The angles A, B and C of a triangle ABC are in A.P.

Question

The angles A, B and C of a triangle ABC are in A.P. and a : b = 1 : $$\sqrt 3 $$. If c = 4 cm, then the area (in sq. cm)
of this triangle is :

Accepted Answer

From the question  2B = A + C \& A + B + C = $$\pi $$

$$ \Rightarrow $$ 3B = $$\pi $$

$$ \Rightarrow $$ B = $${\pi  \over 3}$$

$$ \therefore A + C = {{2\pi } \over 3}\sigma $$

$${a \over b} = {1 \over {\sqrt 3 }}$$

$${{2R\sin A} \over {2R\sin B}} = {1 \over {\sqrt 3 }}$$

sin A = $${1 \over 2}$$

$$ \therefore $$ A = 30o

$$ \therefore $$ a = 2, b = 2$$\sqrt 3$$, c = 4

$$\Delta  = {1 \over 2} \times 2\sqrt 3  \times 2 = 2\sqrt 3 $$

Solutions · 1