Question

$AB$ is vertical pole with $B$ at the ground level and $A$ at the top. A man find that the angle of elevation of the point $A$ from a certain point $C$ on the ground is ${60}^o$. He moves away from the pole along the line $BC$ to a point $D$ such that $CD=7m$. From $D$ the angle of elevation of the point $A$ is ${45}^o$. Then the height of the pole is:

• A. $\frac{7\sqrt{3}}{2}\frac{1}{\sqrt{3}-1}m$
• B. $\frac{7\sqrt{3}}{2}(\sqrt{3}+1)m$
• C. $\frac{7\sqrt{3}}{2}(\sqrt{3}-1)m$
• D. $\frac{7\sqrt{3}}{2}\frac{1}{\sqrt{3}+1}m$

Answer 1

The correct option is B $\frac{7\sqrt{3}}{2}(\sqrt{3}+1)m$

Then in triangle $ABC$

$tan{60}^{\circ }=\frac{AB}{BC}$

$x=\frac{h}{\sqrt{3}}$ ----- (i)

in $\Delta ABD$

$tan{45}^{\circ }=\frac{AB}{BD}$

$1=\frac{h}{x+7}$

$h=x+7$

$h=\frac{h}{\sqrt{3}}+7$ from (i)

$h=\frac{7\sqrt{3}}{\sqrt{3}-1}$

$h=\frac{7\sqrt{3}}{2}(\sqrt{3}+1)$

Therefore the height of the pole is $\frac{7\sqrt{3}}{2}(\sqrt{3}+1)m$