Angle 問題 @ nzphddr的部落格

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Angle 問題

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23. (2n - 4) x 90o = 156o x n 180n - 360 = 156n 24n = 360 n = 15 22. x = (Exterior of a pentagon) + (Exterior of a square) = (360o/5) + (360o/4) = 72o + 90o = 162o 21. BAE + ABE + AEB = 180o ( sum of D) BAE + 30o + 30o = 180o BAE = 120o ABD = AED = 30o (given) BD = ED (given) AB = AE (opp. equal in D) DABD DAED (SAS) BAC = EAC (corr. s of cong. D) But BAC + EAC = BAE = 120o Hence, BAC = 60o BAC + ABC + ACB = 180o ( sum of D) 60o + (30o + 30o) + ACB = 180o ACB = 60o BAC = ABC = ACB = 60o Hence, ABC is an equilateral D. 20. BCF = ACF (given) But BCF + ACF = 60o (int. of equil. D) Hence, BCF = ACF = 30o AEF + DCF = 180o (int. s, EA//CD) AEF + (90o + 30o) = 180o AEF = 60o 19. (a) A = C (given) BC = AB (opp. s equal in D) 3x + 4 = 5x 2x = 4 x = 2 (b) Perimeter = {5(2) + [3(2) + 4] + 3(2)} cm = 26 cm 18. Draw a st. line HCK which // AB and //DE. (H on the left of C, while K on the right.) HCB + ABC = 180o (int. s, HCK // AB) HCB + 120o = 180o HCB = 60o KCD + EDC = 180o (int. s, HCK // DE) KCD + 130o = 180o KCD = 50o HCB + BCD + KCD = 180o (adj. s on a st. line) 60o + c + 50o = 180o c = 70o 17. BAE + EAH = 180o (adj. s on a st. line) BAE + (2x - 10o) = 180o BAE = 190o - 2x DEA = FEG (vert. opp. s) DEA = x + 60o Sum of int. s = (2 x 5 - 4) x 90o (190o - 2x) + 2x + 2x + (3x - 40o) + (x + 60o) = 540o 6x + 210o = 540o x = 55o 16. Let each of the two vertically opposite s be y. Consider the upper (isos.) D: (2x + 15o) + (2x + 15o) + y = 180o y = 180o - 2(2x + 15o) y = 150o - 4x Consider the lower D: The exterior is equal to the sum of the two int. opp. s. y + 72o = 126o (150o - 4x) + 72o = 126o 4x = 96o x = 24o =

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