Question

If in a $△ABC$ the angles $A,B,C$ are in A.P.

Then $\frac{a+c}{\sqrt{a^2-ac+c^2}}=?$

• A. $2\sin \left(\frac{A-C}{2}\right)$
• B. $\sin \left(\frac{A-C}{2}\right)$
• C. $2\cos \left(\frac{A-C}{2}\right)$
• D. $\cos \left(\frac{A-C}{2}\right)$

Answer 1

Answer: (C)

$2\cos \left(\frac{A-C}{2}\right)$

As angles $A,B,C$ are in $A.P.,$

$\therefore B=\frac{A+C}{2}$ $\Rightarrow A+C=2B$

Adding $B$ to both sides

$A+B+C=3B$$\Rightarrow 3B={180}^0$ $\Rightarrow B={60}^0$

$\therefore \cos B=\cos {60}^0$$\Rightarrow \frac{c^2+a^2-b^2}{2ca}=\frac{1}{2}$ [Using cosine formula]

$\Rightarrow c^2+a^2-b^2=ca\Rightarrow a^2-ac+c^2=b^2$

$\therefore \frac{a+c}{\sqrt{a^2+ac+c^2}}=\frac{a+c}{\sqrt{b^2}}=\frac{a+c}{b}$

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

$=\frac{\sin ({90}^0-\frac{B}{2})\cos \frac{A-C}{2}}{\sin \frac{B}{2}\cos \frac{B}{2}}=\frac{\cos \frac{B}{2}\cos \frac{A-C}{2}}{\sin {30}^0\cos \frac{B}{2}}$ $[\because B={60}^0]$

$=\frac{\cos \frac{A-C}{2}}{\frac{1}{2}}=2\cos \frac{A-C}{2}.$