How to prove that this triangle is right-angled?

Letters a, b and c are for the angles of a triangle.

How to prove that

if #cot(a/2) = (sinb + sinc)/(sin a)#, than is the triangle right-angled

2 Answers
Dec 10, 2017

If you have all three side lengths, to be right angled the triangle must obey Pythagorus's theorem.

Explanation:

eg.
if triangle has side lengths of 3, 4 and 5;
#3^2+4^2=5^2#
#9+16=25#
Hence triangle is right angled.
If however, the triangle has side lengths of 3, 4 and 6;
#3^2+4^2!=6^2#
#9+16!=36#
and triangle is NOT right angled.

Hope it helps :)

Dec 10, 2017

Kindly refer to a Proof given in the Explanation.

Explanation:

Given #Delta# has #3# angles #a,b,c.#

# :. a+b+c=pi, or, b+c=pi-a.#

Also, we know that,

#sinb+sinc=2sin((b+c)/2)cos((b-c)/2).#

Now, #cot(a/2)=(sinb+sinc)/sina,#

#rArr cot(a/2)={2sin((b+c)/2)cos((b-c)/2)}/sina,#

#rArr cot(a/2)={2sin((pi-a)/2)cos((b-c)/2)}/sina,#

#={2sin(pi/2-a/2)cos((b-c)/2)}/sina,#

#={2cos(a/2)cos((b-c)/2)}/{2sin(a/2)cos(a/2)}.#

#:. cot(a/2)=cot(a/2){cos((b-c)/2)/cos(a/2)}, i.e.,#

#cot(a/2)-cot(a/2){cos((b-c)/2)/cos(a/2)}=0.#

#cot(a/2)[1-{cos((b-c)/2)/cos(a/2)}]=0.#

#:. cot(a/2)=0, or, 1-{cos((b-c)/2)/cos(a/2)}=0.#

#:. cot(a/2)=0, or, 1={cos((b-c)/2)/cos(a/2)}, i.e.,#

#:. cot(a/2)=0, or, cos((b-c)/2)=cos(a/2),#

#:. cos(a/2)/sin(a/2)=0, or, cos((b-c)/2)=cos(a/2),#

#:. cos(a/2)=0=cos(pi/2), or, cos((b-c)/2)=cos(a/2).#

Taking into a/c that #a,b,c# are angles of a #Delta,# we have,

#:. a/2=pi/2, or, (b-c)/2=a/2.#

#:. a=pi,# which is not possible, or, #b-c=a.#

# b-c=a, &, b+c=pi-a rArr 2b=pi, or, b=pi/2.#

This proves that #Delta# is right-angled at #/_b.#