s =√3/4(a2+B2-C2)=√3/4 * 2 ABC OSC = ab *√3/2 cos = ab * sinC/2。

So √3cosC=sinC, square on both sides plus sin? C+cos？ C= 1, and the solution is sinC=60.

1.2, so a+b = 120. B =120-a. Substitute Sina+sinb = Sina+sin (120-a) = Sina+1/2 * Sina+√ 3/2 * COSA = 3/2 * Sina+√ 3 *.

2. 1

AC/AB=cosB/cosC

b/c=cosB/cosC

sinB/sinC=cosB/cosC

sinBcosC=sinCcosB

sinBcosC-sinCcosB=0

sin(B-C)=0

So B=C

2.2 Xhosa =1/3. cos (180-a) =-1/3 = cos (2b).

So cos(4B)=2cos? (2B)- 1=-7/9

sin(4B)=-4√2/9

Sin(4B+π/3) can be found.