A) \[\frac{3}{7}\]
B) \[\frac{6}{7}\]
C) \[\frac{4}{7}\]
D) \[\frac{5}{7}\]
Correct Answer: B
Solution :
L.C.M. of \[\text{12}0=\text{1}0\text{4}\times \text{1}+\text{16}\]and \[\text{1}0\text{4}=\text{16}\times \text{6}+\text{8}\] \[16=8\times 2+0\] \[\cot \theta =\frac{\cos \theta }{\sin \theta }=\frac{\sqrt{{{\cos }^{2}}\theta }}{\sin \theta }=\frac{\sqrt{1-{{\sin }^{2}}\theta }}{\sin \theta }\]of \[\cot \theta =\frac{\cos \theta }{\sin \theta }=\frac{\sqrt{{{\cos }^{2}}\theta }}{\sin \theta }=\frac{\sqrt{1-{{\sin }^{2}}\theta }}{\sin \theta }\] \[AD=\frac{bc}{\sqrt{{{b}^{2}}+{{c}^{2}}}}\] \[\frac{bc}{\sqrt{{{b}^{2}}+{{c}^{2}}}}\]
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