Question

If the angles A, B, C of a triangle are in A.P.
and sides a, b, c are in G.P, then a?,b2,c2
are in
a. AP
b.HP
c.GP
d.none​

Answer 1

Question:

If the angles A, B, C of a triangle are in A.P. and sides a, b, c are in G.P, then

$a {}^{2} , b {}^{2} , c {}^{2}$

are in?

Theory :

⇒ if a,b,c are in ap

then , 2b= a+c

⇒if $a {}^{2} , b {}^{2} , c {}^{2}$

are in Gp then ,

$b {}^{2} = ac$

${\purple{\boxed{\large{\bold{Cosine \: Formula }}}}}$

Theorem :

In any ∆ABC,

$1)a {}^{2} = b {}^{2} + c {}^{2} - 2bc \cos(A)$

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

$3)c {}^{2} = a {}^{2} + b {}^{2} - 2ab \cos(C)$

Given 3 sides but no angle, this form is more convenient:

$\cos(A) = \frac{b{}^{2} + c {}^{2} - a{}^{2} }{2bc}$

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

$\cos(C) = \frac{a {}^{2} + b{}^{2} - c{}^{2} }{2ab}$

Solution :

it is given that ∠A, ∠B and ∠C are in Ap

⇒∠2B = ∠ A+ ∠C

We know that :

sum of all angles in a triangle = 180°

∠ A + ∠B +∠C = 180°

⇒∠2B +∠B = 180°

⇒ 3∠B = 180°

⇒∠B = 60°

$a {}^{2} , b {}^{2} , c {}^{2}$ are in Gp

⇒$b {}^{2} = ac$

Now apply cosine rule;

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

$\cos(60) = \frac{a {}^{2} + c {}^{2} - b {}^{2} }{2ac}$

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

$ac = a {}^{2} + c{}^{2} - b {}^{2}$

put $ac =b {}^{2}$

_____________________

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

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

⇒ this is the condition for AP

therefore $a {}^{2} , b {}^{2} , c {}^{2}$ are in AP

correct option a) AP