Question

In $△ABC$, with usual notations, if $a,b,c$ are in A.P then $a{\cos }^2\left(\frac{C}{2}\right)+c{\cos }^2\left(\frac{A}{2}\right)=$

• A. $3\frac{a}{2}$
• B. $3\frac{c}{2}$
• C. $3\frac{b}{2}$
• D. $\frac{3abc}{2}$

Answer 1

The correct option is C $3\frac{b}{2}$

Given the lengths of the sides $a,b,c$ are in $A.P$,

$⟹2b=a+c$,

The value of the semiperimeter of the triangle $s=\frac{3b}{2}$,

$\therefore$ the value of ${\cos }^2\frac{A}{2}=\frac{s(s-a)}{bc}=\frac{3(\frac{3b}{2}-a)}{2c}$

Similarly the value of ${\cos }^2\frac{C}{2}=\frac{s(s-c)}{ab}=\frac{3(\frac{3b}{2}-c)}{2a}{\cos }^2\frac{C}{2}=\frac{s(s-c)}{ab}=\frac{3(\frac{3b}{2}-c)}{2a}$

$a{\cos }^2\frac{C}{2}+c{\cos }^2\frac{A}{2}=\frac{3}{2}(3b-a-c)=\frac{3}{2}\left(b\right)$