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Introduction of Trigonometry and its Application — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsIntroduction of Trigonometry and its Application1 Mark
MCQ
Choose the correct answer from the given four options.
Given that $\sin\theta=\frac{\text{a}}{\text{b}},\text{then}\cos\theta$ is equal to:
  • A
    $\frac{\text{b}}{\sqrt{\text{b}^2-\text{a}^2}}$
  • B
    $\frac{\text{b}}{\text{a}}$
  • ✓
    $\frac{\sqrt{\text{b}^2-\text{a}^2}}{\text{b}}$
  • D
    $\frac{\text{a}}{\sqrt{\text{b}^2-\text{a}^2}}$
✓

Answer

Correct option: C.
$\frac{\sqrt{\text{b}^2-\text{a}^2}}{\text{b}}$
$\text{Given, }\sin\theta=\frac{\text{a}}{\text{b}}$ $[\because\sin^2\theta+\cos^2\theta=1\Rightarrow\cos\theta=\sqrt{1-\sin^2\theta}]$
$\therefore\ \cos\theta=\sqrt{1-\sin^2\theta}$
$=\sqrt{1-\Big(\frac{\text{a}}{\text{b}}\Big)^2}=\sqrt{1-\frac{\text{a}^2}{\text{b}^2}}=\frac{\sqrt{\text{b}^2-\text{a}^2}}{\text{b}}$

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