Question

Three angles `A,B` and `C` of a triangle are in A.P. and `sin(2A+B)=sin(C-A)=-sin(B+C)=1/2`. Find the angles?

Answer 1

`A=45^@`, `B=60^@` and `C=75^@`

Explanation:

As `A,B` and `C` are in A.P., we have `2B=A+C`

As `sin(2A+B)=sin(C-A)=-sin(B+C)=1/2`

Hence, `C-A=30^@` and as `C+A=2B` adding the two

`2C=2B+30^@` or `C=B+15^@`

And as they are in A.P. `A=B-15^@`

and therefore `B-15^@+B+B+15^@=180^@`

i.e. `3B=180^@` or `B=60^@`

Hence `A=45^@` and `C=75^@`

Observe that `sin(2A+B)=sin150^@=1/2` and `-sin(B+2C)=-sin210^@=-(-1/2)=1/2`