Question

In a triangle ABC, if A, B, C are in A.P. and b : c $=\sqrt{3}:\sqrt{2}$ then

Answer 1

As $A,B,C$ is in A.P
$A+C=2B$ and $A+B+C=180$
Solving them we get $B=60$
Now as $b:c=\sqrt{3}:\sqrt{2}\Rightarrow \sin C=\frac{c}{b}\sin B=\frac{\sqrt{2}}{\sqrt{3}}\times \frac{\sqrt{3}}{2}=\frac{1}{\sqrt{2}}$
$\Rightarrow C=45$ and $A=75$
A) $\cos (A-C)=\cos 30=\sin 60=\sin B$
B) $\sin (A-C)=\sin 30=\sin \frac{60}{2}=\sin \frac{B}{2}$
C) $\sin (A+C)=\sin 120=\sin 2B$
D) $\sin (2C-A)=\sin 15=\sin \frac{60}{4}=\sin \frac{B}{4}$