Question

A lamp post is situated at the middle point $M$ of the side $AC$ of a triangular plot $ABC$ with $BC=7m$, $CA=8m$ and $AB=9m$. Lamp post subtends an angle $15^{\circ }$ at the point $B$. Determine the height of the lamp post.

• A. $5\left(2-\sqrt{3}\right)m$
• B. $7\left(2-\sqrt{3}\right)m$
• C. $\left(2-\sqrt{3}\right)m$
• D. $4\left(2-\sqrt{3}\right)m$

Answer 1

Answer: (B)

From fig., we have $AB=9=c$,
$BC=7m=a$ and
$AC=8m=b$.



$M$ is the mid-point of the side $AC$ at which lamp post $MP$ of height $h$ (say) is located. Again, it is given that lamp post subtends an angle $\theta$ (say) at $B$ which is $15^{\circ }$.
Applying cosine formulae in $\Delta ABC$, we have
$cosC=\frac{a^2+b^2-c^2}{2ab}$
$=\frac{49+64-81}{2\times 7\times 8}=\frac{2}{7}\dots \left(i\right)$
Similarly using cosine formulae in $\Delta BMC$, we get
$B{M}^2=B{C}^2+C{M}^2-2BC\times CMcosC$.
Here $CM=\frac{1}{2}CA=4$, since $M$ is the mid-point of $AC$.
Therefore, using $\left(i\right)$, we get
$B{M}^2=49+16-2\times 7\times 4\times \frac{2}{7}=49$
$\Rightarrow BM=7$
Thus, from $\Delta BMP$ right angled at $M$, we have
$tan\theta =\frac{PM}{BM}=\frac{h}{7}$
or $\frac{h}{7}=tan\left(15^{\circ }\right)=2-\sqrt{3}$
or $h=7\left(2-\sqrt{3}\right)m$.