D is a point on side BC of a Triangle ABC such that ` AD bot BC ` and E is a point on AD for which `(Ae)/(ED) = (5)/(1)` If `angle BAD = 30^(@) tan angle ACB = 6 tan angle DBE ` . find `angle ACB `

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry
We have triangle ABC with point D on side BC such that AD is perpendicular to BC. This means that angle ADB = 90°. We also know that angle BAD = 30°.

### Step 2: Define the Lengths
Let AE = 5x and ED = x. Therefore, the total length of AD is:
\[ AD = AE + ED = 5x + x = 6x \]

### Step 3: Use the Given Ratio
We are given that:
\[ \tan(\angle ACB) = 6 \tan(\angle DBE) \]
Let’s denote:
- \( \angle ACB = \theta \)
- \( \angle DBE = \phi \)

### Step 4: Write the Tangent Ratios
From triangle ADB:
\[ \tan(\angle DBE) = \frac{ED}{BD} = \frac{x}{BD} \]

From triangle ACB:
\[ \tan(\angle ACB) = \frac{AD}{CD} = \frac{6x}{CD} \]

### Step 5: Set Up the Equation
Using the relationship given:
\[ \tan(\theta) = 6 \tan(\phi) \]
Substituting the expressions we have:
\[ \frac{6x}{CD} = 6 \cdot \frac{x}{BD} \]

### Step 6: Simplify the Equation
We can cancel \( x \) (assuming \( x \neq 0 \)):
\[ \frac{6}{CD} = \frac{6}{BD} \]
This implies:
\[ CD = BD \]

### Step 7: Conclude the Triangle Properties
Since \( CD = BD \), triangle BDC is isosceles. Therefore, angles \( \angle BDC \) and \( \angle DBC \) are equal.

### Step 8: Find the Angles
In triangle ABC, we have:
\[ \angle ADB + \angle BAD + \angle ACB = 180° \]
Substituting the known angles:
\[ 90° + 30° + \theta = 180° \]
\[ \theta = 180° - 120° = 60° \]

### Final Answer
Thus, the angle \( \angle ACB \) is:
\[ \angle ACB = 60° \]

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