In triangle `ABC,/_BAC = 75^@, /_ABC = 45^@. bar(BC)` is produced to D. If `/_ACD = x^@`, then `x/3 %` of `60^@` is

Options

A. A)`30^@`
B. B)`48^@`
C. C)`15^@`
D. D)`24^@`

Answer

To solve the problem step by step, we will follow the reasoning and calculations presented in the video transcript.

### Step 1: Identify the angles in triangle ABC
Given:
- Angle \( \angle BAC = 75^\circ \)
- Angle \( \angle ABC = 45^\circ \)

### Step 2: Calculate the third angle \( \angle ACB \)
Using the property that the sum of angles in a triangle is \( 180^\circ \):
\[ \angle ACB = 180^\circ - \angle BAC - \angle ABC \]
Substituting the known values:
\[ \angle ACB = 180^\circ - 75^\circ - 45^\circ = 60^\circ \]

### Step 3: Determine angle \( \angle ACD \)
Since \( BC \) is produced to point \( D \), we know that:
\[ \angle ACD + \angle ACB = 180^\circ \]
Let \( \angle ACD = x \). Therefore:
\[ x + 60^\circ = 180^\circ \]
Solving for \( x \):
\[ x = 180^\circ - 60^\circ = 120^\circ \]

### Step 4: Calculate \( \frac{x}{3} \) percent of \( 60^\circ \)
Now, we need to find \( \frac{x}{3} \% \) of \( 60^\circ \):
\[ \frac{x}{3} = \frac{120^\circ}{3} = 40^\circ \]
Now, we calculate \( 40^\circ \% \) of \( 60^\circ \):
\[ 40\% = \frac{40}{100} \times 60^\circ \]
Calculating this gives:
\[ \frac{40 \times 60}{100} = \frac{2400}{100} = 24^\circ \]

### Final Answer
Thus, the answer is \( 24^\circ \).

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