Quantitative Aptitude Ques 386

Question: If $ \sin \theta =\frac{a}{\sqrt{a^{2}+b^{2}}}, $ then the value of cot 6 will be

Options:

A) $ \frac{b}{a} $

B) $ \frac{a}{b} $

C) $ \frac{a}{b}+1 $

D) $ \frac{b}{a}+1 $

Show Answer

Answer:

Correct Answer: A

Solution:

  • Given, $ \sin \theta =\frac{a}{\sqrt{a^{2}+b^{2}}} $ … (i) We know that, $ \sin \theta =\frac{Perpendicular}{Hypotenuse} $ Now, $ \Delta ABC, $ $ \sin \theta =\frac{AB}{AC} $ ….(ii) On comparing Eqs. (i) and (ii) we get $ AB=a $ and $ AC=\sqrt{a^{2}+b^{2}} $ Now in $ \Delta ABC, $ by Pythagoras theorem, $ AB^{2}+BC^{2}=AC^{2} $ $ BC^{2}=b^{2} $
    $ \Rightarrow $ $ BC=b $
    $ \Rightarrow $ $ \cot \theta =\frac{b}{a} $
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