In `DeltaABC, /_B = 35^@, /_C = 65^@` and the bisector of `/_BAC` meets BC in D? Then `/_ADB` is :

Options

A. `40^@`
B. `75^@`
C. `90^@`
D. `105^@`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Determine the angles of triangle ABC
We know that in triangle ABC:
- Angle B = 35°
- Angle C = 65°

To find angle A, we use the property that the sum of angles in a triangle is 180°.

**Calculation:**
\[ \text{Angle A} = 180° - \text{Angle B} - \text{Angle C} = 180° - 35° - 65° = 80° \]

### Step 2: Identify the angle bisector
The angle bisector of angle A (which is 80°) divides it into two equal parts. Therefore, each part will be:
\[ \text{Angle BAD} = \text{Angle CAD} = \frac{80°}{2} = 40° \]

### Step 3: Analyze triangle ADB
Now we focus on triangle ADB. We know:
- Angle B = 35°
- Angle ADB (which we need to find) = ?
- Angle BAD = 40°

### Step 4: Use the triangle angle sum property
In triangle ADB, the sum of the angles must equal 180°:
\[ \text{Angle ADB} + \text{Angle BAD} + \text{Angle B} = 180° \]
Substituting the known values:
\[ \text{Angle ADB} + 40° + 35° = 180° \]

### Step 5: Solve for angle ADB
Now we can solve for angle ADB:
\[ \text{Angle ADB} + 75° = 180° \]
\[ \text{Angle ADB} = 180° - 75° = 105° \]

### Conclusion
Thus, the value of angle ADB is:
\[ \text{Angle ADB} = 105° \]