Consider the circle with center O and the diameter PQ
c e n t r a l a n g l e = i n t e r c e p t e d a r c central \ angle=intercepted\ arc ce n t r a l an g l e = in t erce pt e d a rc The measure of each inscribed angle is exactly half the measure of its intercepted arc.
In a circle, the measure of an inscribed angle is half the measure of the central angle with the same intercepted arc.
a. The angle ∠ P Q R \angle PQR ∠ PQR is an inscribed angle
m ∠ P Q R = 1 2 m P R ⏠ m\angle PQR={1\over 2}m \overgroup{PR} m ∠ PQR = 2 1 m PR The angle ∠ P O R \angle POR ∠ POR is a central angle
m ∠ P O R = m P R ⏠ = 2 m ∠ P Q R = 2 ⋅ 25 ° = 50 ° m\angle POR=m \overgroup{PR}=2m \angle PQR=2\cdot25\degree=50\degree m ∠ POR = m PR = 2 m ∠ PQR = 2 ⋅ 25° = 50°
b. We see that OP and OR are radii of the circle. Then Δ P O R \Delta POR Δ POR is the equilateral triangle
O P = O R , m ∠ O P R = m ∠ O R P OP=OR, m\angle OPR=m\angle ORP OP = OR , m ∠ OPR = m ∠ ORP The three interior angles in a triangle will always add up to 180°
m ∠ O P R + m ∠ O R P + m ∠ P O R = 180 ° m\angle OPR+m\angle ORP+m\angle POR=180\degree m ∠ OPR + m ∠ ORP + m ∠ POR = 180°
m ∠ O P R + m ∠ O R P = 180 ° − m ∠ P O R m\angle OPR+m\angle ORP=180\degree -m\angle POR m ∠ OPR + m ∠ ORP = 180° − m ∠ POR
m ∠ O P R = 180 ° − m ∠ P O R 2 m\angle OPR=\dfrac{180\degree -m\angle POR}{2} m ∠ OPR = 2 180° − m ∠ POR
m ∠ O P R = 180 ° − 50 ° 2 = 65 ° m\angle OPR=\dfrac{180\degree -50\degree}{2}=65\degree m ∠ OPR = 2 180° − 50° = 65° c. An angle inscribed in a semicircle is a right angle. Then
m ∠ Q P R = 90 ° m\angle QPR=90\degree m ∠ QPR = 90°
Right Δ Q P R \Delta QPR Δ QPR
cos ∠ P Q R = P Q Q R \cos\angle PQR=\dfrac{PQ}{QR} cos ∠ PQR = QR PQ
P Q Q R = cos 25 ° ≈ 0.91 \dfrac{PQ}{QR}=\cos25\degree\approx0.91 QR PQ = cos 25° ≈ 0.91