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The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles of the triangle.
Given: The angles of a triangle are in A.P. The greatest angle is twice the least.
To do: To find all the angles of the triangle.
Solution:
Let the angles be $a−d,\ a,\ a+d$
As given, $2(a−d)=a+d$
$\Rightarrow 2a−2d=a+d$
$\Rightarrow a=3d\ .....( i)$
$\Rightarrow a−d+a+a+d=180^o$
$\Rightarrow 3a=180^o$
$\Rightarrow a=\frac{180^{o}}{3}=60^o$
$\Rightarrow d=\frac{a}{3}=\frac{60^{o}}{3}=20^o$ [From $( i)\ d=\frac{a}{3}$]
$\Rightarrow a−d=60^o-20^o=40^o$
$\Rightarrow a=60^o$
$\Rightarrow a+d=60^o+20^o=80^o$
Thus, the angles of the triangle are $40^o$, $60^o$ and $80^o$.
Updated on: 10-Oct-2022
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