Question

If in a triangle $ABC,b{\cos }^2\frac{A}{2}+a{\cos }^2\frac{B}{2}=\frac{3}{2}$, then $a,c,b$ are

• A. in A.P
• B. in G.P
• C. in H.P
• D. None

Answer 1

The correct option is A in A.P

We have;

In $\Delta ABC;$

$\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}};\cos \frac{B}{2}=\sqrt{\frac{s(s-b)}{ac}}$

$b{\cos }^2\frac{A}{2}+a{\cos }^2\frac{B}{2}=\frac{3}{2}$

$⟹b\times \frac{s(s-a)}{bc}+a\times \frac{s(s-b)}{ac}=\frac{3}{2}$

$⟹a+b.\frac{a+b+c}{2}\frac{b+c-a}{2}+\frac{a+b+c}{2}=\frac{3}{2}C$

$⟹(b+c{)}^2-a^2+(a+c{)}^2-b^2=6C$

$⟹b^2+c^2+2bc-a^2+a^2+c^2+2ac-b^2=6C$

$⟹2c^2+2c(a+b)=6c$

$⟹C[c+(a+b\left)\right]=3c⟹C[c+(a+b)-3]=0$

$⟹C\ne 0;C=3-(a+b)$

Hence;

$C=3-(a+b)$