Question

1. Consider the figure below where ZV is parallel to WY.

Answer 1

Answer:

Part A:

m∠VHT = 152°

Part B:

m∠QTS = 152°

Part C:

m∠ZHQ = 28°.

Step-by-step explanation:

Part A:

The given parameters are;

m∠HXU = 113°

Segment BQ and segment UD intersect at m∠XAT = 95°

We have that m∠HXU + m∠HXS = 180° (Angles on a straight line)

Therefore;

m∠HXU  = 180° - m∠HXS = 180° - 113° = 67°

m∠HXU = 67°

m∠XAT + m∠XAH = 180° (Angles on a straight line)

m∠XAH  = 180° - m∠XAT = 180° - 95° = 85°

m∠XAH = 85°

In triangle XAH, we have;

m∠XAH + m∠HXU + m∠XHA = 180° (Angle sum property of a triangle)

∴ m∠XHA = 180° - (m∠XAH + m∠HXU) = 180° - (85° + 67°) = 28°

m∠XHA = 28°

m∠VHT + m∠XHA = 180° (Angles on a straight line)

m∠VHT  = 180° - m∠XHA = 180° - 28° = 152°

m∠VHT = 152°

Part B:

m∠QTS ≅ m∠VHT (Corresponding angles are congruent)

∴ m∠QTS = 152° (Substitution property)

Part C:

m∠ZHQ ≅ m∠XHA  (Reflexive property)

∴ m∠ZHQ = 28°.