1 / 9

Inscribed Angles

Inscribed Angles. Geometry CP2 (Holt 12-4) K. Santos. Inscribed angle. Inscribed angle—an angle with its vertex on the circle A C B < C is and inscribed angle is the intercepted arc. Inscribed Angle Theorem 12-4-1.

butch
Télécharger la présentation

Inscribed Angles

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Inscribed Angles Geometry CP2 (Holt 12-4) K. Santos

  2. Inscribed angle Inscribed angle—an angle with its vertex on the circle A C B < C is and inscribed angle is the intercepted arc

  3. Inscribed Angle Theorem 12-4-1 Inscribed angle = half the measure of its intercepted arc. X Y Z m < Y = ½

  4. Example—Inscribed Angle Find the values of a and b. 32 b a a = ½ of a semicircle a = ½ (180) = 90 The inscribed angle has an arc 2(32) = 64 b= 180 – 64 b = 116

  5. Example—Inscribed angles P a Find the values of a and b. T 30 60 S Q b R m < PQT = ½ m 60 = ½ a 120= a m < PRS = ½ m b = ½ (m +m) b = ½ (120 + 30) b = ½ (150) b = 75

  6. Corollary to the Inscribed Angle Theorem 12-4-2 Two inscribed angles that intercept the same arc are congruent. A D B C <A and < D have the same intercepted arcs so, <A <D

  7. Theorem 12-4-3 An angle inscribed in a semicircle is a right angle. M N P O < M has an intercepted arc of this arc is a semicircle So, m < M = 90 (1/2 of 180)

  8. Theorem 12-4-4 Quadrilateral is inscribed in a circle = opposite angles aresupplementary. A B C D < A and <C are opposite angles, so they are supplementary <B and <D are opposite angles, so they are supplementary

  9. Example—Inscribed quadrilateral m < A = 70 and m < B = 120. Find m < C and m < D A B C D < A and < C are supplementary, 180 – 70 =110 so m<C = 110 < B and < D are supplementary 180 – 120 = 60 so m <D = 60

More Related