1.) You are correct.
2.) If ∠KQO = 58° and ∠AMI corresponds to it, then this means [tex]4x - 13y = 58[/tex].
Substitute for x. ⇒ [tex]4(31) - 13y = 58[/tex]
--> [tex]124 - 13y = 58[/tex]
--> [tex]-13y = 58 - 124 --> -13y = -66[/tex]
--> [tex]y = \frac{66}{13}[/tex]
Checking your work: [tex]4(31) - 13(\frac{66}{13}) = 124 - 66 = 58[/tex]
3.) To solve this problem, try to think of triangle ΔQPK. All triangle angles have a sum of 180°. We already have 58. ∠HKQ is an exterior angle with a measure of 113°, so ∠PKQ is supplementary to it. 180 - 113 is 67. Now that we have two angle measures, the sum is 125, meaning that ∠KPQ is 55°. Because ∠MPL corresponds to it, this means it has the same measure of 55°.
4.) ∠KQO and ∠QOD are alternate interior angles, since they share the same transversal. If ∠KQO = 58°, then ∠QOD is also 58°.
5.) Angle ∠MPL is supplementary to ∠LPO. 180 - 55 = 125, so ∠LPO is 125°.
