A) \[\frac{b}{\sqrt{{{b}^{2}}-{{a}^{2}}}}\]
B) \[\frac{b}{a}\]
C) \[\frac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{b}\]
D) \[\frac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}}\]
Correct Answer: C
Solution :
Given, \[\sin \theta =\frac{a}{b}=\frac{AB}{AC}\] |
Let \[AB=ak,\,AC=bk\] |
\[A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}\] [from Pythagoras theorem] |
\[BC=\sqrt{{{b}^{2}}-{{a}^{2}}}\Rightarrow \,\,\cos \theta =\frac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{b}\] |
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