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1) What is the Triangle Exterior Angle Theorem?

1) What is the Triangle Exterior Angle Theorem?. 2) Solve for x . 3) Solve for y . R. (2y - 3)°. A. B. Q. (5x)°. S. ( y - 1)°. C. D. T. B. A. F. E. Secant. Secant is a line that intersects a circle at two points. . Theorem

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1) What is the Triangle Exterior Angle Theorem?

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  1. 1) What is the Triangle Exterior Angle Theorem? 2) Solve for x. 3) Solve for y. R (2y - 3)° A B Q (5x)° S (y - 1)° C D T

  2. B A F E Secant Secant is a line that intersects a circle at two points.

  3. Theorem If two secants or chords intersect in the interior of a circle, then the measure of each angle formed is half the sum of the measures of its intercepted arcs. A Chords AD and BC intersect at E. C 1 E D B m<1 = ½ (mAB +mCD)

  4. Examples S 2. Find m<ABE P Find each angle measure. 1. Find m<SQR A 32° 100° Q T 65° E B R 37° D C 3. Find mMQ M 160° Q R N P 255°

  5. You try • 5. Find mRS • 6. Find m<LPO A 139° • Find m<AEB R L O S B 100° E D 80° P N 65° C 113° M L T N 90°

  6. Theorem If a tangent and a secant, two tangents, or two secants intersect in the exterior of a circle, then the measure of the angle formed is half the difference of the measures of its intercepted arcs. Two Tangents Tangent and Secant Two Secants A E J H K F 2 3 B L M 1 C N G D m<1 = ½ (mAD – mBD) m<2 = ½ (mEHG – mEG) m<3 = ½ (mJN – mKM)

  7. Examples 2. 174° 1. 132° 98° 3. E R J 25° 83° K x° F x° Q L M x° P N G S

  8. 5. 4. R 200° 225° Q 74° x° P S 6. E J 30° 25° x° K x° F L M N G

  9. Summing Up Angles in Circles Where’s the Vertex? What’s the Angle Measure? Examples Vertex on a circle 200° Half of the arc 120° 1 2 m<1=60° m<2=100° Vertex inside a circle Half of the arcs added 44° 1 m<1= ½ (44°+86°) = 65° 86° Vertex outside a circle Half of the arcs subtracted 1 78° 125° 202° 2 45° m<2= ½ (125°-45°) = 40° m<1= ½ (202°-78°) = 62°

  10. Whiteboards • Practice 12-4 Workbook P 491 #1-12 all

  11. Warm Up 2. Find LP L mLR= 100° D M A P N 1. Find mAF 80° 160° E Q F S R 110° 26° 48° B C On the white boards! 3. Find mYZ X Y 49° 113° W Z 68°

  12. Last 4 E F • m<FGJ • m<HJK • Solve for x4. mCE G 31° 52° H J 130° K D C B 150° E J 83° A x 25° G H 48°

  13. P 681 # 5 – 24 all, 36 P 691 # 2-14 even, 20- 24 even Period 1 Due Tuesday Period 2/4 Due Wednesday Exam on Thurs/Fri

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