Question

In a triangle ABC, if acosA=bcosB, show that the triangle is either isosceles or right angled.

Answer 1

here is u r answer

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

hope understand

Answer 2

Hey !!

Given,

      a cos A = b cos B

=> 2R sin A cos A = 2R sin B cos B

=> sin 2A = sin 2B (or) sin (180 - 2B)

=> 2A = 2B (or) 2A = 180 - 2B

=> A = B (or) A = 90 - B

=> A = B (or) A + B = 90°

=> A = B (or) C = 90°

∴ The triangle is isosceles (or) right-angled.

GOOD LUCK !!