| In a \[\Delta ABC,\]\[\angle A+\frac{1}{2}\angle B+\angle C=140{}^\circ ,\] then \[\angle B\] is [SSC (CGL) Mains 2014] |
A) \[50{}^\circ \]
B) \[80{}^\circ \]
C) \[40{}^\circ \]
D) \[60{}^\circ \]
Correct Answer: B
Solution :
| In \[\Delta ABC,\]\[\angle A+\angle B+\angle C=180{}^\circ \] |
| Now, \[\angle A+\frac{\angle B}{2}+\angle C=140{}^\circ \] |
| \[\angle A+\angle B+\angle C=140{}^\circ +\frac{\angle B}{2}\] |
| \[180{}^\circ -140{}^\circ =\frac{\angle B}{2}\]\[\Rightarrow \]\[\angle B=80\] |
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