Question

Observe the following Statements:
$I)$ In $\Delta ABC,b{\cos }^2\frac{C}{2}+c{\cos }^2\frac{B}{2}=s$
$II)$ In $\Delta ABC,\cot \frac{A}{2}=\frac{b+c}{a}\Rightarrow$Triangle is right angled.
Which of the following is correct :

• A. Both $I$ and $II$ are correct.
• B. $I$ is true, $II$ is false.
• C. $I$ is false, $II$ is true.
• D. Both $I$ and $II$ are false.

Answer 1

The correct option is A Both $I$ and $II$ are correct.

Solving statement $I$
We know, ${\cos }^2\frac{C}{2}=\frac{s(s-c)}{ab}$ $\text{and}{\cos }^2\frac{B}{2}=\frac{s(s-b)}{ca}$
$\Rightarrow b{\cos }^2\frac{C}{2}+c{\cos }^2\frac{B}{2}=b\times \frac{s(s-c)}{ab}+c\times \frac{s(s-b)}{ac}=\frac{s(s-c)}{a}+\frac{s(s-b)}{a}$
$\Rightarrow b{\cos }^2\frac{C}{2}+c{\cos }^2\frac{B}{2}=\frac{s(s-c)}{a}+\frac{s(s-b)}{a}=\frac{s(2s-b-c)}{a}=\frac{s\times a}{a}$
$\Rightarrow b{\cos }^2\frac{C}{2}+c{\cos }^2\frac{B}{2}=s$
Solving statement $II$
Now, $\cot \frac{A}{2}=\frac{b+c}{a}$

$\Rightarrow \frac{\cos A/2}{\sin A/2}=\frac{\sin B+\sin C}{\sin A}$

$=\frac{2\sin \left(\frac{B+C}{2}\right)\cdot \cos \left(\frac{B-C}{2}\right)}{2\sin \frac{A}{2}\cdot \cos \frac{A}{2}}$

$=\frac{2\cos \frac{A}{2}\cdot \cos \frac{B-C}{2}}{2\sin \frac{A}{2}\cdot \cos \frac{A}{2}}$

$\Rightarrow \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\cos \frac{B-C}{2}}{\sin \frac{A}{2}}$

$\Rightarrow \cos \frac{A}{2}-\cos \frac{B-C}{2}=0$

$\Rightarrow 2\sin \frac{B-C+A}{4}\cdot \sin \frac{B-C-A}{4}=0$

either $B-C+A=0$ or $B-C-A=0\cdots \left(i\right)$

Also we know, $A+B+C=\pi \cdots \left(ii\right)$
Solving equations $\left(i\right),\left(ii\right)$, we get triangle is right angled at either $B$ or $C$
Clearly, both the statements $I,II$are correct.