Answered

IDNStudy.com, ang iyong destinasyon para sa mabilis at eksaktong mga sagot. Sumali sa aming platform upang makatanggap ng mabilis at eksaktong tugon mula sa mga propesyonal sa iba't ibang larangan.

•••••••••••••••••••••••••••••••••••
PA HELP PO TYTY!

NONSENSE REPORT! ;)
•••••••••••••••••••••••••••••••••••••​

PA HELP PO TYTYNONSENSE REPORT class=

Sagot :

XhionPierreidn

CIRCLES

[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]

#CarryOnLearning

ヾ(〃＾-＾)ﾉ