Question

Mark (✓) against the correct answer in each of the following:
In the given figure, ∆ABC is an equilateral triangle and ∆BDC is an isosceles right triangle, right-angled at D. Then ∠ACD = ?
(a) 60°
(b) 90°
(c) 120°
(d) 105°
Figure

Answer 1

(d) 105°

$$\begin{gathered} ∆ABC\mathrm{is}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}. \\ \mathrm{So},\mathrm{every}\mathrm{angle}\mathrm{of}∆ABC\mathrm{is}60°. \\ \mathrm{Therefore},\angle ACB=60° \\ \mathrm{Now},∆BCD\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}\mathrm{and}\angle BDC=90°. \\ \mathrm{Here},\angle BCD=\angle CBD \\ \mathrm{Now},\angle BCD+\angle CBD+\angle BDC=180°\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\right) \\ \Rightarrow \angle BCD+\angle CBD+90°=180° \\ \Rightarrow 2\angle BCD=90° \\ \Rightarrow \angle BCD=45° \\ \mathrm{From}\text{the}\mathrm{figure}, \\ \angle ACD=\angle ACB+\angle BCD \\ =60°+45° \\ \therefore \angle ACD=105° \end{gathered}$$