The side BC of `Delta` ABC is produced to D. If `angleACD = 108^(@)` and `angleB = 1/2 angleA` then `angleA` is

Options

A. `36^(@)`
B. `72^(@)`
C. `108^(@)`
D. `59^(@)`

Answer

To solve the problem step by step, we will use the information given in the question.

### Step 1: Define the angles
Let angle A = 2θ. Since angle B is half of angle A, we can express angle B as:
\[ \text{Angle B} = \frac{1}{2} \times \text{Angle A} = \frac{1}{2} \times 2θ = θ \]

### Step 2: Use the external angle property
According to the external angle property of a triangle, the external angle (angle ACD) is equal to the sum of the two opposite interior angles (angles A and B). Therefore, we can write:
\[ \text{Angle ACD} = \text{Angle A} + \text{Angle B} \]
Substituting the values we defined:
\[ 108° = 2θ + θ \]

### Step 3: Simplify the equation
Now, simplify the equation:
\[ 108° = 3θ \]

### Step 4: Solve for θ
To find the value of θ, divide both sides by 3:
\[ θ = \frac{108°}{3} = 36° \]

### Step 5: Find angle A
Since angle A was defined as 2θ:
\[ \text{Angle A} = 2θ = 2 \times 36° = 72° \]

### Conclusion
Thus, the value of angle A is:
\[ \text{Angle A} = 72° \]