Question

Mark the correct alternative in each of the following:

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

(a) $\frac{\mathrm{\pi }}{6}$ (b) $\frac{\mathrm{\pi }}{3}$ (c) $\frac{\mathrm{\pi }}{2}$ (d) $\frac{2\mathrm{\pi }}{3}$

Answer 1

Let ∆ABC be the given triangle such that its sides are in the ratio $1:\sqrt{3}:2$.

$\therefore a=k,b=\sqrt{3}k,c=2k$

Now, $a^2+b^2=k^2+3k^2=4k^2=c^2$

So, ∆ABC is a right triangle right angled at C.

$\therefore C=90°$

Using sine rule, we have

$$\begin{gathered} \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ \Rightarrow \frac{k}{\sin A}=\frac{\sqrt{3}k}{\sin B}=\frac{2k}{\sin 90°} \\ \Rightarrow \sin A=\frac{1}{2}\mathrm{and}\sin B=\frac{\sqrt{3}}{2} \\ \Rightarrow A=30°\mathrm{and}B=60° \end{gathered}$$

Thus, the measure of its greatest angle is $\frac{\mathrm{\pi }}{2}$.

Hence, the correct answer is option (c).