Question

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

Answer 1

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 :)

Answer 2

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.`