NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 8: Introduction to Trignometry and Its Applications - Exercise 8.1

Question 4

We are given: \(\sin\theta = \dfrac{a}{b}\).

Step 1: Recall the Pythagorean identity of trigonometry:

\[ \sin^2\theta + \cos^2\theta = 1 \]

Step 2: Substitute the value of \(\sin\theta\):

\[ \left(\dfrac{a}{b}\right)^2 + \cos^2\theta = 1 \]

Step 3: Simplify the square:

\[ \dfrac{a^2}{b^2} + \cos^2\theta = 1 \]

Step 4: Move \(\dfrac{a^2}{b^2}\) to the right side:

\[ \cos^2\theta = 1 - \dfrac{a^2}{b^2} \]

Step 5: Write with a common denominator:

\[ \cos^2\theta = \dfrac{b^2}{b^2} - \dfrac{a^2}{b^2} = \dfrac{b^2 - a^2}{b^2} \]

Step 6: Take the square root on both sides:

\[ \cos\theta = \dfrac{\sqrt{b^2 - a^2}}{b} \]

Note: Since \(\theta\) is an acute angle (between \(0^\circ\) and \(90^\circ\)), cosine is positive. So we take only the positive root.

Final Answer: \(\cos\theta = \dfrac{\sqrt{b^2 - a^2}}{b}\) → Option C.