Question

Tower is observed from two stations $A$and $B$ where $B$ is East of $A$ at a distance $100m$. The tower is to north of $A$ and to North-West of $B$. The angles of elevation of the tower from $A$and $B$ are complementary. The height of the tower (in meters) is:

• A. $100$ $\sqrt{2}$
• B. $100(2{)}^{1/4}$
• C. $100(2{)}^{-1/4}$
• D. $100(2{)}^{1/3}$

Answer 1

The correct option is B $100(2{)}^{1/4}$

$h=AP\tan \theta ,h=BP\tan ({90}^{\circ }-\theta )$

So $h^2=AP.BP\cdots \left(1\right)$

Clearly $AP=AB=100$

$\therefore BP=\sqrt{{100}^2+{100}^2}=100\sqrt{2}$

Putting the values of $AP$ and $BP$ in $\left(1\right),$ we get

$h^2=100\times 100\sqrt{2}$

$⟹h=100(2{)}^{1/4}$