Question

A lamp post is situated at the middle point $M$ of the side $AC$ of a triangular plot $ABC$ with $BC = 7\, m$, $CA = 8\, m$ and $AB = 9 \,m$. 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 $AB = 9 = c$,
$BC = 7 \,m = a$ and
$AC = 8\, m = b$.



$M$ 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
$cos\,C=\frac{a^{2}+b^{2}-c^{2}}{2ab}$
$=\frac{49+64-81}{2 \times 7 \times 8}=\frac{2}{7}\quad\ldots\left(i\right)$
Similarly using cosine formulae in $\Delta BMC$, we get
$BM^{2}=BC^{2}+CM^{2}-2BC \times CM\,cos\,C$.
Here $CM=\frac{1}{2}\,CA=4$, since $M$ is the mid-point of $AC$.
Therefore, using $\left(i\right)$, we get
$BM^{2}=49+16-2 \times 7 \times 4 \times \frac{2}{7}=49$
$\Rightarrow BM=7$
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 $h=7\left(2-\sqrt{3}\right)m$.