For problems 1 - 4, write a two-column proof.

The questions given in the pdf are these pictures, then we can get the solutions.

1) Solution

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.

2) Solution

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.

3) Solution

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.

4) Solution

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.

Solution given:

1:

<5=<6

<5+<4=180°[co interior angle]

Substituting value of<5

<6+<4=180°[it shows a property of co interior angle]

So

l || m

2:

<1=90°[ l is perpendicular to t]

<2=90°[m is perpendicular to t]

since

<1=<2[shows property of corresponding angle]

:.

l || m.

3:

<1=<2

<1=<3

substituting value of<1 in second one

<2=<3[which shows property of alternate Angel]

So

Segment ST || segment UV.

4:

<RSP=<PQR......[I]

<QRS+<PQR=180°.....[ii]

from equation I and ii we get

<RSP+<QRS=180°[which shows property of co interior angle ]

So

Segment PS || segment QR

47 - 90 = 43

43 - 180 = x

x = 137

a

step-by-step explanation: