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

Question 2

Step 1: Recall the definition of sine.

\(\sin A = \dfrac{\text{opposite side}}{\text{hypotenuse}}\).

We are given \(\sin A = \dfrac{1}{2}\).

Step 2: Think about the standard angles (30°, 45°, 60°) where sine values are commonly used.

We know: \(\sin 30^\circ = \dfrac{1}{2}\).

So, \(A = 30^\circ\).

Step 3: Now recall the formula for cotangent.

\(\cot A = \dfrac{\cos A}{\sin A}\).

Step 4: At \(A = 30^\circ\),

\(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \dfrac{1}{2}\).

Step 5: Substitute these values:

\(\cot 30^\circ = \dfrac{\cos 30^\circ}{\sin 30^\circ} = \dfrac{\tfrac{\sqrt{3}}{2}}{\tfrac{1}{2}}\).

Step 6: Simplify the fraction:

\(\dfrac{\tfrac{\sqrt{3}}{2}}{\tfrac{1}{2}} = \sqrt{3}\).

Final Answer: \(\cot A = \sqrt{3}\).