Of the 130 students distance learning math with Mr. Bischof, 63% of them actually work through the material. Approximately, how
1 answer:
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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°.
