Question

If the cosines of the two of the angles of a triangle are proportional to the oppsite sides show that the trangle is right angled.?

Answer 1

Such `triangle`s are isosceles, inclusive of isosceles right angled triangles.

Explanation:

`a / b = sin A/sin B = cos A/cos B`, giving

#sin ( A - B ) = sinA cos B - cos A sin B = 0.So.

`A = B, and so, A + B + C = 2A + C = pi`

`rArr` the`triangle` is isosceles that also could be right angled.

Answer 2

If we consider that the cosines of the two of the angles of a triangle are inversely proportional to the opposite sides then we can get the expected result. Let for `DeltaABC`

`acosA=bcosB`

`=>2RsinAcosA=2RsinBcosB`

`=>sin2A=sin2B`

`=>sin2A=sin(pi-2A)`

`=>2A=pi-2A`

`=>A+B=pi/2`

Hence `C=pi/2`. This means the triangle is right angled when cosine of two angles are inversely proportional to opposite sides.