Question

Let the angles $A,B,C$ of a $△ABC$ be in A.P and let $B:C=4:3$.Find the angle $A$

Answer 1

Let the angles be $a-d,a,a+d$

$\Rightarrow a-d+a+a+d={180}^{\circ }$

$\Rightarrow 3a={180}^{\circ }$

$\Rightarrow a=\frac{{180}^{\circ }}{3}={60}^{\circ }$

Let $C=a-d$

$B:C=4:3$

$\Rightarrow \frac{B}{C}=\frac{4}{3}$

$\Rightarrow \frac{60}{60-d}=\frac{4}{3}$

$\Rightarrow \frac{15}{60-d}=\frac{1}{3}$

$\Rightarrow 60-d=45$

$\Rightarrow d=60-45=15$

$\therefore a={60}^{\circ },d={15}^{\circ }$

$\therefore \angle A=a+d={60}^{\circ }+{15}^{\circ }={75}^{\circ }$