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)cot(a2)=sinb+sincsina, than is the triangle right-angled

2 Answers
Harold Walden
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=5^232+42=52
9+16=259+16=25
Hence triangle is right angled.
If however, the triangle has side lengths of 3, 4 and 6;
3^2+4^2!=6^232+42≠62
9+16!=369+16≠36
and triangle is NOT right angled.

Hope it helps :)

Ratnaker Mehta
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.

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