Question

From the top of a tower, the angles of depression of two objects on the same side of the tower are found to be $\alpha$ and $\beta$ ($\alpha >\beta$). If the distance between the objects is ${}^{\prime }{p}^{\prime }$ metres, determine the height of the tower if $\alpha ={60}^o,\beta ={30}^o,p=150metre$

• A. $h=116.2m$
• B. $h=129.9m$
• C. $h=136.5m$
• D. None of these

Answer 1

The correct option is B $h=129.9m$

Let $OX$ be the horizontal ground and let $AB$ be the tower. Let $C$ and $D$ the given points of observation of hte ground. Then we have:
$\angle KAC=\alpha =\angle ACB$ and $\angle KAD=\beta =\angle ADB$
Where $\alpha >\beta$
$AB$ be the required height of the tower such that $AB=h$ metres
Let $CD=p$ metres and $BC=x$ metres
$BD=CD+BC=(p+x)m$
$ABD$ is right $△$ at $B$:
$\frac{AB}{BD}=\tan \beta$
$\frac{h}{p+x}=\tan \beta$
$p+x=\frac{h}{\tan \beta }$
$x=(\frac{h}{\tan \beta }-p)$......(1)
$ABC$ is right $△$ at $B$:
$\frac{AB}{BC}=\tan \alpha$
$\frac{h}{x}=\tan \alpha$
$x\tan \alpha =h$
$x=\frac{h}{\tan \alpha }......\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$\frac{h}{\tan \beta }-p=\frac{h}{\tan \alpha }$
Hence,
Height of the tower $h=\frac{p\tan \alpha \tan \beta }{\tan \alpha -\tan \beta }$
$=\frac{15\tan {60}^o\tan {30}^o}{\tan {60}^o-\tan {30}^o}=\frac{150.\sqrt{3}.\frac{1}{\sqrt{3}}}{\sqrt{3}-\frac{1}{\sqrt{3}}}m$
$=\frac{150}{\frac{3-1}{\sqrt{3}}}=\frac{150\sqrt{3}}{2}m=75\left(1.732\right)m=129.9m$