Question

The angle of elevation of a certain peak when observed, from each end of a horizontal base line of length 2a is found to be $\theta$. When observed from the mid-point of the base the angle of elevation is $\varphi$. Then the height of the peak is
$\frac{asin\theta sin\varphi }{\sqrt{\left(sin\right(\varphi +\theta \left)sin\right(\varphi -\theta \left)\right)}}$

• A. True
• B. False

Answer 1

The correct option is A True

Let$h$ be the height of the peak

From the figure

$AQ=BQ=h\cot \theta ,CQ=h\cot \varphi$

$CQ$ is perpendicular to the base $AB$ of a isosceless triangle

$⟹B{Q}^2=a^2+C{Q}^2$

$⟹h^2({\cot }^2\theta -{\cot }^2\varphi )=a^2$

$⟹h=\frac{a}{\sqrt{{\cot }^2\theta -{\cot }^2\varphi }}=\frac{a\sin \theta \sin \varphi }{\sqrt{{\cos }^2\theta {\sin }^2\varphi -{\cos }^2\varphi {\sin }^2\theta }}=\frac{a\sin \theta \sin \varphi }{\sqrt{\sin (\theta +\varphi )\sin (\theta -\varphi )}}$