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If $A, B$ and $C$ are in AP and $b: c=\sqrt{3}: \sqrt{2}$, then what is the value of sin C?
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{2}}$
Let $\mathrm{a}-\mathrm{d}$, a and $\mathrm{a}+\mathrm{d}$ be three numbers which are in
A.P. since $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are in $\mathrm{A} . \mathrm{P}$. $\therefore \quad A=a-d, B=a, C=a+d$
$\Rightarrow \quad a-d+a+a+d=180^{\circ}$
$(\because \mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are angles of a triangle $)$ $\Rightarrow \mathrm{a}=60^{\circ}$
$\Rightarrow A=60^{\circ}-d, B=60^{\circ}, C=60^{\circ}+d$
Now by sine rule,
$\frac{b}{c}=\frac{\sin B}{\sin C} \Rightarrow \frac{\sqrt{3}}{\sqrt{2}}=\frac{\sin 60^{\circ}}{\sin C}$
$\Rightarrow \quad \sin C=\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{\sqrt{3}}=\frac{1}{\sqrt{2}}$
A.P. since $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are in $\mathrm{A} . \mathrm{P}$. $\therefore \quad A=a-d, B=a, C=a+d$
$\Rightarrow \quad a-d+a+a+d=180^{\circ}$
$(\because \mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are angles of a triangle $)$ $\Rightarrow \mathrm{a}=60^{\circ}$
$\Rightarrow A=60^{\circ}-d, B=60^{\circ}, C=60^{\circ}+d$
Now by sine rule,
$\frac{b}{c}=\frac{\sin B}{\sin C} \Rightarrow \frac{\sqrt{3}}{\sqrt{2}}=\frac{\sin 60^{\circ}}{\sin C}$
$\Rightarrow \quad \sin C=\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{\sqrt{3}}=\frac{1}{\sqrt{2}}$
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