Question

If one side of a triangle is double the other and the angles opposite to these sides differ by $60°$, then the triangle is

• A. Obtuse-angled
• B. Acute-angled
• C. Isosceles
• D. Right-angled

Answer 1

The correct option is D

Right-angled

Explanation for the correct option.

Step 1: Information required for the solution

Let one side of the triangle be $a$ then the other side will be $2a$.

Suppose the angle opposite to the side $a$ be $\theta$ then the angle opposite to the side $2a$ will be $\left(\theta -60°\right)$.

Now, the sine rule is given by when three sides are $A,B,\text{and}C$ with the angles opposite to these sides are $\alpha ,\beta ,\text{and}\gamma$ respectively

$\frac{\mathbf{A}}{\mathbf{s}\mathbf{i}\mathbf{n}\left(\alpha \right)}\mathbf{=}\frac{\mathbf{B}}{\mathbf{s}\mathbf{i}\mathbf{n}\left(\beta \right)}\mathbf{=}\frac{\mathbf{C}}{\mathbf{s}\mathbf{i}\mathbf{n}\left(\gamma \right)}$

Step 2: Determination of the type of a triangle

Apply this rule for the assumed parameters,

$\begin{array}{rcl}\therefore \frac{a}{\sin \left(\theta -60°\right)}& =& \frac{2a}{\sin \left(\theta \right)}\\ & \Rightarrow & \frac{a}{2a}=\frac{\sin \left(\theta -60°\right)}{\sin \left(\theta \right)}\end{array}$

The numerator of the RHS can be expanded by applying the trigonometric identity $\mathit{s}\mathit{i}\mathit{n}\left(A-B\right)\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\left(A\right)\mathit{c}\mathit{o}\mathit{s}\left(B\right)\mathbf{-}\mathit{s}\mathit{i}\mathit{n}\left(B\right)\mathit{c}\mathit{o}\mathit{s}\left(A\right)$,

$\begin{array}{rcl}\therefore \frac{1}{2}& =& \frac{\sin \left(\theta \right)\cos \left(60°\right)-\sin \left(60°\right)\cos \left(\theta \right)}{\sin \left(\theta \right)}\\ & \Rightarrow & \frac{1}{2}=\frac{\sin \left(\theta \right)\cos \left(60°\right)}{\sin \left(\theta \right)}-\frac{\sin \left(60°\right)\cos \left(\theta \right)}{\sin \left(\theta \right)}\\ & \Rightarrow & \frac{1}{2}=\cos \left(60°\right)-\sin \left(60°\right)\cot \left(\theta \right)\\ & \Rightarrow & \frac{1}{2}=\frac{1}{2}-\frac{\sqrt{3}}{2}\cot \left(\theta \right)\\ & \Rightarrow & 1=1-\sqrt{3}\cot \left(\theta \right)\\ & \Rightarrow & \cot \left(\theta \right)=0\\ & \Rightarrow & \theta =\frac{\pi }{2}\end{array}$

This proves that one angle is a right angle and the two angles will be $30°\text{and}60°$. Therefore, the triangle is right-angled.

Hence, the correct option is (D).