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 $d > 0 \quad...\left(i\right)$
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 $\quad(\because 180^{\circ}=\pi$ 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., $30^{\circ}$, $60^{\circ}$, $90^{\circ}$.