Question

Mark the correct alternative in each of the following :

In the sides of a triangle are in the ratio $1:\sqrt{3}:2,$ then the measure of its greatest angle is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{2\pi }{3}$

Answer 1

The correct option is C

$\frac{\pi }{2}$

$Leta=1,b=\sqrt{3},c=2cosA=\frac{b^2+c^2-a^2}{2bc}\Rightarrow cosA=\frac{{\left(\sqrt{3}\right)}^2+{\left(2\right)}^2-{\left(1\right)}^2}{2\times \sqrt{3}\times 2}\Rightarrow cosA=\frac{3+4-1}{4\sqrt{3}}\Rightarrow cosA=\frac{6}{4\sqrt{3}}\Rightarrow cosA=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}cosC=\frac{a^2+b^2-c^2}{2ab}\Rightarrow cosC=\frac{1^2+{\left(\sqrt{3}\right)}^2-{\left(2\right)}^2}{2\times 1\times \sqrt{3}}\Rightarrow cosC=\frac{1+3-4}{2\sqrt{3}}=0\therefore \text{The greaer angle is}cosC=0=cos{90}^{\circ }=\frac{\pi }{2}$