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.
1
Expert's answer
2020-07-19T15:54:05-0400

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


mPQR=12mPRm\angle PQR={1\over 2}m \overgroup{PR}

The angle POR\angle POR is a central angle


mPOR=mPR=2mPQR=225°=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,mOPR=mORPOP=OR, m\angle OPR=m\angle ORP

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


mOPR+mORP+mPOR=180°m\angle OPR+m\angle ORP+m\angle POR=180\degree

mOPR+mORP=180°mPORm\angle OPR+m\angle ORP=180\degree -m\angle POR

mOPR=180°mPOR2m\angle OPR=\dfrac{180\degree -m\angle POR}{2}

mOPR=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


mQPR=90°m\angle QPR=90\degree

Right ΔQPR\Delta QPR


cosPQR=PQQR\cos\angle PQR=\dfrac{PQ}{QR}

PQQR=cos25°0.91\dfrac{PQ}{QR}=\cos25\degree\approx0.91


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Comments

Assignment Expert
19.07.20, 22:54

The solution has already been answered.

jungkookiee
19.07.20, 11:32

Why not answered?

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