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Question: If the sides of a triangle ABC are in A.P. and a is the smallest side, then cos A equals...

If the sides of a triangle ABC are in A.P. and a is the smallest side, then cos A equals

A

3c−4b2c\frac{3c - 4b}{2c}

B

3c−4b2b\frac{3c - 4b}{2b}

C

4c−3b2c\frac{4c - 3b}{2c}

D

None of these

Answer

4c−3b2c\frac{4c - 3b}{2c}

Explanation

Solution

Since the sides of the triangle are in A.P

i.e. a, b, c are in A. P. and let a < b < c, 2b = a + c.

Now cos A = b2+c2−a22bc\frac{b^{2} + c^{2} - a^{2}}{2bc} =b2+c2−(2b−c)22bc\frac{b^{2} + c^{2} - (2b - c)^{2}}{2bc}

= b2+c2−4b2−c2+4bc2bc=4bc−3b22bc\frac{b^{2} + c^{2} - 4b^{2} - c^{2} + 4bc}{2bc} = \frac{4bc - 3b^{2}}{2bc}= 4c−3b2c\frac{4c - 3b}{2c}