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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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