In `DeltaABC, /_BAC = 90^@` and `AB = 1/2 BC.` Then the measure of `/_ACB` is :

Options

A. `60^@`
B. `30^@`
C. `45^@`
D. `15^@`

Answer

To solve the problem, we need to analyze the given information about triangle ABC, where angle BAC is 90 degrees and side AB is half of side BC.

### Step-by-Step Solution:

1. **Identify the Triangle Configuration**:
We have triangle ABC with angle BAC = 90 degrees. This means triangle ABC is a right triangle with the right angle at A.

2. **Assign Lengths to the Sides**:
Given that AB = 1/2 BC, let's denote:
- BC = x (the length of side BC)
- AB = 1/2 * x (the length of side AB)

3. **Use the Pythagorean Theorem**:
In a right triangle, the Pythagorean theorem states that:
\[ AC^2 + AB^2 = BC^2 \]
Substituting the values we have:
\[ AC^2 + \left(\frac{x}{2}\right)^2 = x^2 \]
Simplifying this:
\[ AC^2 + \frac{x^2}{4} = x^2 \]
Rearranging gives:
\[ AC^2 = x^2 - \frac{x^2}{4} \]
\[ AC^2 = \frac{4x^2}{4} - \frac{x^2}{4} = \frac{3x^2}{4} \]
Therefore:
\[ AC = \frac{\sqrt{3}x}{2} \]

4. **Determine the Angles**:
Now we can find angle ACB using the tangent function:
\[ \tan(ACB) = \frac{AB}{AC} = \frac{\frac{x}{2}}{\frac{\sqrt{3}x}{2}} = \frac{1}{\sqrt{3}} \]
This implies:
\[ ACB = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]
The angle whose tangent is \( \frac{1}{\sqrt{3}} \) is 30 degrees.

5. **Conclusion**:
Therefore, the measure of angle ACB is:
\[ \boxed{30 \text{ degrees}} \]