Question

In a ∆ABC, ∠ABC = 100°, ∠BAC = 35° and BD ⊥ AC meets side AC in D. If BD = 2 cm, find ∠C and length DC.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180°. \\ \mathrm{Therefore},\mathrm{for}\mathrm{the}\mathrm{given}△\mathrm{ABC},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{ABC}+\angle \mathrm{BAC}+\angle \mathrm{ACB}=180° \\ 100°+35°+\angle \mathrm{ACB}=180° \\ \Rightarrow \angle \mathrm{ACB}=180°-135° \\ \Rightarrow \angle \mathrm{ACB}=45° \\ \angle \mathrm{C}=45° \\ \mathrm{If}\mathrm{we}\mathrm{apply}\mathrm{the}\mathrm{above}\mathrm{rule}\mathrm{on}△\mathrm{BCD},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{BCD}+\angle \mathrm{BDC}+\angle \mathrm{CBD}=180° \\ 45°+90°+\angle \mathrm{CBD}=180°(\because \angle \mathrm{ACB}=\angle \mathrm{BCD}\mathrm{and}\mathrm{BD}\perp \mathrm{AC}) \\ \Rightarrow \angle \mathrm{CBD}=180°-135° \\ \Rightarrow \angle \mathrm{CBD}=45° \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sides}\mathrm{opposite}\mathrm{to}\mathrm{equal}\mathrm{angles}\mathrm{have}\mathrm{equal}\mathrm{length}. \\ \mathrm{Thus}, \\ \mathrm{BD}=\mathrm{DC} \\ \mathrm{DC}=2\mathrm{cm} \end{gathered}$$