Question

One angle of a triangle is 50° and the other two are in ratio 6:7.Find the angle

Answer 1

Answer:

angles will be 50 degree, 60 degree and 70 degree

Step-by-step explanation:

let the common ratio be x

50+6x+7x=180

50+13x=180

13x=180-50

13x=130

x=130/13

x=10

6x=60

7x=70

Answer 2

Answer :

$\underline{\boxed{\mathsf{60^{\circ} \: and\: 70^{\circ}}}}$

Step-by-step Explanation :

$\underline{\underline{\textsf{Given that...}}}$

One angle of a circle = 50°

and other two angles are in ratio 6:7

$\underline{\underline{\mathsf{Solution...}}}$

$\underline{\textsf{Let x be the constant of ratio}}$

$\textsf{Therefore we have...}$

$\implies \mathsf{\angle 1 = 6x}$

$\implies \mathsf{\angle 2 = 7x}$

$\textsf{Now,}$

$\textsf{Angle 1 + Angle 2 =}$

$\mathsf{\angle 1 + \angle 2= 6x+7x = 13x}$

According to Angle Sum Property Of triangle we have

$\mathsf{Sum\: of \:all \:interior \:angle\: =\: 180^{\circ}}$

$\underline{\mathsf{\therefore \: By \:using\: Angle\:Sum\: Property\:here\:we\: have }}$

$\mathsf{50^{\circ} + \angle 1 + \angle 2 = 180^{\circ}}$

$\implies \mathsf{50^{\circ} + 6x + 7x = 180^{\circ}}$

$\implies \mathsf{13x = 180^{\circ} - 50^{\circ}}$

$\implies \mathsf{x = \dfrac{130^{\circ}}{13} = 10^{\circ}}$

$\underline{\textsf{Therefore we have value of x = 10 degree }}$

$\underline{\textsf{Now taking value of X = 10 degree we have}}$

$\implies \mathsf{\angle 1 = 6x = 6 \times 10 = 60^{\circ}}$

$\implies \mathsf{\angle 2 = 7x = 7 \times 10 = 70^{\circ}}$

$\underline{\textbf{Therefore required answer are}}$

$\underline{\boxed{\boxed{\mathsf{ 60^{\circ}\: and\: 70^{\circ}}}}}$