Pre-Calculus Homework Solutions 192

# Pre-Calculus Homework Solutions 192 - 55. The smallest...

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55. The smallest angle is opposite the side with the smallest measure, 20. Let x and 20 2 x be the measures of the segments formed by the angle bisector. By the Angle Bisector Theorem, } 20 x 2 x } 5 ¬ } 2 3 4 0 } . } 20 x 2 x } 5 ¬ } 2 3 4 0 } 30 x 5 ¬ 480 2 24 x 54 x 5 ¬ 480 x 5 ¬ } 8 9 0 } or 8 } 8 9 } 20 2 x 5 ¬ } 10 9 0 } or 11 } 1 9 } The segments have measures 8 } 8 9 } and 11 } 1 9 } . 56. By the Exterior Angle Inequality Theorem, m / 8 . m / 6, m / 8 . m / 3 1 m / 4, and m / 8 . m / 2. Thus, the measures of / 6, / 4, / 2, and / 3 are all less than m / 8. 57. By the Exterior Angle Inequality Theorem, m / 1 , m / 5 and m / 1 , m / 7. Thus, the measures of / 5 and / 7 are greater than m / 1. 58. By the Exterior Angle Inequality Theorem, m / 1 , m / 7 and m / 6 , m / 7. Thus, the measures of / 1 and / 6 are less than m / 7. 59. By the Exterior Angle Inequality Theorem, m / 2 . m / 6, m / 7 . m / 6, and m / 8 . m / 6. Thus, the measures of / 2, / 7, and / 8 are all greater than m / 6. 60. y 5 mx 1 b y 5 2 x 1 4 61. m 5 } 2 0 8 2 2 2 0 } 5 } 2 2 8 2 } or 4 y 5 mx 1 b y 5 4 x 1 ( 2 8) y 5 4 x 2 8 62. m 5 } 2 0 1 2 2 6 2 } 5 } 2 2 6 3 } or 2 y 2 y 1 5 ¬ m ( x 2 x 1 ) y 2 6 5 ¬ 2( x 2 2) y 2 6 5 ¬ 2 x 2 4
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## This note was uploaded on 12/19/2011 for the course MAC 1140 taught by Professor Dr.zhan during the Fall '10 term at UNF.

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