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# The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles.

**Given:**

The angles of a triangle are in A.P.

The greatest angle is twice the least.

**To do: **

We have to find the angles of the triangle.

**Solution:**

Let the angles be $a−d,\ a,\ a+d$

According to the question,

$a+d=2(a−d)$

$\Rightarrow a+d=2a-2d$

$\Rightarrow 2a-a=2d+d$

$\Rightarrow a=3d\ .....( i)$

$\Rightarrow a−d+a+a+d=180^o$

$\Rightarrow 3a=180^o$

$\Rightarrow a=\frac{180^{o}}{3}=60^o$

This implies,

$\Rightarrow d=\frac{a}{3}=\frac{60^{o}}{3}=20^o$ [From $( i)\ d=\frac{a}{3}$]

$\Rightarrow a−d=60^o-20^o=40^o$

$\Rightarrow a=60^o$

$\Rightarrow a+d=60^o+20^o=80^o$

**Hence, the angles of the triangle are $40^o$, $60^o$ and $80^o$. **

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