Question

In a $∆ABC,b=\surd 3,c=1\text{and}\angle A=30°,$ then the largest angle of the triangle is

• A. $60°$
• B. $135°$
• C. $90°$
• D. $120°$

Answer 1

The correct option is C

$90°$

Determining the largest angle

Given, $b=\surd 3,c=1\text{and}\angle A=30°.$

Step 1: Determine the value of $a$

By the Cosine rule, we know that :

$\cos A=\frac{b^2+c^2-a^2}{2bc}...\left(1\right)$

Substituting the values in $\left(1\right)$.

$$\begin{gathered} \Rightarrow \cos (30°)=\frac{{\left(\sqrt{3}\right)}^2+1^2-a^2}{2\times \sqrt{3}\times 1} \\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{{\left(\sqrt{3}\right)}^2+1^2-a^2}{2\times \sqrt{3}\times 1}[\because \cos 30°=\frac{\sqrt{3}}{2}] \\ \Rightarrow \frac{\sqrt{3}\times \sqrt{3}\times 2}{2}=3+1-a^2 \\ \Rightarrow a^2=4-3 \\ \Rightarrow a^2=1 \\ \Rightarrow a=1 \end{gathered}$$

Step 2: Determine the value of angles

From the Sine rule, we know that :

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}...\left(2\right)$

Substituting the obtained and given value in the equation $\left(2\right)$.

$$\begin{gathered} \Rightarrow \frac{1}{\sin 30°}=\frac{\sqrt{3}}{\sin B} \\ \Rightarrow \sin B=\frac{\sqrt{3}\times \sin 30°}{1} \\ \Rightarrow \sin B=\sqrt{3}\times \frac{1}{2} \\ \Rightarrow \sin B=\frac{\sqrt{3}}{2} \\ \therefore B=60°[\because \sin 60°=\frac{\sqrt{3}}{2}] \end{gathered}$$

We know that the sum of interior angles of a triangle is equal $180°$, thus, $\angle A+\angle B+\angle C=180°$

$$\begin{gathered} \therefore \angle C=180°-\angle A-\angle B \\ \Rightarrow =180°-30°-60° \\ \Rightarrow \angle C=90° \end{gathered}$$

Step 3: Comparing the angles to obtain the largest angle in the triangles.

$\Rightarrow \angle A=30°,\angle B=60°,\angle C=90°$

Therefore, $\angle C=90°$ is the largest angle in the triangle.

Thus, option (C) is correct.