Question

lf the sides of a triangle are in $A.P$ and difference between the greatest and least angles is $\frac{\pi }{2}$, then Sin $(\frac{B}{2})=$

• A. $\frac{1}{2}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is B $\frac{1}{2\sqrt{2}}$

Since the sides are in A.P

Therefore $b=\frac{a+c}{2}$

$\frac{b}{sinB}.sinB=\frac{1}{2}[\frac{a}{sinA}.sinA+\frac{c}{sinC}.sinC]$

Now, using sine rule, give us

$\frac{b}{sinB}=\frac{a}{sinA}=\frac{c}{sinC}=k$

Therefore

$ksinB=\frac{k(sinA+sinC)}{2}$

$sinA+sinC=2sinB$ ...(i)

Let A be the greatest Angle.

Therefore C will be the Smallest angle.

Hence $A-C={90}^0$ ...(ii)

Now $sinA+sinC=2sinB$

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

$sin({90}^0-\frac{B}{2})cos\left({45}^0\right)=2sin\frac{B}{2}cos\frac{B}{2}$

$cos\left(\frac{B}{2}\right).\frac{1}{\sqrt{2}}=2sin\frac{B}{2}cos\frac{B}{2}$

$\frac{1}{\sqrt{2}}=2sin\frac{B}{2}$

$sin\frac{B}{2}=\frac{1}{2\sqrt{2}}$.