Question

ABCD is a rhombus. If ∠ACB = 40°, find ∠ADB.

Answer 1

$$\begin{gathered} \mathrm{In}\mathrm{a}\mathrm{rhombus},\mathrm{the}\mathrm{diagonals}\mathrm{are}\mathrm{perpendicular}. \\ \therefore \angle \mathrm{BPC}=90° \\ \mathrm{From}∆\mathrm{BPC},\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{angles}\mathrm{is}180°. \\ \therefore \angle \mathrm{CB}P+\angle \mathrm{BP}C+\angle PBC=180° \\ \angle \mathrm{CB}P=180°-\angle \mathrm{BP}C-\angle PBC \\ \angle \mathrm{CBP}=180°-40°-90°=50° \\ \angle ADB=\angle \mathrm{CBP}=50°(alternateangle) \end{gathered}$$