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 :
JEE · Math · previous-year question
- A.2$$\sqrt 3 $$correct
- B.4$$\sqrt 3 $$
- C.$${4 \over {\sqrt 3 }}$$
- D.$${2 \over {\sqrt 3 }}$$
Answer
A. 2$$\sqrt 3 $$
Explanation
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 $$\Delta = {1 \over 2} \times 2\sqrt 3 \times 2 = 2\sqrt 3 $$
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