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