The measures of three angles of a triangle form an AP. If the greatest angle is twice the smallest angle, find all the angles of the triangle.

Given:

The measures of three angles of a triangle form an AP.

The greatest angle is twice the smallest angle.

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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Updated on: 10-Oct-2022

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