Question Number 14438 by ajfour last updated on 31/May/17

Commented by ajfour last updated on 31/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 31/May/17

Commented by ajfour last updated on 01/Jun/17

Answered by ajfour last updated on 01/Jun/17
![cos C=((a^2 +b^2 −c^2 )/(2ab)), s=x+y+z, then s^2 =a^2 +b^2 +2abcos (C−(π/3)) x=((2a)/( (√3)))sin [cos^(−1) (((a^2 +s^2 −c^2 )/(2as)))] =((2a)/( (√3)))sin 𝛉 y=((2b)/( (√3)))sin [cos^(−1) (((b^2 +s^2 −a^2 )/(2bs)))] =((2b)/( (√3)))sin 𝛗 z=((2c)/( (√3)))sin [cos^(−1) (((c^2 +s^2 −b^2 )/(2cs)))] =((2c)/( (√3)))sin 𝛙 . .........................................](https://www.tinkutara.com/question/Q14461.png)
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Jun/17

Commented by ajfour last updated on 01/Jun/17

Commented by ajfour last updated on 01/Jun/17
