Question

The angles of elevation of an aeroplane flying vertically above the ground from two consecutive milestones(1 km) apart are 45${}^o$ and 60${}^o$. The height of the aeroplane from the ground is _____

• A. $\frac{1}{2}$(3 + $\sqrt{3}$) km
• B. $\frac{1}{2}$($\sqrt{3}$ + 1) km
• C. (3 + $\sqrt{3}$) km
• D. ($\sqrt{3}$ + 1) km

Answer 1

The correct option is A $\frac{1}{2}$(3 + $\sqrt{3}$) km

Let A be the position of aeroplane and

B and C be the two milestones

$\therefore$ BC = 1 km.
Let AD = perpendicular meeting BC at D.

$\angle$ABD = 45${}^o$ and $\angle$ACD = 60${}^o$
let AD = h km, CD = x km
Now, tan45${}^o$ = $\frac{h}{x+1}\Rightarrow 1=\frac{h}{x+1}\Rightarrow$ x = h- 1

Also, tan60${}^o$ = $\frac{h}{x}\Rightarrow \sqrt{3}=\frac{h}{x}\Rightarrow x=\frac{h}{\sqrt{3}}$

$\therefore$$\frac{h}{\sqrt{3}}$ = h - 1 $\therefore$ h = $\frac{\sqrt{3}}{\sqrt{3}-1}=\frac{1}{2}(3+\sqrt{3})$ km

$\therefore$ The height of the aeroplane is $\frac{1}{2}$(3 + $\sqrt{3}$) km