# If cot A/2 = (b+c)/a , then the triangle ABC isA) isoscelesB) equilateralC) right angledD) none of these

Arun
25750 Points
6 years ago
(i) cot(A/2) = cos(A/2)/sin(A/2) [By cot relation]

ii) (b + c)/a = k{sin(B) + sin(C)}/ksin(A) [Application of sine law of triangle]

2*sin{(B+C)/2}*cos{(B-C)/2}/2*sin(A/2)...
[Application of sum-product law of triangles & multiple/sub multiple angle identities]

This simplifies as, (b + c)/a = {cos(A/2)*cos(B - C)/2}/sin(A/2)*cos(A/2)
[By angle sum property of triangle, A + B + C = 180°
==> (B + C)/2 = 90° - (A/2)
==> sin{(B + C)/2} = sin{90° - (A/2)} = cos(A/2)]

So, (b + c)/a = {cos(B - C)/2}/sin(A/2)

iii) From (i) & (ii) above, we have

cos(A/2)/sin(A/2) = {cos(B - C)/2}/sin(A/2)

==> cos(A/2) = cos(B - C)/2

==> cos(A/2) - cos(B - C)/2 = 0

So by sum-product of angle relations,

2*sin{(B - C + A)/4}*sin{(B - C - A)/4} = 0

==> Either of the product is zero.
So either (B - C + A) = 0 or (B - C - A) = 0 ------ (1)

But A + B + C = 180° [Angle sum property of triangle] --------- (2)

Solving the above 2 equations taking in pairs,
Either

Either case it is a Right triangl