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[tex]\green{ \underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}[/tex]

#2

If mPR = 45 and mQS = 49, what is m∠PIR? m∠RTS?

[tex]\:[/tex]

[tex] \large{\sf \underline{Givens:}}[/tex]

mPR = 45
mQS = 49

[tex]\:[/tex]

[tex]\large{\sf \underline{Find:}}[/tex]

[tex]\: \: \: \: \: \: \: \: \: \: \: [/tex]m∠PIR and m∠RTS

[tex]\:[/tex]

[tex] \large{\sf \underline{Solution:}}[/tex]

[tex]\:[/tex]

» First, find m∠PTR

[tex]\:[/tex]

m∠PTR = ½ (mPR + mQS)

m∠PTR = ½ (45 + 49)

m∠PTR = ½ (94)

m∠PTR = 47

[tex]\:[/tex]

» Find m∠RTS

m∠RTS = 180 - (m∠PTR)

m∠RTS = 180 - 47

m∠RTS = 133

[tex]\:[/tex]

∴ m∠PTR is 47° and m∠RTS is 133°

[tex]\green{ \underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}[/tex]

#3

[tex]\:[/tex]

If mMKL = 220 and mML = 140, what is m∠MQL?

[tex]\:[/tex]

[tex] \large{\sf \underline{Givens:}}[/tex]

[tex]\:[/tex]

mMKL = 220
mML = 140

[tex]\:[/tex]

[tex]\large{\sf \underline{Find:}}[/tex]

[tex]\:[/tex]

[tex]\: \: \: \: \: \: \: \: \: \: \: [/tex] m∠MQL

[tex]\:[/tex]

[tex] \large{\sf \underline{Solution:}}[/tex]

[tex]\:[/tex]

m∠MQL = ½(ML - MKL)

m∠MQL = ½(220 - 140)

m∠MQL = ½(80)

m∠MQL = 40

[tex]\:[/tex]

∴ m∠MQL is 40°

[tex]\green{ \underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}[/tex]

#4

[tex]\:[/tex]

Suppose mCG = 6x + 5, mAR = 4x + 15 and m∠AEC = 120.

Find: a)x [tex]\: \: \: \: \: \: \: \: \: \: \: [/tex] ; b) mCG [tex]\: \: \: \: \: \: \: \: \: \: \: [/tex] ; c) mAR

[tex]\:[/tex]

[tex] \large{\sf \underline{Givens:}}[/tex]

mCG = 6x + 5
mAR = 4x + 15
m∠AEC = 120

[tex]\:[/tex]

[tex]\large{\sf \underline{Find:}}[/tex]

a) x
b) mCG
c) mAR

[tex]\:[/tex]

[tex] \large{\sf \underline{Solution:}}[/tex]

[tex]\:[/tex]

» Find (x)

[tex]\:[/tex]

mCG = mAR

6x + 5 = 4x + 5

6x - 4x = 15 - 5

2x = 10

[tex] \frac{2x}{2} = \frac{10}{2} [/tex]

x = 5

[tex]\:[/tex]

» Find mCG

mCG

6x + 5

Substitute x = 5

6(5) + 5

30 + 5 = 35

mCG = 35

[tex]\:[/tex]

» Find mAR

mAR

4x + 15

Substitute x = 5

4(5) + 15

20 + 15 = 35

mAR = 35

[tex]\:[/tex]

∴ All the highlighted part/word(s) are the answers.

[tex]\green{ \underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}[/tex]

#5

[tex]\:[/tex]

OK is tangent to ⊚ R at C. Suppose KC = OC, OK = 8 and RC = 3. Find; OR, RS, and KS

[tex]\:[/tex]

[tex]\large{\sf \underline{Find:}}[/tex]

OR
RS
KS

[tex]\:[/tex]

[tex] \large{\sf \underline{Solution:}}[/tex]

[tex]\:[/tex]

» Since OK = 8 and KC = OC, substitute it to the equation and find out KC and OC.

OK = KC + OC

8 = KC + OC

8 = 2(KC + OC)

4 = KC ; 4 = OC

[tex]\:[/tex]

∆ Find OR using Pythagorean Theorem when RC = 3 and KC = 4.

RC² + KC² = OR²

3² + 4² = OR²

9 + 16 = OR²

25 = OR²

[tex] \sqrt{25} = OR[/tex]

[tex]\:[/tex]

∴ OR measures 5 units

[tex]\:[/tex]

∆ Find RS

[tex]\:[/tex]

Since R is the center of the circle and C is the point on the circle, that means RC is the radius. RS = RC being the radius of the circle. Since RC = 3, RS is also 3.

[tex]\:[/tex]

∴ RS measures 3 units

[tex]\:[/tex]

∆ Find KS

[tex]\:[/tex]

» Using segment addition postulate, we get RS + KS = KR. We know that RS = 3 and KR is 5. Substitute it to get KS.

RS + KS = KR

3 + KS = 5

KS = 5 - 3

KS = 2

[tex]\:[/tex]

∴ KS measures 2 units

[tex]\green{ \underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}[/tex]

All highlighted number(s)/word(s)/parts are the answers.

[tex]\green{ \underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}[/tex]

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