To answer the questions, we’ll use some properties of circles, especially related to inscribed angles and central angles.
Question 1
Given:
- Central angle \( \angle QOR = 107^\circ \)
The relationship between a central angle and the inscribed angle that subtends the same arc is such that the inscribed angle is half of the central angle. Therefore: \[ \angle QSR = \frac{1}{2} \times \angle QOR = \frac{1}{2} \times 107^\circ = 53.5^\circ \]
Answer: 53.5 degrees
Question 2
Given:
- \( \angle CBA = 66^\circ \)
The measure of an inscribed angle is half of the measure of the arc it subtends. Therefore: \[ \text{Arc } AC = 2 \times \angle CBA = 2 \times 66^\circ = 132^\circ \]
Answer: 132 degrees
Question 3
Given:
- Arc \( mFG = 60^\circ \)
To find \( \angle FDG \), we note that the angle \( FDG \) subtends the same arc \( FG \). Thus, using the inscribed angle theorem: \[ \angle FDG = \frac{1}{2} \times mFG = \frac{1}{2} \times 60^\circ = 30^\circ \]
Answer: 30 degrees
Question 4
Given:
- \( m\angle A = 27^\circ \)
In a triangle inscribed in a circle with one side as the diameter, angle \( C \) (the inscribed angle) is a right angle, and the sum of angles in a triangle is \( 180^\circ \). Thus: \[ m\angle A + m\angle B + 90^\circ = 180^\circ \implies m\angle B = 180^\circ - 90^\circ - 27^\circ = 63^\circ \]
Answer: 63 degrees
Question 5
Given:
- \( m\angle A = 15^\circ \)
For an inscribed angle, the measure of the arc it intercepts is twice the angle: \[ \text{Arc } AC = 2 \times m\angle A = 2 \times 15^\circ = 30^\circ \]
Answer: 30 degrees
This provides succinct answers and derivations based on the properties of inscribed and central angles in circle geometry.