In triangle ABC, if cotA/2=b+c/a,then the triangle is
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Answer:
cot
2
A
=
a
b+c
sinA/2
cosA/2
=
sinA
sinBsinC
=
2sin
2
A
.cos
2
A
2sin(
2
B+C
).cos(
2
B−C
)
=
2sin
2
A
.cos
2
A
2cos
2
A
.cos
2
B−C
sin
2
A
cos
2
A
=
sin
2
A
cos
2
B−C
cos
2
A
−cos
2
B−C
=0
2sin
4
B−C+A
.sin
4
B−C−A
=0
either,nB−C+A=0 or B−C−A=0
∵A+B+C=π
Solving the above equation
The triangle is Right angled triangle
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