Question

The angles of a triangle are in $A.P.$ and the least angle is ${30}^o;$ find the greatest angle of the triangle.( in degrees )

Answer 1

Let the angles be, $\theta -d,\theta ,\theta +d$

As we know, same of Ls of a $\Delta$ is ${180}^o$

$\therefore \theta -d+\theta +\theta +d={180}^o$

$\Rightarrow 3\theta =180$

$\Rightarrow \theta ={60}^o$

ATS

Given condition Least L is : $\theta -d={30}^o$

$\Rightarrow 60-d=30$

$\Rightarrow d={90}^o/3$

$\Rightarrow d={30}^o$

Hence, Ls one :-

$30,60,90$

Therefore the longest L is ${90}^o$