Class 10 Maths Chapter 8 Introduction to Trigonometry Notes & NCERT Solutions

1. Trigonometric Ratios

In a right-angled triangle, we define ratios based on the Perpendicular (P), Base (B), and Hypotenuse (H):

• $\sin \theta = P/H$ (Pandit)
• $\cos \theta = B/H$ (Badri)
• $\tan \theta = P/B$ (Prasad)
• $\text{cosec } \theta = H/P$ (Inverse of $\sin$)
• $\sec \theta = H/B$ (Inverse of $\cos$)
• $\cot \theta = B/P$ (Inverse of $\tan$)

2. Standard Trigonometric Table

Ratio	0°	30°	45°	60°	90°
$\sin \theta$	0	1/2	$1/\sqrt{2}$	$\sqrt{3}/2$	1
$\cos \theta$	1	$\sqrt{3}/2$	$1/\sqrt{2}$	1/2	0
$\tan \theta$	0	$1/\sqrt{3}$	1	$\sqrt{3}$	Not Defined

3. Trigonometric Identities

1. $\sin^2 \theta + \cos^2 \theta = 1$
2. $1 + \tan^2 \theta = \sec^2 \theta$
3. $1 + \cot^2 \theta = \text{cosec}^2 \theta$

✏️ Complete NCERT Solutions Class 10 Trigonometry

Exercise 8.1

Q1. In $\Delta ABC$, right-angled at $B$, $AB = 24\text{ cm}$, $BC = 7\text{ cm}$. Determine $\sin A, \cos A$.

Step 1: Find Hypotenuse (AC)
Using Pythagoras: $AC = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\text{ cm}$

Step 2: Calculate Ratios
$\sin A = \text{Opposite/Hypotenuse} = BC/AC = \mathbf{7/25}$
$\cos A = \text{Adjacent/Hypotenuse} = AB/AC = \mathbf{24/25}$

Exercise 8.2

Q1. Evaluate: $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$

Step 1: Put values from table
$(\sqrt{3}/2 \times \sqrt{3}/2) + (1/2 \times 1/2)$