Angles.Triangles .polygons

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°


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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.