Answer:
75°
Step-by-step explanation:
1. Consider right triangle ABC. In this triangle
m∠C = 90º and CM - angle bisector.
If CM is angle C bisector, then
m∠ACM = m∠MCB = 45º
2. Consider triangle CMB. The sum of the measures of all interior angles of the triangle is always 180°, then
m∠MCB + m∠CBM + m∠BMC = 180°,
m∠BMC = 180° - 45° - 30°
m∠BMC = 105°
3. Angles AMC and BMC are supplementary angles, so
m∠AMC + m∠BMC = 180°
m∠AMC = 180° - 105°
m∠AMC = 75°
Answer:
Its True
Step-by-step explanation:
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One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle. Then the lines are parallel
<h3><u>Solution:</u></h3>
Given that, One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle.
We have to prove that the lines are parallel.
If they are parallel, sum of the described angles should be equal to 180 as they are same side exterior angles.
Now, the 1st angle will be 1/6 of right angle is given as:

And now, 15 degrees is 11 times smaller than the other
Then other angle = 11 times of 15 degrees

Now, sum of angles = 15 + 165 = 180 degrees.
As we expected their sum is 180 degrees. So the lines are parallel.
Hence, the given lines are parallel
Answer: second option.
Step-by-step explanation:
You must apply 
Then, given the right triangle and the angle D, the opposite side and the hypotenuse of the triangle are the following:

Therefore, when you substitute values, you have:

Reduce the fraction, then you obtain:

System: x + y = 44; x = 2y + 2
Solve with substitution: x = 30; y = 14
x + y = 44
x = 2y + 2
Substitute 2y + 2 for x
2y + 2 + y = 44
3y + 2 = 44
Subtract 2 on both sides.
3y = 42
y = 14
Now substitute back in.
x + 14 = 44
Subtract 14 on both sides.
x = 30