Question

in triangle ABC Prove that; cos B=c^ 2 +a^ 2 -b^ 2 / 2a c​

Answer 1

Answer:

From the question

Consider that for ΔABC,∠B is in a standard position i.e. vertex B is at the origin and the side BC is along positive x-axis. As ∠B is an angle of a triangle ∠B can be acute or B ∠B can be obtuse.

Using the Cartesian co-ordinate system in both figure (1) and figure (2)

we get B ≡ (0,0)A ≡ (c cos B, c sin B) and C ≡ (a,0)

Now consider l(CA) =b

ℴ∴b2=(a-osB)2+(0-csinB)2 , by distance formula

b2=a2-2accosB+c2cos2B+c2sin2B

b2=a2-2accosB+c2(sin2B+cos2B)

b2=a2+c2-2accosB