A. Chords Arc Theorem

Given: UPT and SPT are central angles and congruent.
mUPQ = 86, mQPR = 154, mRPS = 60
Find m ST and m TU

Solutions:
mUPQ + mQPR +
mRPS + mUPS = 3600 by sum of all angle
measures in a circle
860 + 1540 + 600 + mUPS = ___0 by substitution
3000 + mUPS = 360
3000 - ____0 + mUPS = 3600 -___0 by APE
mUPS = 600
mUPS = mUPT + mSPT Angle Addition Postulate
What’s More

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DISCIPLINE • GOOD TASTE • EXCELLENCE
mUPT = mSPT Given
mUPT = mUPT Identity Property
mUPS = mUPT + mUPT
mUPS = 2(mUPT)
1⁄2 ( 600 = 2(mUPT)) 1⁄2
___0 = mUPT
Since mUPT = mSPT
Therefore mSPT = ___0
m ST = m TU Converse of the Chords Central

Angle Theorem

B. Chords Central Angle Theorem
Given: Chords PQ and SR are congruent
PC, QC, RC and SC are radii of the same circle
mRCS = 600
Find mQCP.
Solution: If PQ and SR are chords and equal,
PC = QC, SC = RC (radii of the same circle
are equal and mRCS = __0, then
QCP  RCS by SSS Postulate
PCQ  QCP by CPCTC
So mQCP = ___0

C. Chord Distance to Center Theorem
Given:

Find the value of y.

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DISCIPLINE • GOOD TASTE • EXCELLENCE

Solutions :
A perpendicular drawn from the center of the circle,
bisects the chord.
AE = EB = 15/2 = 7.5 cm
In triangle OEB, by using the Pythagorean Theorem
OB2 = OE2 + EB2
OB2 = (__)2 + (___)2
OB2 = ___+ 56.25
OB2 = 105.25
√OB2 = √105.25
OB = 10.25 cm
OB = OD = radius of the circle = 10.25 cm

In triangle OFD, by Phytagorean Theorem
OD2 = OF2 + FD2
(10.25)2 = (__)2 + FD2
105.0625 = (__) + FD2
105.0625 – (__) = (__) – (__)+ FD2
56.0625 = FD2
√56.0265 = √FD2
7.49 ≈ FD
CD ≈ 2(7.49)
CD ≈ 14.98 cm
So, y ≈ 14.96 cm.