in triangle abc if acosA= bcosB then how to prove the triangle is isosceles or right angled
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
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
not useful
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
To prove that triangle ABC is isosceles or right-angled using the given equation a*cos(A) = b*cos(B), we can use the properties of trigonometric ratios and the laws of cosines. Here's how you can approach the proof:
1. Start by assuming that triangle ABC is not isosceles.
- If triangle ABC is not isosceles, then sides AB and AC are not equal in length (a ≠ b) and angles A and B are not equal.
- Suppose a = b. This will lead us to the conclusion that triangle ABC is isosceles.
2. Use the law of cosines to express sides AB and AC in terms of side BC and the angles of the triangle.
- According to the law of cosines, a² = b² + c² - 2bc * cos(A).
- Similarly, b² = a² + c² - 2ac * cos(B).
3. Substitute the given equation into the law of cosines expressions.
- Substitute a * cos(A) for b * cos(B) in the second equation to get b² = a² + c² - 2ac * cos(B).
- Simplify the equation to obtain a² - b² = 2ac(cos(B) - cos(A)).
4. Rearrange the equation.
- Factor out a² - b² = 2ac(cos(B) - cos(A)) to obtain (a + b)(a - b) = 2ac(cos(B) - cos(A)).
5. Analyze the equation:
- If a + b = 0, then a = -b, and triangle ABC is isosceles.
- If a - b = 0, then a = b, and triangle ABC is isosceles.
- If cos(A) = cos(B), then angle A = angle B, and triangle ABC is isosceles.
- If cos(A) = cos(B) = 0, then angle A = angle B = 90 degrees, and triangle ABC is right-angled.
Thus, you can conclude that if a*cos(A) = b*cos(B) holds true, then triangle ABC is either isosceles or right-angled.