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# 1 STANDARD 21 A secant and a tangent Intersecting at a point of tangency Two secants intersecting at interior of circle Two tangents, two secants, and.

## Presentation on theme: "1 STANDARD 21 A secant and a tangent Intersecting at a point of tangency Two secants intersecting at interior of circle Two tangents, two secants, and."— Presentation transcript:

1 STANDARD 21 A secant and a tangent Intersecting at a point of tangency Two secants intersecting at interior of circle Two tangents, two secants, and a tangent and a secant, intersecting at an exterior pt. PROBLEM 1aPROBLEM 1b PROBLEM 2aPROBLEM 2b PROBLEM 3PROBLEM 4 END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2 Standard 21: Students prove and solve problems regarding relationships among chords, secants, tangents, inscribed angles, and inscribed and circumscribed polygons of circles. Los estudiantes prueban y resuelven problemas relacionados con cuerdas, secantes, tangentes, ángulos inscritos y polígonos inscritos y circunscritos a círculos. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3 KLM m = K M L N O m KOL 1 2 KLN m = If a secant and a tangent intersect at the point of tangency, then the measure of each angle formed is one-half the measure of its intercepted arc. Si una secante y una tangente intersecan en el punto de tangencia, entonces la medida de cada ángulo formado es la mitad de su arco intersecado. m KL 1 2 STANDARD 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4 K M L N O If LOK m = 255°findKLM= m ? 255° 360°-255°=105° 105° KLM m = m KL 1 2 KLM= m 1 2 ( ) 105° KLM= m 52.5° KLm= STANDARD 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5 P O R N S If RSP m = 202°findORP = m ? 202° 360°-202°=158° 158° ORP m = m RP 1 2 ORP = m 1 2 ( ) 158° ORP = m 79° RPm= STANDARD 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6 Q R P S T QRP m = ST m + QP m 1 2 ( ) If two secants intersect in the interior of a circle, then the measure of an angle formed is one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. Si dos secantes intersecan en el interior de un círculo, entonces la medida de un ángulo formado es la mitad de la suma de las medidas de los arcos intersecados por el ángulo y su ángulo vertical. STANDARD 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7 L R J N G If LN m = 70° and JG m = 210° findNRG = m ? First we findLRN= m ? LRN m = 1 2 ( ) LRN= m 1 2 ( ) + LRN= m 1 2 ( ) 280° LRN= m 140° LRN m + NRG m = 180° 140° +NRG= m 180° -140° NRG= m 40° JG m LN m + 70° 210° STANDARD 21 We have a linear pair: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8 D R H E I If DE m = 90° and HI m = 200° findERI = m ? First we findDRE= m ? DRE m = 1 2 ( ) DRE= m 1 2 ( ) + DRE= m 1 2 ( ) 290° DRE= m 145° DRE m + ERI m = 180° 145° +ERI= m 180° -145° ERI= m 35° HI m DE m + 90° 200° STANDARD 21 We have a linear pair: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9 Two tangents Secant and a tangent Two secants M L K A B C D E Q R S T LNM m - LM m SR m - TR m 1 2 ( ) DE m - BC m 1 2 ( ) 1 2 N If two secants, a secant and a tangent, or two tangents intersect in the exterior of a circle, then the measure of the angle formed is one-half the possitive difference of the measures of the intercepted arcs. Si dos secantes, una secante y una tangente, o dos tangentes intersecan en el exterior de un círculo, entonces la medida de el ángulo formado es la mitad de la diferencia positiva de las medidas de los arcos intersecados. LKM m = TQR m = BAC m = STANDARD 21

10 M L K LNM m - LM m N LKM m = 1 2 ( ) If LNM m = 220° LKM m = 1 2 ( ) LMm=360°- 220°= 140° 140° 220°140°- LKM m = 1 2 (80°) LKM m = 40° find LKM m =? and we know that 220° There are two tangents, so: STANDARD 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11 We have a secant and a tangent, so: Q R S T SR m - TR m 1 2 ( ) TQR m = If TQR m = 55° andSRm=130°findTVSm V 55°= 1 2 ( ) TRm130° - 55° = 1 2 ( ) TRm130° - (2) 110° = 130° -TRm 110° = 130° -TRm -130° TR-m=-20 TRm=20° TVS m + TRmSRm+= 360° TVS m + 20° + 130° = 360° TVS m +150° = 360° -150° TVS m = 210° Now in a circle we have 360°, so: 130°55° 210° STANDARD 21 (-1) PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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