Question

In Fig., measures of some angles are indicated. Find the value of x.

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ \angle \mathrm{AED}+120°=180°\left(\text{Linear pair}\right) \\ \Rightarrow \angle \mathrm{AED}=180°-120°=60° \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180°. \\ \mathrm{Therefore},\mathrm{for}∆\mathrm{ADE},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{ADE}+\angle \mathrm{AED}+\angle \mathrm{DAE}=180° \\ \Rightarrow 60°+\angle \mathrm{ADE}+30°=180° \\ \mathrm{Or}, \\ \angle \mathrm{ADE}=180°-60°-30°=90° \\ \mathrm{From}\mathrm{the}\mathrm{given}\mathrm{figure},\mathrm{we}\mathrm{can}\mathrm{also}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{FDC}+90°=180°\left(\text{Linear pair}\right) \\ \Rightarrow \angle \mathrm{FDC}=180°-90°=90° \\ \mathrm{Using}\mathrm{the}\mathrm{above}\mathrm{rule}\mathrm{for}∆\mathrm{CDF},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{CDF}+\angle \mathrm{DCF}+\angle \mathrm{DFC}=180° \\ \Rightarrow 90°+\angle \mathrm{DCF}+60°=180° \\ \angle \mathrm{DCF}=180°-60°-90°=30° \\ \mathrm{Also}, \\ \angle \mathrm{DCF}+\mathrm{x}=180°\left(\text{Linear pair}\right) \\ \Rightarrow 30°+\mathrm{x}=180° \\ \mathrm{Or}, \\ \mathrm{x}=180°-30°=150° \end{gathered}$$