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1 PROBLEM 1A PROBLEM 2A PROBLEM 1B PROBLEM 2B PROBLEM 3A PROBLEM 4BPROBLEM 4A PROBLEM 3B STANDARD 13 ANGLES: INTRODUCTION PROBLEM 5BPROBLEM 5A ANGLES:

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Presentation on theme: "1 PROBLEM 1A PROBLEM 2A PROBLEM 1B PROBLEM 2B PROBLEM 3A PROBLEM 4BPROBLEM 4A PROBLEM 3B STANDARD 13 ANGLES: INTRODUCTION PROBLEM 5BPROBLEM 5A ANGLES:"— Presentation transcript:

1 1 PROBLEM 1A PROBLEM 2A PROBLEM 1B PROBLEM 2B PROBLEM 3A PROBLEM 4BPROBLEM 4A PROBLEM 3B STANDARD 13 ANGLES: INTRODUCTION PROBLEM 5BPROBLEM 5A ANGLES: CLASSIFICATION ADJACENT ANGLES ANGLE ADDITION POSTULATE SUPPLEMENTARY ANGLES LINEAR PAIR COMPLEMENTARY ANGLES VERTICAL ANGLES END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2 2 STANDARD 13: Students prove relationships between angles in polygons using properties of complementary, supplementary, vertical and exterior angles. ESTÁNDAR 13: Los estudiantes prueban relaciones entre ángulos en polígonos usando propiedades de ángulos complementarios, suplementarios, verticales y ángulos exteriores. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3 3 STANDARD 13 What is an angle? How many ways may we call this angle? ABC CBA B 1 A B C 1 A B C Common endpoint called: VERTEX Non-collinear ray Non-collinear Ray INTERIOR EXTERIOR A B C PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4 4 STANDARD 13 ANGLES A C F B D E 1 2 3 4 How else may we call 2? DFC or CFD May be called F? No, because point F has more than two rays departing from it. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5 5 STANDARD 13 ANGLES CLASSIFICATION ACUTE ANGLE: OBTUSE ANGLE: RIGHT ANGLE: 0° < ANGLE < 90° ANGLE = 90° 90° < ANGLE < 180° STRAIGHT ANGLE: ANGLE = 180° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6 6 STANDARD 13 ANGLE RELATIONSHIPS A C F B D E 1 2 3 4 2 and 3 are ADJACENT ANGLES because they have a COMMON RAY FC ADJACENT ANGLES: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7 7 STANDARD 13 ANGLE RELATIONSHIPS A C F B D E 1 2 3 4 ADJACENT ANGLES: Can you see other adjacent angles in the figure? EFD DFC and ; common ray FD PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8 8 STANDARD 13 ANGLE RELATIONSHIPS A C F B D E 1 2 3 4 ADJACENT ANGLES: Can you see other adjacent angles in the figure? CFB AFB and ; common ray FB PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9 9 STANDARD 13 ANGLE RELATIONSHIPS A C F B D E 1 2 3 4 ADJACENT ANGLES: Can you see other adjacent angles in the figure? EFC AFC and ; common ray FC PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10 10 STANDARD 13 ANGLE RELATIONSHIPS A C F B D E 1 2 3 4 ADJACENT ANGLES: Are and adjacent? EFD BFA NO, because they don’t have a common ray. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11 11 STANDARD 13 ANGLE RELATIONSHIPS A C F B D E 1 2 3 4 2 and 3 are ADJACENT ANGLES because they have a COMMON RAY FC ADJACENT ANGLES: Can you see other adjacent angles in the figure? EFD DFC and CFB BFA and EFC AFC and Are and adjacent? EFD BFA NO, because they don’t have a common ray. ; common ray FD ; common ray FB ; common ray FC PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12 12 STANDARD 13 ANGLE RELATIONSHIPS: ANGLE ADDITION POSTULATE (Simplified) L N M K KNL m + LNM m =MNK m Now may you find EFB? m A C F B D E EFB= m EFD m + DFC m +CFB m It means: MEASURE of angle KNL PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

13 13 STANDARD 13 SUPPLEMENTARY ANGLES R S T 75° What kind of angles are these two angles? F H G 105° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14 14 STANDARD 13 SUPPLEMENTARY ANGLES What kind of angles are these two angles? 75° F H G 105° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

