Question

The angels of a triangle are in A.P and the number of degrees in the least is to the number of radians in the greatest is 60 to ${\pi }^c$ Find the angles in degrees

• A. The three angles are ${40}^{\circ },{30}^{\circ }$ and ${70}^{\circ }$
• B. The three angles are ${30}^{\circ },{60}^{\circ }$ and ${90}^{\circ }$
• C. The three angles are ${20}^{\circ },{20}^{\circ }$ and ${100}^{\circ }$
• D. The three angles are ${60}^{\circ },{70}^{\circ }$ and ${90}^{\circ }$

Answer 1

The correct option is B The three angles are ${30}^{\circ },{60}^{\circ }$ and ${90}^{\circ }$

The three angles in A.P are $(x-y{)}^{\circ },x^{\circ }$ and $(x+y{)}^{\circ }$

$\therefore x+y+x+x-y={180}^{\circ }$ [sum of three angles ${180}^{\circ }$]
$\Rightarrow x={60}^{\circ }$

According to the question
$\frac{(x-y)}{(x+y)\frac{\pi }{180}}=\frac{60}{\pi }$

$\pi (x-y)=(x+y)\frac{\pi }{{180}^{\circ }}\times {60}^{\circ }$

$\Rightarrow 3(x-y)=x+y$
$=>4y=2x$

$\Rightarrow y=\frac{2x}{4}$

$\Rightarrow y=\frac{x}{2}$

$\Rightarrow y=\frac{{60}^{\circ }}{2}$

$\therefore y={30}^{\circ }$

$\therefore$ The angles are ${30}^{\circ }$, ${60}^{\circ }$ and ${90}^{\circ }$