Question

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

Answer 1

Answer:

ANSWER

cosA=2bcb2+c2−a2

b=2c+a ∵ a,b,c are in A.P

a=(2b−c)

=2bcb2+c2−(4b2+c2−4bc)

=2bc4bc−3b2

=2c4c−3b.

MARK AS BRAINLIEST

Answer 2

Answer:

( 4 c - 3 b ) / 2 c

Step-by-step explanation:

Since all three sides of triangle are in A.P.

Let a < b < c

= > 2 b = a + c

a = 2 b - c ... ( i )

Now we have :

cos A = b² + c² - a² / 2 b c

From ( i ) we have value of a :

cos A = b² + c² - (2 b - c )² / 2 b c

cos A = ( b²  + c²  - 4 b² - c² + 4 b c ) / 2 b c

cos A = 4 b c - 3 b² / 2 b c

cos A = 4 c - 3 b / 2 c

Therefore , the value of cos A is  ( 4 c - 3 b ) / 2 c .