The angle of elevation of the top of a lighthouse 60 m high, from two points on the ground on its opposite sides are 45° and 60°. What is the distance between these two points?
A. 45 m B. 30 m C. 103.8 m D. 94.6 m Answer: Option DShow Answer
Solution(By Apex Team)
Let BD be the lighthouse and A and C be the two points on ground.
Then, BD, the height of the lighthouse = 60 m
$\begin{aligned}\angle BAD&=45^{\circ},\angle BCD=60^{\circ}\\
\tan45^{\circ}&=\frac{BD}{BA}\\
\Rightarrow1=&\frac{60}{BA}\\
\Rightarrow BA&=60\mathrm{~m}\ldots\ldots\ldots(\mathrm{i})\\
\tan60^{\circ}&=\frac{BD}{BC}\\
\Rightarrow\sqrt{3}&=\frac{60}{BC}\\
\Rightarrow BC&=\frac{60}{\sqrt{3}}\\
&=\frac{60\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}\\
&=\frac{60\sqrt{3}}{3}\\
&=20\sqrt{3}\\
&=20\times1.73\\
&=34.6\mathrm{~m}\ldots\ldots\ldots\text{ (ii) }\end{aligned}$
Distance between the two points A and C
= AC = BA + BC
= 60 + 34.6 [∵ Substituted value of BA and BC from (i) and (ii)]
= 94.6 m
$\begin{aligned}\angle BAD&=45^{\circ},\angle BCD=60^{\circ}\\
\tan45^{\circ}&=\frac{BD}{BA}\\
\Rightarrow1=&\frac{60}{BA}\\
\Rightarrow BA&=60\mathrm{~m}\ldots\ldots\ldots(\mathrm{i})\\
\tan60^{\circ}&=\frac{BD}{BC}\\
\Rightarrow\sqrt{3}&=\frac{60}{BC}\\
\Rightarrow BC&=\frac{60}{\sqrt{3}}\\
&=\frac{60\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}\\
&=\frac{60\sqrt{3}}{3}\\
&=20\sqrt{3}\\
&=20\times1.73\\
&=34.6\mathrm{~m}\ldots\ldots\ldots\text{ (ii) }\end{aligned}$
Distance between the two points A and C
= AC = BA + BC
= 60 + 34.6 [∵ Substituted value of BA and BC from (i) and (ii)]
= 94.6 m
