Question

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

• A. is acute angled
• B. is right angled
• C. is obtuse angled
• D. has one angle $=\pi /3$

Answer 1

Answer: (C)

is obtuse angled

Let $2(\frac{\sqrt{3}}{2}.\cos \theta +\frac{1}{2}.\sin \theta )$
$=2\sin (\theta +{60}^0)$ $=k$

Hence
$\sin (\theta +{60}^0)=\frac{k}{2}$

If $k=1$

we get
$\theta +{60}^0={30}^0,{150}^0$

$\theta =-{30}^0,{90}^0$

$\theta ={330}^0,{90}^0$

And if $k=\sqrt{3}$

We get,
$\theta +{60}^0={60}^0,{120}^0$

Hence
$\theta ={360}^0,{60}^0$

Hence it will always form an obtuse-angled triangle.