Question

$OAB$ is a triangle in the horizontal plane through the foot $P$ of the tower at the middle point of the side $OB$ of the triangle. If $OA=2m,OB=6m,AB=5m$ and $\angle AOB$ is equal to the angle subtended by the tower at $A$ then the height of the tower is:

• A. $\sqrt{\frac{11\times 39}{25\times 3}}$
• B. $\sqrt{\frac{11\times 39}{25\times 2}}$
• C. $\sqrt{\frac{11\times 25}{39\times 2}}$
• D. $\sqrt{\frac{11\times 25}{39\times 7}}$

Answer 1

The correct option is B $\sqrt{\frac{11\times 39}{25\times 2}}$

$\angle AOB=\theta ,\tan \theta =\frac{h}{AP}$
$cos\theta =\frac{2^2+6^2-5^2}{2\times 2\times 6}$
$=\frac{5}{8}$
length $=\frac{1}{2}\sqrt{2b^2+2c^2-a^2}$
$=\frac{1}{2}\sqrt{2\times 2^2+2\times 5^2-6^2}$
$AP=\frac{\sqrt{22}}{2}$
$\tan \theta =\frac{\sqrt{8^2-5^2}}{5}$
$\frac{\sqrt{39}}{5}=\frac{h}{\sqrt{\frac{11}{2}}}$
$h=\sqrt{\frac{11\times 39}{25\times 2}}$