A) \[120{}^\circ \]
B) \[40{}^\circ \]
C) \[20{}^\circ \]
D) \[90{}^\circ \]
Correct Answer: A
Solution :
| Let \[\angle A=x\]and \[\angle B=y\] |
| \[\because \] \[\angle C=3\angle B\] \[\Rightarrow \]\[\angle C=3y\] |
| But \[\angle A+\angle B+\angle C=180{}^\circ \] |
| \[x+4y=180{}^\circ \] ?(i) |
| Also, \[\angle C=2\,\,(\angle A+\angle B)\] |
| \[3y=2\,\,(x+y)\] |
| \[3y=2x+2y\]\[\Rightarrow \]\[y=2x\] ?(ii) |
| Put the value of y in Eq. (i). |
| \[x+4\,\,(2x)=180{}^\circ \] |
| \[9x=180{}^\circ \]\[\Rightarrow \]\[x=20{}^\circ \]\[\Rightarrow \]\[y=40{}^\circ \] |
| \[\therefore \] \[\angle C=3y=3\times 40{}^\circ =120{}^\circ \] |
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