# In the given figure, ΔABC is an equilateral triangle and ΔBDC is an isosceles right triangle, right-angled at D. Then ∠ACD = ? A. 60°B. 90°C. 120°D. 105°

Since we know all the angles in an equilateral triangle is of 60°.

So, ABC = ACB = CAB = 60° …(i)

Also for an isosceles triangle, the angles opposite to equal sides are equal.

So, DBC = DCB = x (lets say)

Also sum of all angles in a triangle = 180°.

So, in ΔBDC,

DBC + DCB + BDC = 180°

x + x + 90 = 180 {since BDC = 90°}

2x = 90

x = 45°

so DCB = 45 (ii)

And ACD = ACB + DCB = 60° + 45° = 105° {from (i) and (ii)}

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