A) \[48{}^\circ ,\,\,112{}^\circ \]
B) \[132{}^\circ ,\,\,\text{60}{}^\circ \]
C) \[48{}^\circ ,\,\,\text{132}{}^\circ \]
D) \[45{}^\circ ,\,\,\text{135}{}^\circ \]
Correct Answer: C
Solution :
\[x:y=4:11\] Let \[x=4z\] and \[y=11z\] then \[x+y=180{}^\circ \] (by linear pair) \[4z+11z=180{}^\circ \] \[15z=180{}^\circ \] \[z=\frac{180{}^\circ }{15}=12{}^\circ \] then \[x=4\times 12{}^\circ =48{}^\circ \] \[y=11\times 12=132{}^\circ \] Now, \[x=a=48{}^\circ \] (by vertically opposite angles) \[y=b=132{}^\circ \](by vertically opposite angles)
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