Angle ABC = ½(angle BAC) since angle ACD and angle BCD are equal and length BC = length AD + length AC
It implies;
a +½(100)+100 = 180
a = 180-150
a = 30°
a is angle ACB.
b = ½(a)
b = 15°
b is angle BCD = angle ACD
Therefore, the required angle, alpha is;
Let it be c.
It implies;
100+15+c = 180
c = 180-115
c = 65°
Again, c is the required angle alpha.
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