question_answer

If \[\sin \theta =\frac{a}{b}\], then \[\cos \theta \] is equal to

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]

\[BC=\sqrt{{{b}^{2}}-{{a}^{2}}}\Rightarrow \,\,\cos \theta =\frac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{b}\]