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Starrysoul100

[tex]\bold{\huge{\underline{ Solution }}}[/tex]

Given :-

  • We have given two parallel lines and two lines intersecte each other and act as a transverse
  • Due to the intersection between two parallel and two traverse lines a triangle is formed.
  • The measurements of the given triangle are 60° , 20° and C

To Find :-

  • We have to find the values of angles A, B and C

Let's Begin :-

Here, we have

  • 1 triangle whose measures are 60° , 20° and C

Therefore,

By using Angle sum property

  • ASP states that the sum of all angles of triangles are equal to 180°

That,

[tex]\bold{\angle{X + }}{\bold{\angle{Y + }}}{\bold{\angle{Z = 180}}}{\degree}[/tex]

Subsitute the required values,

[tex]\sf{ 60{\degree} + 20{\degree} + C = 180{\degree}}[/tex]

[tex]\sf{ 80{\degree} + C = 180{\degree}}[/tex]

[tex]\sf{ C = 180{\degree} - 80{\degree}}[/tex]

[tex]\sf{ C = 100}{\degree}[/tex]

Thus, The value of C is 180°

Now,

We have to find the measurement of Angles A and B

Here,

  • Angle B , unknown angle and Angle C lie on a straight line and we know that the angle formed by straight line is 180°
  • Let the unknown angle be x which is equal to 20° ( Alternative interior angles)

Therefore,

[tex]\sf{\angle{B + }}{\sf{\angle{ x + }}}{\sf{\angle{C = 180}}}{\degree}[/tex]

[tex]\sf{\angle{ B + 20{\degree}+ 100{\degree} = 180{\degree}}}[/tex]

[tex]\sf{\angle{ B + 20{\degree} = 180{\degree}- 100{\degree}}}[/tex]

[tex]\sf{\angle{ B = 80{\degree} - 20{\degree}}}[/tex]

[tex]\sf{\angle{ B = 60{\degree}}}[/tex]

Now,

  • A, B and 100° lie on a straight line and we know that the angle formed by straight line is equal to 180°

Therefore ,

[tex]\sf{\sf{A + }}{\sf{\angle{B + }}}{\sf{ 100{\degree} = 180}}{\degree}[/tex]

[tex]\sf{\angle{ A + 60{\degree} = 180{\degree} - 100{\degree}}}[/tex]

[tex]\sf{\angle{ A = 80{\degree} - 60{\degree}}}[/tex]

[tex]\sf{\angle{ A = 20{\degree}}}[/tex]

Hence, The measure of Angles A, B and C are 20° , 60° and 100° .

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