in triangle abc if acosA= bcosB then how to prove the triangle is isosceles or right angled

Answer 1

we know that a/sinA = b/sinB
if acosA = bcosB, then
a/cosB = b/cosA

b = acosA/cosB
b = asinB/sinA

so, acosA/cosB = asinB/sinA
cosAsinA = sinBcosB
sin2A = sin2B

means A=B and the triangle is isoceles

Answer 2

After sin2A=sin2B
Sin2A-sin2B=0
2cos(A+B)sin(A-B)=0
Cos(Pi-C)=0 or sin(A-B)=0
Angle C=pi/2 or A=B
Hence triangle. ABC is isosceles or right angled

Answer 3

not useful

Answer 4

First, according to the formula, a / Sina = B / SINB = C / sinc = 2R
The original formula sinacosa = sinbcosb
sin2A=sin2B
According to trigonometric function line
2A + 2B = 180 or 2A = 2B
A + B = 90
Or A=B