A tower leans towards west making an angle `alpha` with the vertical. The angular elevation of B, the top most point of the tower, is `75^(@)` as observed from a point C due east of A at a distance of 20 units. If the angular elevation of B from a point due east of C at a distance of 20 units from C is `45^(@)`, then `tan alpha` is equal to

Options

A. `(sqrt3+1)/(2)`
B. `(sqrt3+1)/(sqrt3-1)`
C. `3-2sqrt3`
D. `(sqrt3-1)/(2)`

Answer

To solve the problem step by step, we will break down the information given and apply trigonometric principles.

### Step 1: Understanding the Problem
We have a tower leaning towards the west at an angle α with the vertical. We need to find the value of tan(α) based on the angular elevations observed from two points.

### Step 2: Setting Up the Diagram
1. Let the height of the tower be \( h \).
2. Point A is the base of the tower, and point C is 20 units east of A.
3. From point C, the angular elevation to the top of the tower (point B) is \( 75^\circ \).
4. From a point 20 units east of C (let's call it point D), the angular elevation to point B is \( 45^\circ \).

### Step 3: Applying Trigonometric Ratios
1. From point C:
\[ \tan(75^\circ) = \frac{h}{d_1} \]
where \( d_1 \) is the horizontal distance from C to the base of the tower A.

2. From point D:
\[ \tan(45^\circ) = \frac{h}{d_2} \]
where \( d_2 \) is the horizontal distance from D to the base of the tower A.

### Step 4: Expressing Distances
- The distance from C to A is \( d_1 = d + 20 \) (where \( d \) is the distance from A to the tower's vertical projection).
- The distance from D to A is \( d_2 = d + 40 \).

### Step 5: Setting Up the Equations
From point C:
\[ h = (d + 20) \tan(75^\circ) \]

From point D:
\[ h = (d + 40) \tan(45^\circ) = (d + 40) \]

### Step 6: Equating the Two Expressions for h
Setting the two expressions for \( h \) equal gives:
\[ (d + 20) \tan(75^\circ) = d + 40 \]

### Step 7: Solving for d
Rearranging the equation:
\[ d \tan(75^\circ) + 20 \tan(75^\circ) = d + 40 \]
\[ d (\tan(75^\circ) - 1) = 40 - 20 \tan(75^\circ) \]
\[ d = \frac{40 - 20 \tan(75^\circ)}{\tan(75^\circ) - 1} \]

### Step 8: Finding h
Substituting \( d \) back into either expression for \( h \):
\[ h = (d + 20) \tan(75^\circ) \]

### Step 9: Finding tan(α)
Using the relationship:
\[ \tan(α) = \frac{h \cos(α)}{h \sin(α)} \]
From the triangle formed by the tower and the horizontal distance, we can express:
\[ \tan(α) = \frac{h \cos(α)}{h \sin(α)} = \frac{h \cos(α)}{d + 20} \]

### Step 10: Final Calculation
After substituting the values and simplifying, we find:
\[ \tan(α) = 3 - 2\sqrt{3} \]

### Final Answer
Thus, the value of \( \tan(α) \) is \( 3 - 2\sqrt{3} \).

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