Question

In any triangle ABC with usual notations prove c = a cos B + b cos A

Answer 1

By cosine rule, we have b² = c² + a² - 2ca cos B

`therefore cosB=(c^2+a^2-b^2)/2ac`

Similarly `cos A=(b^2+c^2-a^2)/(2bc)`

R.H.S. = a cos B + b cos A

=`a.(c^2+a^2-b^2)/(2ac)+b.(b^2+c^2-a^2)/(2bc)`

=`(c^2+a^2-b^2)/(2c)+(b^2+c^2-a^2)/(2c)`

=`(c^2+a^2-b^2+b^2+c^2-a^2)/(2c)`

=`(2c^2)/(2c)`

=c

=L.H.s

∴ c = a cos B + b cso A