Search any question & find its solution
Question:
Answered & Verified by Expert
If the sides of a triangle are in the ratio $2: \sqrt{6}: 1+\sqrt{3}$, then what is the smallest angle of the triangle?
Options:
Solution:
2881 Upvotes
Verified Answer
The correct answer is:
$45^{\circ}$
Let $\mathrm{ABC}$ be a triangle with sides $\mathrm{a}=1+\sqrt{3}, \mathrm{~b}=2$ and
$c=\sqrt{6}$
So, $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{(1+\sqrt{3})^{2}+(\sqrt{6})^{2}-4}{2(1+\sqrt{3})(\sqrt{6})}$
$=\frac{2 \sqrt{3}+6}{2 \sqrt{6}+\sqrt{18}}=\frac{3+\sqrt{3}}{\sqrt{6}+3 \sqrt{2}}$
$=\frac{\sqrt{3} \sqrt{3}+\sqrt{3}}{(\sqrt{3}+\sqrt{3} \sqrt{3}) \sqrt{2}}=\frac{1}{\sqrt{2}}$
$\Rightarrow B=45^{\circ}$ is the smallest angle.
$(\because$ smallest side is $b=2$ )
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.