# In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :

(i) sin A, cos A (ii) sin C, cos C

**Solution:**

We use the basic formulas of trigonometric ratios to solve the question.

Applying the Pythagoras theorem for ∆ABC, we can find the hypotenuse (side AC). Once hypotenuse is known, we can find sine and cosine angles using trigonometric ratios.

In ΔABC, we obtain.

AC^{2} = AB^{2} + BC^{2}

= (24 cm)^{2 } + (7 cm)^{2}

= (576 + 49) cm^{2}

= 625 cm^{2}

∴ Hypotenuse AC = √625 cm = 25 cm

(i)

sin A = side opposite to ∠A/hypotenuse = BC/AC

sin A = 7 cm/25 cm = 7/25

sin A = 7/25

cos A = side adjacent to ∠A/hypotenuse = AB/AC

= 24 cm/25 cm = 24/25

cos A = 24/25

(ii)

sin C = side opposite to ∠C/hypotenuse = AB/AC

sin C = 24cm/25 cm = 24/25

sin C = 24/25

cos C = side adjacent to ∠C//hypotenuse = BC/AC

= 7 cm/25 cm = 7/25

cos C = 7/25

**Video Solution:**

## In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm, determine: (i) sin A, cos A (ii) sin C, cos C

### Maths NCERT Solutions Class 10 - Chapter 8 Exercise 8.1 Question 1:

In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine : (i) sin A, cos A (ii) sin C, cos C

In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm, then sin A, cos A is 7/25 and 24/25 respectively and sin C, cos C is 24/25 and 7/25 respectively