Compute the sum \[p+q+r\]
A) \[180{}^\circ \]
B) \[70{}^\circ \]
C) \[190{}^\circ \]
D) \[80{}^\circ \]
Correct Answer: C
Solution :
(c): From the figure. \[150-p+70-p+p=180{}^\circ \] \[\Rightarrow \,\,\,p=220{}^\circ -180{}^\circ =40{}^\circ \] Since \[AE||BD,\text{ }q=p\] as they are alternate angles. In \[\Delta BCD,\text{ }\angle BDC=p\] (Alternate angles) \[70-p+p+r=180{}^\circ \Rightarrow r=110{}^\circ \] \[\therefore \] The required sum \[=\angle p+\angle q+\angle r=40{}^\circ +40{}^\circ +110{}^\circ =190{}^\circ \]
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