Question

Two flagstaffs stand on a horizontal plane. $A$ and $B$ are two points on the line joining the bases of the flagstaffs and between them. The angle of elevation of the tops of the flagstaffs as seen from $A$ are ${30}^{\circ }$ and ${60}^{\circ }$ and , as seen from $B$, they are ${60}^{\circ }$ and ${45}^{\circ }$. If the length $AB$ be $30feet$, find the height of the flagstaffs and the distance between them.

Answer 1

$AP=x\cot {60}^o$

$\cot {60}^o=\frac{AP}{x}$

$\Rightarrow AP=x\cot {60}^o$

$AP=\frac{x}{\sqrt{3}}$

$AQ=y\cot {30}^o=y\sqrt{3}$

$BP=x\cot {45}^o=x$

$BQ=y\cot {60}^o=\frac{y}{\sqrt{3}}$

$AB=BP-AP=x-\frac{x}{\sqrt{3}}$

$AB=\frac{\sqrt{3}x-x}{\sqrt{3}}\Rightarrow 30=\frac{(\sqrt{3}-1)}{\sqrt{3}}$

$\Rightarrow 30\left(\sqrt{3}\right)=(\sqrt{3}-1)x$

$x=15(3+\sqrt{3})$

$AB=AQ-BQ=y(\sqrt{3}-\frac{1}{\sqrt{3}})\Rightarrow y=15\sqrt{3}$

$\therefore PQ=BP+BQ=x+y/\sqrt{3}$

$=15(3+\sqrt{3})+15$

$PQ=60+15\sqrt{3}$