In the given figure, ΔABC is an equilateral triangle and ΔBDC is an isosceles right triangle, right-angled at D. Then ∠ACD = ?

Question

In the given figure, ΔABC is an equilateral triangle and ΔBDC is an isosceles right triangle, right-angled at D. Then ∠ ACD = ?

Philoid · Accepted Answer

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 (let’s 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)}