15 15 STANDARD 13 SUPPLEMENTARY ANGLES What kind of angles are these two angles? TSR m + FGH m = 180° Two angles that added together are 180°, are called: SUPPLEMENTARY ANGLES. 75° 105° TSR FGH Angles and are SUPPLEMENTARY! F H G 105° R S T 75° and each one is the SUPPLEMENT of the other. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

16 16 STANDARD 13 SUPPLEMENTARY ANGLES 64° Find the measure of the supplement of the angle below: 180° – 64° = 116° 116° Supplement: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

17 17 STANDARD 13 SUPPLEMENTARY ANGLES Find the measure of the supplement of the angle below: 180° –145° = 35° 145° Supplement: 35° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

18 18 STANDARD 13 SUPPLEMENTARY ANGLES 70° Find the measure of the supplement of the angle below: 180° – 70° = 110° 110° Supplement: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

19 19 STANDARD 13 105° 75° Let’s take a closer look at the angle formation we got before: They are supplementary: 105° + 75° = 180° They are adjacent: have a common ray. They are LINEAR PAIR: because they are all of the above! PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

20 20 32X + 108X +10+= 180 40X +20 = 180 -20 40X = 160 40 X= 4 This is a linear pair, so both angles are supplementary: m RST = 32X + 10 = 32( )+10 = 128 + 10 = 138° m UST =180-138 = 42° 4 R S U T 32X + 10 8X + 10 STANDARD 13 FIND: RST and UST PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

21 21 20X + 105X +20+= 180 25X +30 = 180 -30 25X = 150 25 X= 6 This is a linear pair, so both angles are supplementary: m KLN = 20X + 10 = 20( )+10 = 120 + 10 = 130° m MLN =180-130 = 50° 6 K L M N 20X + 10 5X + 20 STANDARD 13 FIND: KLN and MLN PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

22 22 STANDARD 13 COMPLEMENTARY ANGLES M L K 60° A B C 30° What kind of angles are these two angles? PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

23 23 STANDARD 13 COMPLEMENTARY ANGLES A B C 30° What kind of angle are these two angles? 60° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

24 24 STANDARD 13 COMPLEMENTARY ANGLES What kind of angles are these two angles? 60° 30° KLM m + ABC m = 90° Two angles that added together are 90°, are called: COMPLEMENTARY ANGLES. KLM ABC Angles and are complementary! M L K 60° A B C 30° and each one is the COMPLEMENT of the other. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

25 25 STANDARD 13 SUPPLEMENTARY ANGLES Find the measure of the COMPLEMENT of the angle below: 90° – 35° = 55° Complement 55° 35° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

26 26 STANDARD 13 COMPLEMENTARY ANGLES Find the measure of the COMPLEMENT of the angle below: 90° – 67° = 23° 67° 23° Complement PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

27 27 STANDARD 13 COMPLEMENTARY ANGLES Find the measure of the COMPLEMENT of the angle below: 90° – 39° = 51° Complement 51° 39° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

28 28 9X + 1020X-7+= 90° 29X +3 = 90 -3 29X = 87 29 Both angles are complementary: m TRS = 9X+10 = 9( ) + 10 = 27 + 10 m TRU =90-37 3 X = 3 = 37° = 53° R S T U 9X + 10 20X - 7 STANDARD 13 FIND: TRS and TRU PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

29 29 7X + 1213X-2+= 90° 20X+10 = 90 -10 20X = 80 20 Both angles are complementary: m FGT = 7X+12 = 7( ) + 12 = 28+ 12 m HGT =90-40 4 X = 4 = 40° = 50° G F T H 7X + 12 13X - 2 STANDARD 13 FIND: FGT and HGT PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

30 30 STANDARD 13 q s Y U V X Z VERTICAL ANGLES UZYVZX UZY= m VZX m UZVYZX UZV= m YZX m VERTICAL ANGLES are congruent, that is they have the same measure. Observe they are straight lines Means congruence VERTICAL ANGLES are nonadjacent angles formed by two intersecting lines. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

