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

Suppose the triangle is as follows, 

Using Pythagoras theorem, we get

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2 = {\left(24\right)}^2+7^2$

$A{C}^2 = (576+49)$

$A{C}^2 = 625c{m}^2$

$AC = \sqrt{625}= 25$

Therefore, $AC = 25 cm$

(i) To find Sin (A), Cos (A)

It is known that sine (or) Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side. So it becomes

$Sin \left(A\right) =\frac{Opposite side}{Hypotenuse}= \frac{BC}{AC}=\frac{7}{25}$

Cosine or Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side and it becomes,

$Cos \left(A\right) =\frac{Adjacent side}{Hypotenuse}= \frac{AB}{AC}= \frac{24}{25}$

(ii) To find Sin (C), Cos (C)

$Sin \left(C\right) =\frac{Opposite side}{Hypotenuse}= \frac{AB}{AC}=\frac{24}{25}$

$Cos \left(C\right) =\frac{Adjacent side}{Hypotenuse}= \frac{BC}{AC}= \frac{7}{25}$