In `triangleABC, BD bot AC`. E is a point on BC such that `angleBEA` = x . If `angleEAC` = 38 and `angleEBD = 40^@`, then the value of x is:

Options

A. `88^@`
B. `68^@`
C. `78^@`
D. `72^@`

Answer

To solve the problem step by step, we will analyze the given triangle and the angles involved.

### Step 1: Understand the Triangle and Given Angles
We have triangle ABC where BD is perpendicular to AC. We know:
- Angle EAC = 38°
- Angle EBD = 40°
- Angle BEA = x (which we need to find)

### Step 2: Identify Angles in Triangle BDC
In triangle BDC, we know that:
- Angle BDC = 90° (since BD is perpendicular to AC)
- Let angle DCB = C (unknown)
- Angle CBD = Angle EBD = 40°

Using the property that the sum of angles in a triangle is 180°, we can write:
\[ \text{Angle BDC} + \text{Angle DCB} + \text{Angle CBD} = 180° \]
Substituting the known values:
\[ 90° + C + 40° = 180° \]
This simplifies to:
\[ C + 130° = 180° \]
Thus:
\[ C = 50° \]

### Step 3: Identify Angles in Triangle AEC
Next, we consider triangle AEC. We know:
- Angle EAC = 38° (given)
- Angle ECA = C = 50° (found in Step 2)
- Let angle AEC = A (unknown)

Again, using the sum of angles in a triangle:
\[ \text{Angle AEC} + \text{Angle ECA} + \text{Angle CAE} = 180° \]
Substituting the known values:
\[ A + 50° + 38° = 180° \]
This simplifies to:
\[ A + 88° = 180° \]
Thus:
\[ A = 92° \]

### Step 4: Relate Angles AEC and BEA
Now, we know:
- Angle AEC = 92°
- Angle BEA = x

Since angle AEC and angle BEA are linear pairs (they form a straight line), we can write:
\[ \text{Angle AEC} + \text{Angle BEA} = 180° \]
Substituting the known value:
\[ 92° + x = 180° \]
This simplifies to:
\[ x = 180° - 92° \]
Thus:
\[ x = 88° \]

### Conclusion
The value of x is 88°.