A) \[\frac{1}{8}\]
B) \[\frac{1}{2}\]
C) 2
D) \[\frac{3}{5}\]
Correct Answer: A
Solution :
Given, \[\sin \,A=\frac{4}{5}=\frac{BC}{AC}\] |
Let \[BC=4k,\,AC=5k\] |
Apply Pythagoras theorem, |
\[A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}\] |
\[\Rightarrow \,\,A{{B}^{2}}+{{\left( 4k \right)}^{2}}={{\left( 5k \right)}^{2}}\] |
\[\Rightarrow \,AB=\sqrt{25-16}=3\] |
\[\Rightarrow \frac{1-\sin \,A}{1+\cos A}=\frac{1-\frac{4}{5}}{1+\frac{3}{5}}\] |
\[=\frac{\left( 5-4 \right)}{5}\times \frac{5}{\left( 5+3 \right)}=\frac{1}{8}\] |
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