In `DeltaABC`, right - angled at B , AC = 20 cm and `tan angleACB = 3/4` , calculate the measures of AB and BC.

Answer

To solve the problem step by step, we will use the information given in the question about triangle ABC, which is right-angled at B.

### Step 1: Understand the triangle and the given information
We have a right triangle ABC where:
- Angle B is the right angle.
- AC (the hypotenuse) = 20 cm.
- \( \tan(\angle ACB) = \frac{3}{4} \).

### Step 2: Define the sides of the triangle
Let:
- BC = X (the base)
- AB = Y (the perpendicular)

### Step 3: Use the tangent ratio
From the definition of tangent:
\[ \tan(\angle ACB) = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{BC} = \frac{Y}{X} \]
Given that \( \tan(\angle ACB) = \frac{3}{4} \), we can write:
\[ \frac{Y}{X} = \frac{3}{4} \]
Cross-multiplying gives:
\[ Y = \frac{3}{4}X \]

### Step 4: Apply the Pythagorean theorem
According to the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Substituting the known values:
\[ 20^2 = Y^2 + X^2 \]
This simplifies to:
\[ 400 = Y^2 + X^2 \]

### Step 5: Substitute Y in the Pythagorean theorem
Now, substitute \( Y \) from step 3 into the equation:
\[ 400 = \left(\frac{3}{4}X\right)^2 + X^2 \]
Calculating \( \left(\frac{3}{4}X\right)^2 \):
\[ \left(\frac{3}{4}X\right)^2 = \frac{9}{16}X^2 \]
Now, the equation becomes:
\[ 400 = \frac{9}{16}X^2 + X^2 \]
Convert \( X^2 \) to have a common denominator:
\[ X^2 = \frac{16}{16}X^2 \]
So, we have:
\[ 400 = \frac{9}{16}X^2 + \frac{16}{16}X^2 = \frac{25}{16}X^2 \]

### Step 6: Solve for X
Multiply both sides by 16 to eliminate the fraction:
\[ 6400 = 25X^2 \]
Now, divide by 25:
\[ X^2 = \frac{6400}{25} = 256 \]
Taking the square root gives:
\[ X = 16 \text{ cm} \]

### Step 7: Find Y using the value of X
Now, substitute \( X \) back into the equation for \( Y \):
\[ Y = \frac{3}{4}X = \frac{3}{4} \times 16 = 12 \text{ cm} \]

### Final Answer
Thus, the measures of the sides are:
- \( BC = X = 16 \text{ cm} \)
- \( AB = Y = 12 \text{ cm} \)