Question #126886

QR is a diameter of a circle,centre O,and P is a point on its circumference.

Given that PQR= 25°, calculate

a, POR

b, OPR

c, the value, correct to 2 significant figures of PQ/QR.

Expert's answer

Consider the circle with center O and the diameter PQ


central angle=intercepted arccentral \ angle=intercepted\ arc

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 ∠PQR\angle PQR is an inscribed angle


m∠PQR=12mPR⏠m\angle PQR={1\over 2}m \overgroup{PR}

The angle ∠POR\angle POR is a central angle


m∠POR=mPR⏠=2m∠PQR=2⋅25°=50°m\angle POR=m \overgroup{PR}=2m \angle PQR=2\cdot25\degree=50\degree



b. We see that OP and OR are radii of the circle. Then ΔPOR\Delta POR is the equilateral triangle


OP=OR,m∠OPR=m∠ORPOP=OR, m\angle OPR=m\angle ORP

The three interior angles in a triangle will always add up to 180°


m∠OPR+m∠ORP+m∠POR=180°m\angle OPR+m\angle ORP+m\angle POR=180\degree

m∠OPR+m∠ORP=180°−m∠PORm\angle OPR+m\angle ORP=180\degree -m\angle POR

m∠OPR=180°−m∠POR2m\angle OPR=\dfrac{180\degree -m\angle POR}{2}

m∠OPR=180°−50°2=65°m\angle OPR=\dfrac{180\degree -50\degree}{2}=65\degree

c. An angle inscribed in a semicircle is a right angle. Then


m∠QPR=90°m\angle QPR=90\degree

Right ΔQPR\Delta QPR


cos⁡∠PQR=PQQR\cos\angle PQR=\dfrac{PQ}{QR}

PQQR=cos⁡25°≈0.91\dfrac{PQ}{QR}=\cos25\degree\approx0.91


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