Question : In the figure BCDE is a square and ABC is equilateral triangle then $\angle \mathrm{ADC}$ is:
Option 1: 45°
Option 2: 30°
Option 3: 60°
Option 4: 15°
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Correct Answer: 15°
Solution :
According to the question, $\angle BCD = 90° $and $\angle ACB = 60°$ $\angle ACD = (90 + 60)° = 150°$ Now, AB = BC = AC (∵ ABC is an equilateral triangle) Also, BC = CD = DE = EB (∵ BCDE is a square) ⇒ AC = CD = BC So, ΔACD is an isosceles triangle where AC = CD and $\angle CAD = \angle ADC$ ⇒ $\angle ADC = \frac{180°−150°}{2}$ = 15° Hence, the correct answer is 15°.
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