Math Tutor Online – In any triangle ABC, prove that

- A cosA + b cosB + c cosC = 4RsinA sinB sinC
- sinA + sinB + sinC = S/R

Let’s see the solution of the following

- lets first have left hand side equation which is

A cosA + b cosB + c cosC

= 2R sin A cosA + 2R sin B cos B + 2R sinC cos C

This is because as we know a = 2R sinA, b=2RsinB, c = 2RsinC

So now taking R common and converting into formula that is 2 sin A cosA = sin2A

= R (sin2A + sin2B + sin2C)

Again by applying formula

= R (2sin (A+B) cos(A-B) + 2 sinC cos C )

= R (2sin (π –C)cos(A-B)+ 2 sinC cos C )

= R (2sinC cos(A-B)+ 2 sinC cos C )

Now taking 2 sin C as common we have

= 2RsinC (cos(A-B)+ cos C )

= 2RsinC (cos(A-B)+ cos(π – (A+B ))

= 2RsinC (cos(A-B)- cos (A+B ))

= 2RsinC (2sinA sin B)

= 4R sin A sin B sin C = right hand side

Hence proved

- sinA + sinB + sinC = S/R

here we should know things like

R = a/2sinA= b/2sinB = c/2sinC

sinA = a/2R, sinB = b/2R, sinC= c/2R

Just by putting these values in left hand side we get

a/2R +b/2R+c/2R = a+b+c/2R = 2S/2R = S/R = right hand side

Hence proved.

It’s not necessary that perpendicular bisector passes through vertex. It is tge case of equilateral or isosceles triangle which is very easy to prove. Prove it using scalene triangle.