Quantitative Aptitude Ques 2162

Question: If $ \sin \theta =\frac{a}{\sqrt{a^{2}+b^{2}}}, $ then the value of $ \cot \theta $ 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 in $ \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, we have $ AC^{2}=AB^{2}+BC^{2} $ $ {{(\sqrt{a^{2}+b^{2}})}^{2}}={{(a)}^{2}}+{{(BC)}^{2}} $ $ {{(BC)}^{2}}=a^{2}+b^{2}-a^{2} $

$ \Rightarrow $ $ BC^{2}=b^{2}\Rightarrow BC=b $

$ \Rightarrow $ $ \cot \theta =\frac{Base}{Perpendicular}=\frac{BC}{AB} $ On putting values, $ \cot \theta =\frac{b}{a} $

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