Sec. 12.3 Use Inscribed Angles and Polygons p. 780 Objective: To use inscribed angles of circles. Vocabulary: inscribed angle, intercepted arc, inscribed polygon, circumscribed circle A, B, and C represent soccer players facing the goal. Which player has the greatest kicking angle for the goal? Goal

A C B After this lesson, you be able to answer the question? USING INSCRIBED ANGLES An inscribed angle is an angle whose vertex is on a circle and whose sides contain chords of the circle. The arc that lies in the

interior of an inscribed angle and has endpoints on the angle is called the intercepted arc of the angle. inscribed angle intercepted arc USING INSCRIBED ANGLES THEOREM THEOREM 10.8 Measure of an Inscribed Angle

If an angle is inscribed in a circle, then its measure is half the measure of its intercepted arc. A m ADB = 1 m AB 2

C D mAB = 2m ADB B Finding Measures of Arcs and Inscribed Angles

Find the measure of the blue arc or angle. W R S C C Z

M 115 T Q N Y X

mQTS = 180 mZWX = 230 M NM = 50 100 C P Use inscribed angles

Find the indicated measure in a. m T a. m T = 24o mQR = 80o b. mQR

P. Find the measure of the red arc or angle. m G = 45o Find the measure of the red arc or angle. mTV = 76

o Finding Measures of Arcs and Inscribed Angles Find each measure. mPRU Finding Measures of Arcs and Inscribed Angles Find each measure. mSP Find each measure.

Find each measure. mDAE USING INSCRIBED ANGLES THEOREM A THEOREM 10.9 If two inscribed angles of a circle intercept the same arc, then the angles are congruent.

B C D C D Find the measure of the red arc or angle.

N ZXN 72 Find the measure of an intercepted arc Find mRS and m STR. mRS = 62o. m STR 31o

Finding Angle Measures in Inscribed Triangles Find mLJM. mLJM = mLKM 5b 7 = 3b 2b 7 = 0 2b = 7 b = 3.5 mLJM = 5(3.5) 7 = 10.5 USING PROPERTIES OF INSCRIBED POLYGONS

THEOREMS ABOUT INSCRIBED POLYGONS THEOREM 10.10 If a right triangle is inscribed in a circle, then the hypotenuse is a diameter of the circle. Conversely, if one side of an inscribed triangle is a diameter of the circle, then the triangle is a right triangle and the angle opposite the diameter is the right angle. B is a right angle if and only if AC is a diameter of the circle.

A B C Finding Angle Measures in Inscribed Triangles Find a. WZY is a right angle mWZY = 90 5a + 20 = 90 5a = 70 a = 14

ABC is a rt. angle mABC = 90 8z 6 = 90 8z = 96 z = 12 Find z. Find mEDF.

mEDF = mEGF 2x + 3 = 75 2x 4x = 72 x = 18 mEDF = 2(18) + 3 = 39 USING PROPERTIES OF INSCRIBED POLYGONS If all of the vertices of a polygon lie on a circle, the polygon is inscribed in the circle and the circle is circumscribed about the polygon. The polygon is an inscribed polygon and the

circle is a circumscribed circle. A typical parallelogram will not fit inside a circle with the vertices on the circle! So, the opposite angles cannot be congruent. They must be supplementary! USING PROPERTIES OF INSCRIBED PLOYGONS THEOREMS ABOUT INSCRIBED POLYGONS THEOREM 10.11 F

E A quadrilateral can be inscribed in a circle if and only if its opposite angles are supplementary. C D .

D, E, F, and G lie on some circle, C, if and only if m D + m F = 180 and m E + m G = 180. G Find the value of each variable. y = 112 x = 98 Find the value of each variable.

y = 105 x = 100 Find the value of each variable. c = 62 x = 10 Find the value of each variable.

o o 2a + 2a = 180 o 4bo + 2bo = 180o 4a = 180

6b = 180 a = 45 b = 30 Finding Angle Measures in Inscribed Quadrilaterals Find the angle Measures of GHJK. Step 1 Find the value of b.

mG + mJ = 180 3b + 25 + 6b + 20 = 180 9b + 45 = 180 9b = 135 b = 15 Continued Step 2 Find the measure of each angle. mG = 3(15) + 25 = 70 mJ = 6(15) + 20 = 110

mK = 10(15) 69 = 81 mH + mK = 180 mH + 81 = 180 mH = 99 Find the angle measures of JKLM. mM + mK = 180 4x 13 + 33 + 6x = 180 10x + 20 = 180 10x = 160

x = 16 Find the angle measures of JKLM. mM = 4(16) 13 = 51 mK = 33 + 6(16) = 129 mJ = 360 252 = 108 EXAMPLE 4Use a circumscribed circle Photograp hy

Your camera has a 90o field of vision and you want to photograph the front of a statue. You move to a spot where the statue is the only thing captured in your picture, as shown. You want to change your position. Where else can you stand so that the statue is perfectly framed in this way? Use a circumscribed circle SOLUTION From Theorem 10.9, you know that if a right triangle is inscribed in a circle, then the hypotenuse of the triangle is a diameter of the circle. So, draw the circle

that has the front of the statue as a diameter. The statue fits perfectly within your cameras 90o field of vision from any point on the semicircle in front of the statue. Using the Measure of an Inscribed Angle When you go to the movies, you want to be close to the movie screen, but you dont want to have to move your eyes too much to see the edges of the picture. If E and G are the ends of the screen and you are at F, m EFG is called your viewing angle. THEATER DESIGN

E movie screen You decide that the middle of the sixth row has the best viewing angle. If someone is sitting there, where else can you sit to have the same viewing angle? F G

Using the Measure of an Inscribed Angle SOLUTION Draw the circle that is determined by the endpoints of the screen and the sixth row center seat. Any other location on the circle will have the same viewing angle. Lesson Quiz: Part I Find each measure.

1. RUS 2. a 3 25 Lesson Quiz: Part II 3. Find the angle measures of ABCD. m m

m m A = 95 B = 85 C = 85 D = 95 Using an Inscribed Quadrilateral In the diagram, ABCD is inscribed in Find the measure of each angle.

P. ABCD is inscribed in a circle, so opposite angles are supplementary. 3x + 3y = 180 3y = 180 3x 5x + 2y = 180 To solve this system of linear equations,

you can solve the first equation for y to get y = 60 x. Substitute this expression into the second equation. A 2y D 3y P 3x 5x

C B Using an Inscribed Quadrilateral 3x + 3y = 180 5x + 2y = 180 A

To solve this system of linear equations, you can solve the first equation for y. 3x + 3y = 180 3y = -3x + 180 y = -x + 60 or y = 60 x Substitute this expression into the second equation. 2y

D 3y P 3x 5x C 5x + 2y = 180 5x + 2(60 x) = 180 5x + 120 2x = 180

3x = 60 x = 20 y = 60 20 = 40 x = 20 and y = 40, B