[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° .
