Question

# The angles of a triangle are in AP and the number of degrees in the least is to the number of radians in the greatest as $$60 : \pi$$. Find the angles in degrees.

Solution

## Let the angles be $$a-d$$, $$a$$, $$a+d$$ (as in A.P.)$$(a-d)+a+(a+d)=180^o$$$$\Rightarrow a=60^o$$$$\dfrac{a-d}{a+d}=\dfrac{60^o}{\pi rad}=\dfrac{60}{180}=\dfrac{1}{3}$$$$\Rightarrow 3a-3d=a+d$$$$\Rightarrow a=2d$$$$\Rightarrow d=30^o$$$$\therefore$$ Angles are $$30, 60, 90$$.Mathematics

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