MCQ
If $\sin\theta=\frac{\text{a}}{\text{b}}$ then $\cos\theta=?$
  • A
    $\frac{\text{a}}{\sqrt{\text{b}^2-\text{a}^2}}$
  • $\frac{\sqrt{\text{b}^2-\text{a}^2}}{\text{b}}$
  • C
    $\frac{\text{a}}{\sqrt{\text{b}^2-\text{a}^2}}$
  • D
    $\frac{\text{b}}{\text{a}}$

Answer

Correct option: B.
$\frac{\sqrt{\text{b}^2-\text{a}^2}}{\text{b}}$
$\cos^2\theta=1-\sin^2\theta$
$=1-\Big(\frac{\text{a}}{\text{b}}\Big)^2$
$=1-\frac{\text{a}^2}{\text{b}^2}$
$=\frac{\text{b}^2-\text{a}^2}{\text{b}^2}$
$\Rightarrow\cos\theta=\frac{\sqrt{\text{b}^2-\text{a}^2}}{\text{b}}$

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