Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Inscribed Angles Challenge Problem Given: πΊπ» and the tangent l intersecting at π» on circle β¨πΈ Prove: πβ πΊπ»πΌ = 1 2 ππΊπΉπ» F G l I E H D F G l I E 1. πΊπ» and tangent line l intersect at π» on β¨ πΈ (given) 2. Construct diameter π»π· intersecting β¨πΈ at π·. 3. β π·π»πΌ is a right angle (tangent and radius are perpendicular) 4. π·πΊπ» is a semicircle of measure 180 (def of semicircle) 5. πβ π·π»πΊ +πβ πΊπ»πΌ = πβ π·π»πΌ (angle addition) 6. ππ·πΊ + ππΊπΉπ» = ππ·πΊπ» (arc addition) 7. 90 = πβ π·π»πΊ + πβ πΊπ»πΌ (substitution) 8. 180 = ππ·πΊ + ππΊπΉπ» (substitution) 1 9. 90 = 2 (ππ·πΊ + ππΊπΉπ») (division) 10. πβ π·π»πΊ + πβ πΊπ»πΌ = 1 2 ππ·πΊ + 1 ππΊπΉπ» 2 (substitution and distribution) 11. πβ π·π»πΊ = 12. πβ πΊπ»πΌ = 1 ππ·πΊ (Inscribed angle 2 1 ππΊπΉπ» (subtraction) 2 theorem) H Objectives β’ Use inscribed angles to solve problems. β’ Use properties of inscribed polygons. Using Inscribed Angles β’ 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. intercepted arc inscribed angle Measure of an Inscribed Angle A β’ If an angle is inscribed in a circle, then its measure is one half the measure of its intercepted arc. ποπ΄π·π΅ = ½π AB ο© C D B Finding Measures of Arcs and Inscribed Angles β’ Find the measure of the blue arc or angle. S R ο© m QTS = 2mοQRS = 2(90°) = 180° T Q Finding Measures of Arcs and Inscribed Angles W β’ Find the measure of the blue arc or angle. ο© Z Y X m ZWX = 2mοZYX = 2(115°) = 230° Ex. 1: Finding Measures of Arcs and Inscribed Angles β’ Find the measure of the blue arc or angle. ο© ο© N 100° M m NMP = ½ m NP P ½ (100°) = 50° Ex. 2: Comparing Measures of Inscribed Angles A β’ Find mοACB, mοADB, and mοAEB. E The measure of each angle is half the measure of AB m AB = 60°, so the measure of each angle is 30° ο© B ο© D C Theorem A β’ If two inscribed angles of a circle intercept the same arc, then the angles are congruent. β’ οC ο οD D B C Ex. 3: Finding the Measure of an Angle G β’ It is given that mοE = 75°. What is mοF? ο© β’ οE and οF both intercept GH , so οE ο οF. So, mοF = mοE = 75° E 75° F H Using Properties of Inscribed Polygons β’ 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. Theorem β’ 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 B diameter is the right angle. β’ οB is a right angle if and only if AC is a diameter of the circle. A C Theorem β’ A quadrilateral can be inscribed in a circle if and E only if its opposite angles are supplementary. β’ D, E, F, and G lie on some circle, C, if and only if ποπ· + ποπΉ = 180° D β’ πππ ποπΈ + ποπΊ = 180° F C G Ex. 5: Using Theorems β’ Find the value of each variable. β’ AB is a diameter. So, οC is a right angle and mοC = 90° β’ 2π₯° = 90° β’ π₯ = 45 B Q A 2x° C Ex. 5: Using Theorems β’ Find the value of each variable. β’ DEFG is inscribed in a circle, so opposite angles are supplementary. β’ ποπ· + ποπΉ = 180° β’ π§ + 80 = 180 β’ π§ = 100 D E z° 120° 80° y° G F Ex. 5: Using Theorems β’ Find the value of each variable. β’ DEFG is inscribed in a circle, so opposite angles are supplementary. β’ ποπΈ + ποπΊ = 180° β’ π¦ + 120 = 180 β’ π¦ = 60 D E z° 120° 80° y° G F Ex. 6: Using an Inscribed Quadrilateral β’ In the diagram, ABCD is inscribed in circle P. Find the measure of each angle. A 2y° D 3y° 3x° B β’ ABCD is inscribed in a circle, so opposite 2x° angles are C supplementary. To solve this system of linear equations, β’ 3x + 3y = 180 you can solve the first equation for y to β’ 5x + 2y = 180 get y = 60 β x. Substitute this expression into the second equation. Ex. 6: Using an Inscribed Quadrilateral β’ β’ β’ β’ β’ β’ 5x + 2y = 180. 