#### Question

In any ΔABC, with usual notations, prove that `b^2=c^2+a^2-2ca cosB`.

#### Solution

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

`therefore b^2=(a-ccosB)^2+(0-csinB)^2` , by distance formula

`b^2=a^2-2ac cosB+c^2cos^2B+c^2sin^2B`

`b^2=a^2-2ac cosB+c^2(sin^2B+cos^2B)`

`b^2=a^2+c^2-2ac cosB`

Is there an error in this question or solution?

#### APPEARS IN

Solution In any ΔABC, with usual notations, prove that b2=c2+a2−2cacosB Concept: Trigonometric Functions - Solution of a Triangle.