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

(i) sin A, cos A

(ii) sin C, cos C

ABC is a right angled triangle as shown below:

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 24² + 7²

⇒ AC² = 576 + 49

⇒ AC² = 625

⇒ AC = $\sqrt{625}$ = 25 cm.

(i) sin A = $\dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}} = \dfrac{BC}{AC}$

Substituting values we get,

sin A = $\dfrac{7}{25}$.

cos A = $\dfrac{\text{Side adjacent to ∠A}}{\text{Hypotenuse}} = \dfrac{AB}{AC}$

Substituting values we get,

cos A = $\dfrac{24}{25}$.

Hence, sin A = $\dfrac{7}{25}$ and cos A = $\dfrac{24}{25}$.

(ii) sin C = $\dfrac{\text{Side opposite to ∠C}}{\text{Hypotenuse}} = \dfrac{AB}{AC}$

Substituting values we get,

sin C = $\dfrac{24}{25}$.

cos C = $\dfrac{\text{Side adjacent to ∠C}}{\text{Hypotenuse}} = \dfrac{BC}{AC}$

Substituting values we get,

cos C = $\dfrac{7}{25}$.

Hence, sin C = $\dfrac{24}{25}$ and cos C = $\dfrac{7}{25}$.