1 markMCQSection A

Given that $\sin \theta = a/b$ , then cos θ is equal to :

(a) $b/\sqrt{b^2 - a^2}$ (b) $b/a$ (c) $\sqrt{b^2 - a^2}/b$ (d) $a/\sqrt{b^2 - a^2}$

Q: How do you find $\cos \theta$ if only $\sin \theta$ is given?

Use $\sin^2\theta + \cos^2\theta = 1$, so $\cos \theta = \sqrt{1 - \sin^2\theta}$, or use the right-triangle method: if sin θ = opposite/hypotenuse, find the adjacent side via Pythagoras and take cos θ = adjacent/hypotenuse.

Introduction to Trigonometry

Trigonometric Ratios - Relation between sin and cos

Options

(A)

b/\sqrt{b^2 - a^2}

(B)

b/a

(C)

\sqrt{b^2 - a^2}/b

(D)

a/\sqrt{b^2 - a^2}

Official Answer

(c) $\sqrt{b^2 - a^2}/b$ — with opposite = a and hypotenuse = b, $\text{adjacent} = \sqrt{b^2 - a^2}$ by Pythagoras, so cos θ = adjacent/hypotenuse.

trigonometric ratiossin thetacos thetaPythagoras theoremright triangleopposite adjacent hypotenuse

Marking Scheme

11 mark: correct option (c) $\sqrt{b^2 - a^2}/b$ ; full credit reasoning: opposite = a, hypotenuse = b, $\text{adjacent} = \sqrt{b^2 - a^2}$ by Pythagoras, so cos θ = adjacent/hypotenuse.

Hint

Draw a right triangle with opposite = a, hypotenuse = b, find adjacent using Pythagoras, then cos θ = adjacent/hypotenuse.

Quick Oral Answer

If $\sin \theta = a/b$ , then in a right triangle opposite = a and hypotenuse = b, so $\text{adjacent} = \sqrt{b^2 - a^2}$ by Pythagoras, giving $\cos \theta = \sqrt{b^2 - a^2}/b$ .

Analysis & Explanation

Tests trigonometric ratios via a right-triangle picture rather than rote formula recall.

Concept

Draw a right triangle where sin θ = opposite/hypotenuse = $a/b$ , so opposite = a, hypotenuse = b.
By the Pythagorean theorem, adjacent = √(hypotenuse² − opposite²) = √(b² − a²).

Key points

cos θ = adjacent/hypotenuse = $\sqrt{b^2 - a^2}/b$ — option (c).

Common mistakes

Writing $b/a$ (that is cosec θ, the reciprocal of sin θ, not cos θ).
Mixing up opposite and adjacent, giving $a/\sqrt{b^2 - a^2}$ instead of the correct expression.
Trying to recall the formula from memory instead of sketching the triangle, which causes numerator/denominator swaps.

Common Mistakes

1Confusing cos θ with cosec θ (reciprocal of sin θ), giving answer $b/a$ .
2Forgetting to apply the Pythagorean theorem to find the third side before writing the ratio.
3Mixing up which side (opposite vs adjacent) goes in the numerator for $\cos \theta$ .

Interesting Facts

The trigonometric identity $\sin^2\theta + \cos^2\theta = 1$ (used implicitly here) was known in essentially modern form to Indian mathematician Bhaskaracharya (12th century) in his work Siddhanta Shiromani.

This type of 'given one ratio, find another' question is one of the most frequently repeated MCQ formats in CBSE Class 10 trigonometry across multiple years.

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Frequently Asked Questions

How do you find $\cos \theta$ if only $\sin \theta$ is given?

Use $\sin^2\theta + \cos^2\theta = 1$ , so $\cos \theta = \sqrt{1 - \sin^2\theta}$ , or use the right-triangle method: if sin θ = opposite/hypotenuse, find the adjacent side via Pythagoras and take cos θ = adjacent/hypotenuse.

What is the difference between $\cos \theta$ and $\text{cosec}\ \theta$ ?

cos θ = adjacent/hypotenuse, while cosec θ = hypotenuse/opposite = $1/\sin \theta$ ; they are different ratios and should not be confused.