=xfromfigure\tan {45}^o=1]$

$\iff 1=\frac{h}{x}\iff h=x\to$ equation $\left(2\right)$

Let us now substitute equation $\left(2\right)$ in equation $\left(1\right)$

$\iff \sqrt{3}x=x+20[\because fromequation(2)h=x]$

$\iff \sqrt{3}x-x=20\iff x(\sqrt{3}-1)=20\iff x=\frac{20}{\sqrt{3}-1}\to$ equation $\left(3\right)$

Let us nows $M/D$ by $\sqrt{3}+1$ for equation $\left(3\right)$ we get,

$\iff x=\frac{20}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\iff \frac{20(\sqrt{3}+1)}{(\sqrt{3}{)}^2-(1{)}^2}[\because (a+b\left)\right(a-b)=a^2-b^2]$

$\iff \frac{20(\sqrt{3}+1)}{3-1}\iff \frac{20(\sqrt{3}+1)}{2}\Rightarrow 10(\sqrt{3}+1)=h$

$\therefore height\left(h\right)=10(\sqrt{3}+1)mt$