A) \[160{}^\circ \]
B) \[180{}^\circ \]
C) \[190{}^\circ \]
D) \[170{}^\circ \]
E) None of these
Correct Answer: B
Solution :
| [b] Let the angles of the quadrilateral be and \[3x,\]\[4x,\]\[5x\] and \[6x\]respectively. |
| Then, \[3x+4x+5x+6x=360{}^\circ \] |
| \[\Rightarrow \] \[18x=360{}^\circ \]\[\Rightarrow \]\[x=20{}^\circ \] |
| \[\therefore \] Smallest angle of the triangle |
| \[=3\times 20\times \frac{2}{3}=40{}^\circ \] |
| \[\therefore \] Largest angle of the triangle \[=40{}^\circ \times 2=80{}^\circ \] |
| \[\therefore \] Second largest angle of triangle |
| \[=180{}^\circ -(40{}^\circ +80{}^\circ )=60{}^\circ \] |
| and largest angle of the quadrilateral |
| \[=6x=6\times 20{}^\circ \] |
| \[=120{}^\circ \] |
| Hence, required sum \[=60{}^\circ +120{}^\circ =180{}^\circ \] |
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