In `DeltaABC` , right - angled at B , AB = 24 cm , BC = 7 cm .
`sinA, cos A`

Answer

To solve for \( \sin A \) and \( \cos A \) in triangle \( \Delta ABC \) where \( B \) is the right angle, \( AB = 24 \, \text{cm} \), and \( BC = 7 \, \text{cm} \), we can follow these steps:

### Step 1: Identify the sides of the triangle
In triangle \( \Delta ABC \):
- \( AB \) is the side opposite angle \( C \) (perpendicular).
- \( BC \) is the side opposite angle \( A \) (base).
- \( AC \) is the hypotenuse.

### Step 2: Use the Pythagorean theorem to find the hypotenuse \( AC \)
According to the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Substituting the values:
\[ AC^2 = 24^2 + 7^2 \]
Calculating the squares:
\[ AC^2 = 576 + 49 = 625 \]
Taking the square root:
\[ AC = \sqrt{625} = 25 \, \text{cm} \]

### Step 3: Calculate \( \sin A \)
The sine of angle \( A \) is defined as the ratio of the length of the side opposite angle \( A \) (which is \( BC \)) to the length of the hypotenuse \( AC \):
\[ \sin A = \frac{BC}{AC} \]
Substituting the known values:
\[ \sin A = \frac{7}{25} \]

### Step 4: Calculate \( \cos A \)
The cosine of angle \( A \) is defined as the ratio of the length of the adjacent side (which is \( AB \)) to the length of the hypotenuse \( AC \):
\[ \cos A = \frac{AB}{AC} \]
Substituting the known values:
\[ \cos A = \frac{24}{25} \]

### Final Answers
Thus, the values are:
\[ \sin A = \frac{7}{25}, \quad \cos A = \frac{24}{25} \]

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