Question

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

Answer 1

Since A,B and C are A.P

$A+C=2B...\left(1\right)$

but $A+B+C=\pi$

$2B+B=\pi$

$B=\pi /3$

$A+C=\pi -B=\pi -\pi /3$

$A+C=2\pi /3$

$C=\frac{2\pi }{3}-A$

By sine rule :- $\frac{sinB}{b}=\frac{sinC}{c}$

$\frac{sinB}{b}=\frac{sin(2\frac{\pi }{3}-A)}{c}$

$sin(\frac{2\pi }{3}-A)=\frac{c}{b}+sinB.$

$sin(\frac{2\pi }{3}-A)=\frac{\sqrt{2}}{\sqrt{3}}.sin\frac{\pi }{3}$

$[\because b:c=\sqrt{3}:\sqrt{2}]$

$sin(\frac{2\pi }{3}-A)=\frac{\sqrt{2}}{\sqrt{3}}\times \frac{\sqrt{3}}{2}=\frac{1}{\sqrt{2}}$

$=\frac{1}{\sqrt{2}}=\pi /4$

$\frac{2\pi }{3}-A=\pi /4$

$A=\frac{2\pi }{3}-\frac{\pi }{4}=\frac{5\pi }{12}$

$C=\frac{2\pi }{3}-\frac{5\pi }{12}=\frac{\pi }{12}.$