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Pre-Calculus Homework Solutions 191

# Pre-Calculus Homework Solutions 191 - 44 AD CD 12 CD CD BD...

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44. } C AD D } 5 ¬ } C B D D } } C 1 D 2 } 5 ¬ } C 4 D } ( CD ) 2 5 ¬ 48 CD 5 ¬ Ï 48 w or 4 Ï 3 w ( CB ) 2 5 ( CD ) 2 1 ( DB ) 2 ( CB ) 2 5 (4 Ï 3 w ) 2 1 4 2 ( CB ) 2 5 48 1 16 ( CB ) 2 5 64 CB 5 Ï 64 w or 8 n CED , n ACB ,so } C A D B } 5 } D CB E } . } 1 4 2 Ï 1 3 w 4 } 5 ¬ } D 8 E } 16 DE 5 ¬ 32 Ï 3 w DE 5 ¬ 2 Ï 3 w 45. Given: / PQR is a right angle. Q w S w is an altitude of n PQR. Prove: n PSQ , n PQR n PQR , n QSR n PSQ , n QSR Proof: 46. Given: / ADC is a right angle. D w B w is an altitude of n ADC . Prove: } D AB B } 5 } D CB B } Proof: It is given that / ADC is a right angle and D w B w is an altitude of n ADC. n ADC is a right triangle by the definition of a right triangle. Therefore, n ADB , n DCB ,because if the altitude is drawn from the vertex of the right angle to the hypotenuse of a right triangle, then the two triangles formed are similar to the given triangle and to each other. So } D AB B } 5 } D CB B } by definition of similar polygons. 47. Given: / ADC is a right angle. D w B w is an altitude of n ADC.
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