Question

If the angles A, B, and C of a triangle ABC are in AP and the sides a, b and c opposite to these angles are in GP; then a${}^2$, b${}^2$ and c${}^2$ are related as

• A. $b^2=a^2{c}^2$
• B. $b^2=a^2+c^2$
• C. $2b^2=a^2+c^2$
• D. $\text{none}$

Answer 1

Answer: (C)

Since, $A,B$ and $C$ are in $AP$

$2B-A+C$

$A+B+C={180}^o$

$2B+B=180$ $B={60}^o$

$a,b,c$ are in $GP$ then

$b^2=ac$

$\cos B=\frac{c^2+a^2-b^2}{2ac}$

$\Rightarrow \cos {60}^o=\frac{c^2+a^2-b^2}{2b^2}$

$\frac{1}{2}=\frac{c^2+a^2-b^2}{2b^2}$

$c^2+a^2-b^2=b^2$

$2b^2=c^2+a^2$