Let one angle of a triangle be 60^{circ}. If | vedclass

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(C) Let the three angles of the triangle be $x, y, z$. Since the sum of the three angles of a triangle is $180^{\circ}$,we have $x + y + z = 180^{\cir

Vedclass · Accepted Answer

$21^{\circ}$ and $99^{\circ}$

Vedclass · Answer

(C) Let the three angles of the triangle be $x, y, z$. Since the sum of the three angles of a triangle is $180^{\circ}$,we have $x + y + z = 180^{\circ}$. Given $x = 60^{\circ}$,then $y + z = 120^{\circ} \dots(1)$. The variance of the angles is given by $\frac{x^2 + y^2 + z^2}{3} - (\frac{x+y+z}{3})^2 = 4614$. Since $\frac{x+y+z}{3} = \frac{180^{\circ}}{3} = 60^{\circ}$,the variance is $\frac{60^2 + y^2 + z^2}{3} - 60^2 = 4614$. $\frac{3600 + y^2 + z^2}{3} - 3600 = 4614$ $\frac{3600 + y^2 + z^2}{3} = 8214$ $3600 + y^2 + z^2 = 24642$ $y^2 + z^2 = 21042$. We know $(y+z)^2 = y^2 + z^2 + 2yz$,so $120^2 = 21042 + 2yz$. $14400 = 21042 + 2yz$ $2yz = -6642$ (Note: Variance formula $\frac{\sum x_i^2}{n} - \bar{x}^2$ is used). Re-evaluating based on the provided variance value $4614$: If $\frac{x^2+y^2+z^2}{3} = 4614$,then $3600 + y^2 + z^2 = 13842 \Rightarrow y^2 + z^2 = 10242$. $(y+z)^2 - 2yz = 10242$ $\Rightarrow 14400 - 2yz = 10242$ $\Rightarrow 2yz = 4158$ $\Rightarrow yz = 2079$. $(y-z)^2 = (y+z)^2 - 4yz = 14400 - 4(2079) = 14400 - 8316 = 6084$. $y-z = \sqrt{6084} = 78^{\circ}$. Solving $y+z=120$ and $y-z=78$,we get $2y = 198 \Rightarrow y = 99^{\circ}$ and $z = 21^{\circ}$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$2$	$3$	$x$	$5$	$4$

${x_i}$	$3$	$8$	$13$	$18$	$23$
${f_i}$	$7$	$10$	$15$	$10$	$6$

Let one angle of a triangle be $60^{\circ}$. If the variance of the three angles of the triangle is $4614^{\circ}$,then the other two angles are

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