Let an angle of a triangle is 60 ° . If the variance of the angles of the triangle is 4614 ° ,…

Let three angles of triangle be $x, y \& z$.

As sum of three angles of triangle$=180°$.

$\Rightarrow x+y+z=180°$

$\Rightarrow 60°+y+z=180°$ (Given one angle is ${60}^{\circ }$)

$\Rightarrow y+z=120° \dots \left(1\right)$

Now, it is given that variance of the angles of triangle is equal to $4614°$.

$\therefore \frac{x^2+y^2+z^2}{3}=4614°$

$\Rightarrow \frac{{\left({60}^{\circ }\right)}^2+y^2+z^2}{3}=4614°$

$\Rightarrow y^2+z^2=10242°$

$\Rightarrow {\left(y+z\right)}^2-2yz=10242°$

$\Rightarrow 14400°-2yz=10242°$  (Using $\left(1\right)$)

$\Rightarrow yz=2079° \dots \left(2\right)$

Now, ${\left(y+z\right)}^2={\left(y-z\right)}^2+4yz$

$\Rightarrow \left(120\right)°={\left(y-z\right)}^2+4\times 2079°$  (Using $\left(1\right) \& \left(2\right)$)

$\Rightarrow {\left(y-z\right)}^2=14400°-8316°=6084°$

$\Rightarrow y-z=78° \dots \left(3\right)$

Solving equations $\left(1\right) \& \left(3\right)$, we get,

$y=99° \& z=21°$

Thus, other two angles are $99° \& 21°$.