Question

triangle ABC has BC=1,and

Answer 1

\[\sin^{23}\frac{ A }{ 2 }\cos^{48}\frac{ B }{ 2 }=\sin^{23} \frac{ B }{ 2 }\cos ^{48}\frac{ A }{ 2 }\]
find AC

Answer 2

startless-i dont know where to start

Answer 3

\[AB^2=1+AC^2-2(AC)\cos A\]
\[A+B+C=180\]
\[\frac{ \sin A }{ AB }=\frac{\sin B }{1}=\frac{ \sin C }{ AC }\]

Answer 4

@Callisto ,@dpaInc @ParthKohli

Answer 5

@dpaInc

Answer 6

Something's lacking, since the given trigonometry condition can be fulfilled by e.g. any isosceles triangle where ∠A = ∠B, hence AC is not determined.

Sorry for late response.