In triangle \[ABC\], right-angled at \[B\], \[AB = 24{\text{ cm}}\] and \[BC = 7{\text{ cm}}\]. Determine
(i) \[\sin A\], \[\cos A\]
(ii) \[\sin C\], \[\cos C\]

Hint: Here, we need to find the value of the sine and cosine of angles \[A\] and \[C\]. First, we will find the length of the hypotenuse using the Pythagoras’s theorem. Then, using the lengths of the sides of the right angled triangle, we can find the required trigonometric ratios using the formulae \[\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}\] and \[\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}\].
Formula Used: We will use the following formula to solve the question:
1.The Pythagoras’s theorem states that the square of the hypotenuse of a right angled triangle is equal to the sum of squares of the other two sides.
2.The sine of an angle \[\theta \] of a right angled triangle is given by \[\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}\].
3.The cosine of an angle \[\theta \] of a right angled triangle is given by \[\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}\].

Complete step-by-step answer:
(i)
First, we will draw the diagram using the given information.

Here, \[\angle B = 90^\circ \], \[AB = 24{\text{ cm}}\] and \[BC = 7{\text{ cm}}\].
We know that the side opposite to the right angle of a right angled triangle is the hypotenuse.
Therefore, \[AC\] is the hypotenuse.
Now, we will use the Pythagoras’s theorem to get the length of the hypotenuse.
The Pythagoras’s theorem states that the square of the hypotenuse of a right angled triangle is equal to the sum of squares of the other two sides, that is \[{\text{Hypotenuse}}{{\text{e}}^2} = {\text{Bas}}{{\text{e}}^2} + {\text{Perpendicular}}{{\text{r}}^2}\].
Therefore, in triangle \[ABC\], we get
\[A{C^2} = A{B^2} + B{C^2}\]
Substituting \[AB = 24\] cm and \[BC = 7\] cm in the equation, we get
 \[ \Rightarrow A{C^2} = {24^2} + {7^2}\]
Applying the exponents on the bases, we get
\[ \Rightarrow A{C^2} = 576 + 49\]
Adding 576 and 49, we get
\[ \Rightarrow A{C^2} = 625\]
Taking the square root of both sides, we get
\[\Rightarrow AC = \sqrt {625} \\
   \Rightarrow AC = 25{\text{ cm}} \\\]
Therefore, the length of the hypotenuse \[AC\] is 25 cm.
Now, we will use the sides of the triangle to find the required trigonometric ratios.
The sine of an angle \[\theta \] of a right angled triangle is given by \[\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}\].
Therefore, we get the sine of \[A\] as
\[ \Rightarrow \sin A = \dfrac{{BC}}{{AC}}\]
Substituting \[BC = 7\] cm and \[AC = 25\] cm in the equation, we get
\[ \Rightarrow \sin A = \dfrac{7}{{25}}\]
The cosine of an angle \[\theta \] of a right angled triangle is given by \[\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}\].
Therefore, we get the cosine of \[A\] as
\[ \Rightarrow \cos A = \dfrac{{AB}}{{AC}}\]
Substituting \[AB = 24\] cm and \[AC = 25\] cm in the equation, we get
\[ \Rightarrow \cos A = \dfrac{{24}}{{25}}\]
Therefore, we get the values of \[\sin A\] and \[\cos A\] as \[\dfrac{7}{{25}}\] and \[\dfrac{{24}}{{25}}\] respectively.
(ii)
We will use the sides of the triangle to find the required trigonometric ratios.
In triangle \[ABC\], we get the sine of \[C\] as
\[ \Rightarrow \sin C = \dfrac{{AB}}{{AC}}\]
Substituting \[AB = 24\] cm and \[AC = 25\] cm in the equation, we get
\[ \Rightarrow \sin C = \dfrac{{24}}{{25}}\]

Also, in triangle \[ABC\], we get the cosine of \[C\] as
\[ \Rightarrow \cos C = \dfrac{{BC}}{{AC}}\]
Substituting \[BC = 7\] cm and \[AC = 25\] cm in the equation, we get
\[ \Rightarrow \cos C = \dfrac{7}{{25}}\]
\[\therefore \] We get the values of
\[\sin C\] and \[\cos C\] as \[\dfrac{{24}}{{25}}\] and \[\dfrac{7}{{25}}\] respectively.

Note: We can also use the trigonometric ratios of complementary angles to find the sine and cosine of \[C\].
Using the angle sum property of a triangle, we get
\[\angle A + \angle B + \angle C = 180^\circ \]
Therefore, we get
\[\Rightarrow \angle A + 90^\circ + \angle C = 180^\circ \\
   \Rightarrow \angle A + \angle C = 90^\circ \\
   \Rightarrow \angle C = 90^\circ - \angle A \\\]
Therefore, the angles \[A\] and \[C\] are complementary angles.
We know that the sine and cosine of complementary angles can be written as \[\sin \left( {90^\circ - \theta } \right) = \cos \theta \] and \[\cos \left( {90^\circ - \theta } \right) = \sin \theta \].
Rewriting the sine of angle \[C\], we get
\[\Rightarrow \sin C = \sin \left( {90^\circ - A} \right) \\
   \Rightarrow \sin C = \cos A \\
  \therefore \sin C = \dfrac{{24}}{{25}} \\\]
Similarly, rewriting the cosine of angle \[C\], we get
\[\Rightarrow \cos C = \cos \left( {90^\circ - A} \right) \\
   \Rightarrow \cos C = \sin A \\
   \Rightarrow \cos C = \dfrac{7}{{25}} \\\]
\[\therefore \] We get the values of \[\sin C\] and \[\cos C\] as \[\dfrac{{24}}{{25}}\] and \[\dfrac{7}{{25}}\] respectively.