A) \[48{}^\circ \]
B) \[96{}^\circ \]
C) \[36{}^\circ \]
D) None of these
Correct Answer: A
Solution :
As, AB|| CD and BD cuts them \[\angle ABD=\angle BDC,\] so \[\angle BDC=x\] So, \[x+y+z=180{}^\circ \] In \[\Delta BCD,\]\[\frac{4}{3}y+y+z=180{}^\circ \]\[\Rightarrow \]\[\frac{7y}{3}+z=180{}^\circ \] \[\frac{7}{3}\left( \frac{3}{8}z \right)+z=180{}^\circ \]\[\Rightarrow \]\[\frac{7}{8}z+z=180{}^\circ \] \[\frac{15z}{8}=180{}^\circ \]\[\Rightarrow \]\[z=96{}^\circ \] So,\[y=\frac{3}{8}\times 96{}^\circ =36{}^\circ \]and\[x=\frac{4}{3}\times 36{}^\circ =48{}^\circ \]
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