| Consider the figure & find \[\alpha +\beta +\gamma \] if \[\frac{\alpha }{2}=\frac{\beta }{3}=\frac{\gamma }{2}\] |
 |
A) \[100{}^\circ \]
B) \[120{}^\circ ~\]
C) \[140{}^\circ \]
D) \[210{}^\circ \]
Correct Answer: C
Solution :
| \[\angle F=\alpha +\gamma \] & \[\angle D=\angle \beta +\gamma \] |
| Quadrilateral AFED is cyclic |
| \[\Rightarrow \angle F+\angle D=180{}^\circ \] \[\Rightarrow \alpha +\beta +2\gamma =180{}^\circ \] \[\Rightarrow 2k+3k+4k=180{}^\circ \left\{ let\frac{\alpha }{2}=\frac{\beta }{3}=\frac{\gamma }{2}=k \right\}\] |
| \[\Rightarrow k=20{}^\circ \] \[\Rightarrow \alpha +\beta +\gamma =7k=140{}^\circ \] |
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