5x + 2 (60 β x) = 180 5x + 120 β 2x = 180 3x = 60 x = 20 y = 60 β 20 = 40 Write the second equation. Substitute 60 β x for y. Distributive Property. Subtract 120 from both sides. Divide each side by 3. Substitute and solve for y. ο―x = 20 and y = 40, so mοA = 80°, mοB = 60°, mοC = 100°, and mοD = 120° Angle Measures and Segment Lengths in Circles Objectives: 1. To find the measures of οs formed by chords, secants, & tangents. 2. To find the lengths of segments associated with circles. Secants F B Secant β A line that intersects a circle in exactly 2 points. β’πΈπΉ or π΄π΅ are secants β’π΄π΅ is a chord A E Theorem. The measure of an ο formed by 2 lines that intersect inside a circle is 1 πο1 = (π₯ + π¦) 2 π° Measure of intercepted arcs π π° Theorem. The measure of an ο formed by 2 lines that intersect outside a circle is Smaller Arc π ποπ = (π β π) π 3 cases: Larger Arc 2 Secants: Tangent & a Secant 1 1 1 y° y° y° x° 2 Tangents x° x° Ex.1 & 2: Find the measure of arc π₯. Find the ποπ₯. π° π° 92° 104° 68° 94° 112° 1 πο1 = (π₯ + π¦) 2 1 94 = (112 + π₯) 2 188 = (112 + π₯) 76° = π₯ 268° ποπ₯ = ½(π₯ β π¦) ποπ₯ = ½(268 β 92) ποπ₯ = ½(176) ποπ₯ = 88° Lengths of Secants, Tangents, & Chords 2 Chords 2 Secants Tangent & Secant π¦ π π π‘ π π§ π₯ π π§ πβ’π = πβ’π π€ π‘2 = π¦(π¦ + π§) π¦ π€(π€ + π₯) = π¦(π¦ + π§) Ex. 3 & 4 Find length of π₯. Find the length of π. 8 3 15 π₯ π 7 5 πβ’π = πβ’π (3) β’ (7) = (π₯) β’ (5) 21 = 5π₯ 4.2 = π₯ π‘2 = π¦(π¦ + π§) 152 = 8(8 + π) 225 = 64 + 8π 161 = 8π 20.125 = π Ex.5: 2 Secants β’ Find the length of π₯. 20 π€(π€ + π₯) = π¦(π¦ + π§) 14(14 + 20) = 16(16 + π₯) (34)(14) = 256 + 16π₯ 476 = 256 + 16π₯ 220 = 16π₯ 13.75 = π₯ 14 16 π₯ Ex.6: A little bit of everything! Solve for π first. π€(π€ + π₯) = π¦(π¦ + π§) 12 π 175° 9(9 + 12) = 8(8 + π) 186 = 64 + 8π 9 8 60° π = 15.6 π° π Next solve for π Lastly solve for ποπ π‘2 = π¦(π¦ + π§) πο1 = ½(π₯ β π¦) π2 = 8(8 + 15.6) ποπ = ½(175 β 60) π2 = 189 ποπ = 57.5° π = 13.7 What have we learned?? β’ When dealing with angle measures formed by intersecting secants or tangents you either add or subtract the intercepted arcs depending on where the lines intersect. β’ There are 3 formulas to solve for segments lengths inside of circles, it depends on which segments you are dealing with: Secants, Chords, or Tangents. Challenge #1 D 6 C 30° A 3 Q 5 2 1 4 E 25° π΄πΉ is a diameter, ππ΄πΊ = 100°, ππΆπΈ = 30° ππΈπΉ = 25° F πΉπππ π‘βπ ππππ π’ππ ππ πππ ππ’ππππππ ππππππ 100° G Challenge #2 Find the value of the variable. π§ 8 6 7 16 π¦ 8 Challenge #3 Find the value of the variable. 20 14 6.5 3 π 16 π₯ 7 D 6 C 30° E A 3 100° Q 2 1 5 4 G 25° F mο1 ο½ mFG ο½ 80 mο2 ο½ m AG ο½ 100 55 1 ο½ 22.5 mο3 ο½ (mCE ο« mEF ) ο½ 2 2 80 ο« 155 1 ο½ 117.5 mο4 ο½ (mGF ο« m ACE ) ο½ 2 2 mο5 ο½ 180 ο 117.5 ο½ 62.5 1 1 mο6 ο½ (m AG ο mCE ) ο½ (100 ο 30) ο½ 35 2 2 Find the value of the variable. 6 7 π§ 8 π¦ (6 + 8) 6 = (7 + π¦) 7 84 = 49 + 7π¦ 35 = 7π¦ π¦ = 5 16 8 π§2 = (16 + 8) 8 π§2 = 192 π§ = 13.9 Find the value of the variable to the nearest tenth. 6.5 20 3 14 π 16 7 π₯ (20 + 14) 14 476 220 π₯ = = = = (π₯ + 16) 16 16π₯ + 256 16π₯ 13.8 6.5π = 3 β 7 6.5π = 21 π = 3.2