31 31 STANDARD 13 VERTICAL ANGLES M K V X L Are and vertical angles?KLM VLX No, although they look to be formed by two intersecting lines, in reality LK and LX are two rays that don’t lie in the same line and therefore the angles aren’t vertical. Note: two red lines in one angle means that such angle is different to the other with just one red line. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

32 32 Both angles are vertical : 20 + 5X = 8X + 5 - 20 -20 5X = 8X -15 -8X -3X = -15 -3 X= 5 m ABC = 20 + 5X = 20 + 5( ) m CBE =180-45 5 = 45° = 135° m CBE m ABC and are linear pair C A D E B 20 + 5X 8X + 5 STANDARD 13 FIND: ABC and CBE = 20 + 25 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

33 33 Both angles are vertical : 10 + 4X = 6X + 4 - 10 -10 4X = 6X -6 -6X -2X = -6 -2 X= 3 m SKR = 10 + 4X = 10 + 4( ) m TKS =180-22 3 = 22° = 158° m TKS m SKR and are linear pair R S T U K 10 + 4X 6X + 4 STANDARD 13 FIND: SKR and TKS = 10 + 12 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

34 34 40° STANDARD 13 m KJO + m NJO = 180° m KJO + 40° = 180° - 40 m KJO = 140° m MJN + m NJO = 90° m MJN + 40° = 90° - 40 m MJN = 50° m MJO = 90° Two perpendicular lines form 4 right angles, so: m NJO = m LJK If by vertical angles: m MJK = 90° + 40° = 130° 140° 50° then 130° FIND: K L M N O J KJO, MJN,MJO MJK and 40° By supplementary angles: By complementary angles: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

35 35 STANDARD 13 m KJO + m NJO = 180° m KJO + 35° = 180° - 35 m KJO = 145° m MJN + m NJO = 90° m MJN + 35° = 90° - 35 m MJN = 55° m MJO = 90° Two perpendicular lines form 4 right angles, so: m NJO = m LJK If by vertical angles: m MJK = 90° + 35° = 125° 145° 55° then 125° FIND: K L M N O J KJO, MJN,MJO MJK and 35° By supplementary angles: By complementary angles: 35° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

36 36 STANDARD 13 A E B C D (12x + 10)° (5x + 8)° (3x + 2)° Find and DEC, CEB,AEB,if rays ED and EA are opposite at point E. DEC m AEB m + + CEB m DEA= m m 180° by definition of a STRAIGHT ANGLE. Now using all the above information and the figure: (3x + 2) + (12x + 10) + (5x + 8) = 180° 3x + 12x + 5x + 2 + 10 + 8 = 180 20x + 20 = 180 -20 20x = 160 20 x = 8 Let’s find the measure of the angles: m DEC = 3x + 2 = 3( ) + 2 8 = 24 + 2 = 26° m CEB = 12x + 10 = 12( ) + 10 8 = 96 + 10 = 106° m AEB = 5x + 8 = 5( ) + 8 8 = 40 + 8 = 48° Verifying solution: 26° + 106° + 48° = 180° 180° = 180° DEA, If rays ED and EA are opposite at point E then is a STRAIGHT ANGLE and the following holds true by Angle Addition Postulate: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

37 37 STANDARD 13 B F C D E (13x + 12)° (3x + 5)° (4x + 3)° Find and EFD, DFC,BFC,if rays FE and FB are opposite at point F. EFD m BFC m + + DFC m EFB= m m 180° by definition of a STRAIGHT ANGLE. Now using all the above information and the figure: (4x + 3) + (13x + 12) + (3x + 5) = 180° 4x + 13x + 3x + 3 + 12 + 5 = 180 20x + 20 = 180 -20 20x = 160 20 x = 8 Let’s find the measure of the angles: m EFD = 4x + 3 = 4( ) + 3 8 = 32 + 3 = 35° m DFC = 13x + 12 = 13( ) + 12 8 = 104 + 12 = 116° m BFC = 3x + 5 = 3( ) + 5 8 = 24 + 5 = 29° Verifying solution: 35° + 116° + 29° = 180° 180° = 180° EFB, If rays FE and FB are opposite at point F then is a STRAIGHT ANGLE and the following holds true by Angle Addition Postulate: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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