Question

The angles of a triangle are in $A$.$P$. and the number of degrees in the least to the number of radians in the greatest is $60:\pi$, find the angles in degrees.

• A. $30^{\circ }$, $60^{\circ }$, $90^{\circ }$
• B. $40^{\circ }$, $60^{\circ }$, $90^{\circ }$
• C. $30^{\circ }$, $30^{\circ }$, $120^{\circ }$
• D. $20^{\circ }$, $130^{\circ }$, $30^{\circ }$

Answer 1

Answer: (A)

Let the angles of the triangle be
${\left(a-d\right)}^{\circ }$, ${\left(a+d\right)}^{\circ }$, where $d>0...\left(i\right)$
then $\left(a-d\right)+a+\left(a+d\right)=180$
$\Rightarrow a=60$
$\therefore$ From $\left(i\right)$, the angles are ${\left(60-d\right)}^{\circ }$, $60^{\circ }$, ${\left(60+d\right)}^{\circ }$
Now, the least angle $={\left(60-d\right)}^{\circ }$
and the greatest angle $={\left(60+d\right)}^{\circ }$
$=\left(60+d\right)\times \frac{\pi }{180}$ radian $(\because 180^{\circ }=\pi$ radian$)$
By the given condition, we have
$\frac{60-d}{\frac{\pi }{180}\left(60+d\right)}=\frac{60}{\pi }$
$\Rightarrow \frac{180\left(60-d\right)}{\left(60+d\right)}=60$
$\Rightarrow 180-3d=60+d$
$\Rightarrow 4d=120$
$\Rightarrow d=30$
$\therefore$ From $\left(i\right)$, the angles are ${\left(60-30\right)}^{\circ }$, $60^{\circ }$, ${\left(60+30\right)}^{\circ }$
i.e., $30^{\circ }$, $60^{\circ }$, $90^{\circ }$.