Question

In a $△ABC,$ the angles $A$ and $B$ are two different values of $\theta$ satisfying $\sqrt{3}\cos \theta +\sin \theta =k,\left|k\right|<2.$ The triangle:

• A. is an acute angled
• B. is aright angled
• C. is an obtuse angled
• D. has one angle $=\frac{\pi }{3}$

Answer 1

The correct option is C is an obtuse angled

$\sqrt{3}\cos \theta +\sin \theta =k,\left|k\right|<2$

$\Rightarrow 2\sin (\theta +\frac{\pi }{3})=k\Rightarrow \sin (\theta +\frac{\pi }{3})=\frac{k}{2}\in (-1,1)$ ...(1)

Now $A$ and $B$ are two different values of $\theta$ satisfying (1)

$\Rightarrow \sin (A+\frac{\pi }{3})=\sin (B+\frac{\pi }{3})=\frac{k}{2}$

$\Rightarrow \sin (A+\frac{\pi }{3})=\sin (B+\frac{\pi }{3})\Rightarrow A+\frac{\pi }{3}=n\pi +{(-1)}^n(B+\frac{\pi }{3})$

$\Rightarrow A+\frac{\pi }{3}=B+\frac{\pi }{3}$ at $n=0$

or $A+\frac{\pi }{3}=\pi -B-\frac{\pi }{3}$ at $n=1$

$\Rightarrow A=B$ or $A+B=\frac{\pi }{3}$

But $A$ and $B$ are different

$\Rightarrow A+B=\frac{\pi }{3}\Rightarrow C=\frac{2\pi }{3}$

$\Rightarrow △$ is obtuse angled