Triangle Exterior Angle Sum Theorem

Name Key PD: Date: October 10, 2023 2.a.2 Triangle Exterior Angle Sum Theorem Given AABC with m angle B = 62 ^ n and side AC extended to D, as shown below. 2) 1) In the diagram below, RCBT and AABC are shown with m angle A = 60 and m angle ABT = 125 . 60° X + 60 = 125 x = 65 deg 3) In A.4BC shown below, side AC is extended to point D with m∠DAB (180-3x). m angle B = (6x - 40) deg . and m angle C = (x + 20) deg . B 62 180 - 6Z = 118 119/2=59° 159 59 X A C D T Which value of x makes AB = CB? 115/x R C What is m/ACR? MCACR=180-65115 125 B x = 121 deg In the diagram below, A.4BC is shown with AC extended through point D. B (-40) 5) In the diagram of APQR shown below, PR is extended to S, m angle P = 110 , m angle Q = 4x . and m angle QRS = x ^ 2 + 5x . x² x ^ 2 + 5x = 110 + 4x (x + 10)(x - 10) = 0 X = 10% (180-3) A 6x - 1/0 + x + 20 = 180 - 3x 4x = 200 ↓ C 7x - 20 = 180 - 3x 10x = 200 x = 20 ) V4x P 110 What is mZQ? m< Q=410=40 (x ^ 2 + 5x) ^ 2 R S. XF-11 in geonetry C 6x + 2 D m angle BCD = 6x + 2 . m angle BAC = 3x + 15 . and m angle ABC = 2x - 1 . what ? 3x + 15 + 2x - 1 = 6x + 2 5x + 14 = 6x + 2 In the diagram below of AABC, BC is extended to D. x ^ 2 - 6x + 2x - 3 = 9x + 27 x ^ 2 - 13x - 30 = (x - 15)(x + 3) = 0 x = 19 (Not drawn to scale) C If m angle A = x ^ 2 - 6x . m angle B = 2r - 3 , and m angle ACD = 9x + 27 . what is the value of x? D X = 15X≠-2 in geo (x + 20) ^ 2 7) In the diagram below of triangle HQP . side overline HP * is extended through P to T, m angle QPT = 6r + 20 , m angle HQP = x + 40 . and m angle PHQ = 4x - 5 . Find m angle QPT 6x + 20 = x + 40 + 4x - 5 x = 15 MLQPT=110° angle( 2PT = 6(rs) + 20 T Q (x + 40) deg (4x - 5) deg P H 8) AB is extended to point D. If m angle CAB = x + 40 , m angle ACB = 3x + 10 ,m angle CBD=6x, what is m∠CAB? D B X+40 A 2x = 50 x = 25 6x = 3x + 10 + x + 40 angle CAB = 2514O = 65 deg Solve for x. 9) T 5x+10 58 A U S 11x + 2 = 5x + 10 + 58 6x = 66 11) Find m∠FAB. 10) C 46 B W D 18x + 5 = 46 + (- 1) + 8x 10x = 40 x = 4 12) Find m∠YDC. C 55 13x - 3 = 3x + 2 + 55 10x = 60 D F M angle FAB = 13(6) - 3 m angle FAB = 75 deg c / 5 * x + 5 = 6x + 6 + 80 9x = 81 x = 9 MLYTC=15(9)+5 m subset YDC = 140 deg x = 11 (6x + 20) deg 18x + 5 X = 6 11x + 2 ) - 1 + 8x 13x - 3y 5x + 6 3x + 2