Angles $A,\ B,\ C$ of a $\vartriangle ABC$ are in AP and $b:c= \sqrt{3} :\sqrt{ 2}$ , then find the $\angle A$

Given: Angles $A,\ B,\ C$ of a $\vartriangle ABC$ are in AP and $b:c= \sqrt{3} :\sqrt{2}$ 

To do: To find the $\angle A$.

Solution:

Let the angles be $a,\ a+d,\ a+2d$.

Then, $a+a+d+a+2d=180$

$\Rightarrow 3( a+d)=180^o$

$\Rightarrow a+d=\frac{180}{3}$

$\Rightarrow a+d=60^o$

So angle $B=60^o$

Now, $\frac{b}{c}=\frac{sinB}{sinC}=\sqrt{\frac{3}{2}}$

$\Rightarrow\ sinC=\sqrt{\frac{2}{3}}sinB$

$\Rightarrow sinC=\sqrt{\frac{2}{3}}\times sin60^o$

$\Rightarrow sinC=\sqrt{\frac{2}{3}}\times\frac{\sqrt{3}}{2}$

$\Rightarrow sinC=\frac{1}{\sqrt{2}}$

$\Rightarrow sinC=sin45^o$

$\Rightarrow C=45^o$

$\therefore\ \angle A=180^o-\angle B-\angle C$

$\Rightarrow\ \angle A=180^o-60^o-45^o=75^o$

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Updated on: 10-Oct-2022

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