# Bell work 1 T R TU = 92 º U Find the measure of the inscribed angles, R, given that their common intercepted TU = 92º

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Bell work 1 T R TU = 92 º U Find the measure of the inscribed angles, R, given that their common intercepted TU = 92º

Bell work 1 Answer T R TU = 92 º U Angles R = ½ the intercepted arc TU since their intercepted Arc TU = 92º, then Angle R = 46º

Bell work 2 A quadrilateral WXYZ is inscribed in circle P, if _ X = 130º and _ Y = 106º, Find the measures of _ X = 130º and _ Y = 106º, Find the measures of _ W = ? and _ Z = ? _ W = ? and _ Z = ? X Y P Z The Quadrilateral WXYZ is inscribed in the circle iff / X + / Z = 180º, and / W + / Y = 180º W 130 º 106 º

Bellwork 2 Answer From Theorem 10.11 _ W = 180º – 106º = 74º and _ W = 180º – 106º = 74º and _ Z = 180º – 130º = 50º _ Z = 180º – 130º = 50º X Y P Z The Quadrilteral WXYZ is inscribed in the circle iff / X + / Z = 180º, and / W + / Y = 180º W 130 º 106 º

Unit 3 : Circles: 10.4 Other Angle Relationships in Circles Objectives: Students will: 1.Use angles formed by tangents and chords to solve problems related to circles 2.Use angles formed by lines intersecting on the interior or exterior of a circle to solve problems related to circles

(p. 621) Theorem 10.12 If a tangent and a chord intersect at a point on a circle, then the measure of each angle formed is ½ the measure of its intercepted arc P A B Angle 1 C m _ 1 = ½ m minor AC m _ 2 = ½ m Major ABC Angle 2 m

(p. 621) Theorem 10.12 Example 1 Find the measure of Angle 1 and Angle 2, if the measure of the minor Arc AC is 130 º P A B Angle 1 C º m minor AC = 130º Angle 2 m

(p. 621) Theorem 10.12 Example 1 Answer The measure of Angle 1 = 65 º and Angle 2 = 115 º P A B 65 º = Angle 1 C º m minor AC = 130º Angle 2 = 115 º m

(p. 621) Theorem 10.12 Example 2 Find the measure of Angle 1, if Angle 1 = 6x º, and the measure of the minor Arc AC is (10x + 16) º P A B Angle 1= 6x º C º m minor AC = (10x + 16)º m

(p. 621) Theorem 10.12 Example 2 Answer Angle 1 = 6xº = ½ Arc AC = ½ (10x + 16)º 6xº = ½ (10x + 16)º 6xº = 5x + 8 x = 8º thus, Angle 1 = 48º P A B Angle 1= 6x º C º m minor AC = (10x + 16)º m

Intersections of lines with respect to a circle There are three places two lines can intersect with respect to a circle. On the circle In the circle Outside the cirlce

(p. 622) Theorem 10.13 If two chords intersect in the interior of a circle, then the measure of each angle is ½ the sum of the measures of the arcs intercepted by the angle and its vertical angle. P D C Angle 1 B m _ 1 = ½ (m AB + m CD) m _ 2 = ½ (mBC + mAD) Angle 2 A

m CD = 16 º (p. 622) Theorem 10.13 Example Find the value of x. P D C Angle 1 B m AB = 40 º A xºxº

m CD = 16 º (p. 622) Theorem 10.13 Example Answer x = ½ (m AB + m CD) = ½ (40º + 16º) x = ½ (56º) x = 28 º P D C Angle 1 B m AB = 40 º A xºxº

3 2 1 (p. 622) Theorem 10.14 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 ½ the difference of the intercepted arcs. B m _ 1 = ½ (m BC – m AC) m _ 2 = ½ (m PQR – m PR) 1 Tangent and 1 Secant 2 Tangents2 Secants B A C m _ 3 = ½ (m XY – m WZ) P Q R W X Y Z

xºxº (p. 622) Theorem 10.14 Example 1 Find the value of x m _ x = ½ (m PQR - mPR) Major Arc PQR = 266 º P Q R

xºxº (p. 622) Theorem 10.14 Example 1 Answer m PR = (360º - m PQR) = (360º - 266º) = 94º x = ½ (m PQR - m PR) = ½ (266º - 94º) = ½ (172º) x = 86 º m _ x = ½ (m PQR - mPR) Major Arc PQR = 266 º P Q R

(p. 622) Theorem 10.14 Example 2 Find the value of x, GF. The m EDG = 210 º The m angle EHG = 68 º m _ EHG = 68 º = ½ (m EDG – m GF) E H G xºxº F 68 º D Major Arc EDG = 210 º

(p. 622) Theorem 10.14 Example 2 Answer m _ EHG = 68º = ½ (m EDG – m GF) 68º = ½ ( 210º - xº ) 136º = 210º - xº xº = 210º - 136º xº = 74º m _ EHG = 68 º = ½ (m EDG – m GF) E H G xºxº F 68 º D Major Arc EDG = 210 º

Home work PWS 10.4 A P. 624 (8 -34) even

Journal Write two things about the intersections of chords, secants, and/or tangents related to circles from this lesson.

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