Question

In the given figure, in a ∆ABC, BE ⊥ AC, ∠EBC = 40° and ∠DAC = 30°. Find the values of x, y and z.
Figure

Answer 1

$$\begin{gathered} \mathrm{In}∆\mathrm{BCE}, \\ \angle \mathrm{CBE}+\angle \mathrm{CEB}+\angle \mathrm{BCE}=180°\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\right) \\ \Rightarrow 40°+90°+x°=180° \\ \Rightarrow x°+130°=180° \\ \Rightarrow x°=180°-130° \\ \Rightarrow x°=50° \\ \mathrm{Therefore}, \\ x=50 \\ In∆ADC, \\ \angle ADC+\angle ACD+\angle CAD=180°\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\right) \\ \Rightarrow \angle ADC+50°+30°=180° \\ \Rightarrow \angle ADC=180°-80° \\ \Rightarrow \angle ADC=100° \\ \mathrm{Now}, \\ \angle ADB+\angle ADC=180°(\mathrm{Linear}\mathrm{pair}) \\ \Rightarrow y°+100°=180° \\ \Rightarrow y°=180°-100° \\ \Rightarrow y°=80° \\ \mathrm{Therefore}, \\ y=80 \end{gathered}$$

$$\begin{gathered} \mathrm{Let},\text{l}\mathrm{ine}\mathrm{segment}sAD\mathrm{and}BE\mathrm{intersect}\mathrm{at}F. \\ \mathrm{Therefore},s\mathrm{ide}EF\mathrm{of}△AEF\text{is}\mathrm{produced}\mathrm{to}B. \\ \angle AFB=\angle FAE+\angle AEF(\mathrm{Exterior}\mathrm{angle}\mathrm{property}) \\ \Rightarrow z°=30°+90° \\ \Rightarrow z°=120° \\ \mathrm{Therefore},z=120. \end{gathered}$$