The questions given in the pdf are these pictures, then we can get the solutions.
<u>1) Solution</u>
It is given that,
→ <5 = <6
Then the co interior angles,
→ <5+ <4 = 180°
Now substituting value of <5,
→ <6+ <4 = 180°
This shows property of co interior angle.
Therefore, L II m.
<u>2) Solution</u>
Take it as,
→ <1= 90°
In above eq. L is perpendicular to t.
→ <2 = 90°
In above eq. m is perpendicular to t.
Then it will be,
→ <1 = <2
It shows property of corresponding angle.
Therefore, L II m.
<u>3) Solution</u>
It is given that,
→ <1 = <2 and <1 = <3
Now substitute,
The value of <1 in second one,
→ <2 = <3
This shows property of alternate angle.
Therefore, ST II UV.
<u>4) Solution</u>
It is given that,
→ <RSP = <PQR --- (1)
→ <QRS + <PQR = 180° --- (2)
Now from the equation (1) and (2),
→ <RSP + <QRS = 180°
It shows property of co interior angle.
Therefore, PS II QR.
