Question

Sides a , b & c of the triangle ABC are in arithmetic progression. Angle A is twice the angle C.

What's a+b+c if the sides are in the ratio a : b : c?

Answer 1

here is your answer OK sir.................... ☺☺☺

since the angle A , B and C of the triangle ABC is given to be A. B , therefore , we have....

A+ B + C = 180 and A + C = 2B . Solving these two equations, we get angle B = 60° . Now by use of Cosine formula, we can obtain,

b^2 = a^2 + c^2 - 2acCos(60°) = 4 +16-2 x 2 x 4 (1/3)

= 12 which in turn implies b = 2 sqrt(3) .

Answer 2

Answer:

Step-by-step explanation:

Here,sine rule is applied.

Also,we know:

sin2x=2sinxcosx

cos2x=2cos^2(x)-1

sinx+siny=2sin((x+y)/2)cos((x-y)/2)

Also, when a,b,c are in AP,

b-a=c-b

And sum of angles in a triangle is 180°.

The above concepts are applied,and we get the answer a+b+c=15