Which of the following is the same as cos15°
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°.
<h2>Explanation:</h2><h2></h2>
A circle is simply a curve made up of all the points that are the same distance from the center. This distance is called the radius of the circle. It's important to know that <em>all circles are similar to each other</em>, but what is similarity? It's simply, shapes are similar if we can turn one into the other by moving, rotating, flipping, or scaling. So every circle can match any other circle by moving it, rotating it, flipping it, or scaling it.
Then circles O and Q shown below are similar no matter what's the radii of them.