NCERT Solutions
Class 10 - Mathematics - Chapter 8: INTRODUCTION TO TRIGONOMETRY - Exercise 8.1

Question 1

Given: In \(\triangle ABC\), \(\angle B = 90^\circ\), \(AB = 24\) cm, \(BC = 7\) cm.

Goal: Find \(\sin A, \cos A\) and \(\sin C, \cos C\).

Step 1: Identify the hypotenuse

In a right-angled triangle, the side opposite the right angle is the hypotenuse. Since \(\angle B = 90^\circ\), the hypotenuse is \(AC\).

Step 2: Find \(AC\) using Pythagoras theorem

Pythagoras theorem says:

\(AC^2 = AB^2 + BC^2\)

Substitute values:

\(AC^2 = 24^2 + 7^2 = 576 + 49 = 625\)

So, \(AC = \sqrt{625} = 25\) cm.

Student Note: Always compute the hypotenuse first in such questions, because \(\sin\) and \(\cos\) need the hypotenuse in the denominator.

Step 3: Use definitions of \(\sin\) and \(\cos\)

For any angle \(\theta\) in a right triangle:

\(\sin\theta = \dfrac{\text{opposite side}}{\text{hypotenuse}}\),   \(\cos\theta = \dfrac{\text{adjacent side}}{\text{hypotenuse}}\)

(i) For angle \(A\):

Look at angle \(A\):

Opposite to \(A\) is \(BC = 7\) cm (the side across from angle \(A\)).

Adjacent to \(A\) is \(AB = 24\) cm (the side touching angle \(A\) other than the hypotenuse).

Hypotenuse is \(AC = 25\) cm.

So,

\(\sin A = \dfrac{BC}{AC} = \dfrac{7}{25}\)

\(\cos A = \dfrac{AB}{AC} = \dfrac{24}{25}\)

Student Note: For angle \(A\), remember: opposite is \(BC\), adjacent is \(AB\). (Opposite changes with the angle you choose!)

(ii) For angle \(C\):

Now look at angle \(C\):

Opposite to \(C\) is \(AB = 24\) cm.

Adjacent to \(C\) is \(BC = 7\) cm.

Hypotenuse is still \(AC = 25\) cm.

So,

\(\sin C = \dfrac{AB}{AC} = \dfrac{24}{25}\)

\(\cos C = \dfrac{BC}{AC} = \dfrac{7}{25}\)

Quick Check (helps avoid mistakes):

Since \(\angle A\) and \(\angle C\) are complementary in a right triangle, \(\sin A = \cos C\) and \(\cos A = \sin C\). Here, \(\sin A = \dfrac{7}{25}\) matches \(\cos C = \dfrac{7}{25}\), and \(\cos A = \dfrac{24}{25}\) matches \(\sin C = \dfrac{24}{25}\). So the answers are consistent.