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Angles.Triangles .polygons

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28. (a) ΔCDE is equilateral. Hence, ∠EDC = 60° ∠ADC = 90° (int. ∠ of square) ∠ADE + ∠EDC = 90° ∠ADE + 60° = 90° ∠ADE = 30° ΔDEA is an isos. Δ with DE = DA Hence, ∠DEA = ∠DAE = x In ΔDEA : ∠DEA + ∠DAE + ∠ADE= 180° (∠sum of Δ) x + x + 30° = 180° x = 75° (b) ∠BAD = 90° (int. ∠ of square) ∠DAE + ∠EAB = 90° 75° + ∠EAB = 90° ∠EAB = 15° ===== 29. (a) Let n be the number of sides of regular polygon I. (n - 2) x 180° / n = 2 x (360°/n) 180n - 360 = 720 180n = 1080 n = 6 The number of sides of polygon I = 6 (b) An interior angle of polygon I = (6 - 2) x 180° / 6 = 120° (An exterior angle of polygon II) + (An external angle of polygon III) = 120° A possible solution is : 90° + 30° = 120° (360°/4) + (360°/12) = 120 A pair of possible number of sides for polygon II and polygon III is 4 and 12 respectively.

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