The angle of elevation of a tower PQ at a point A due north of it is `30^(@)` and at another point B due east of A is `18^(@)`. If AB =a. Then the height of the tower is

Options

A. `a(2+sqrt(5))`
B. `a(2-sqrt(5))`
C. `a/sqrt(2+2sqrt(5))`
D. None of these

Answer

To solve the problem step by step, we will follow the geometric relationships and trigonometric ratios involved in the angles of elevation.

### Step 1: Draw the Diagram
Draw a vertical tower PQ and mark point A due north of the tower and point B due east of A. Label the height of the tower as \( h \).

### Step 2: Identify the Angles
From point A, the angle of elevation to the top of the tower (point P) is \( 30^\circ \). From point B, the angle of elevation to point P is \( 18^\circ \).

### Step 3: Set Up the Right Triangle from Point A
In triangle PAQ (where Q is the base of the tower), we can use the tangent function:
\[ \tan(30^\circ) = \frac{h}{AQ} \]
From trigonometric values, we know:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Thus, we can write:
\[ \frac{1}{\sqrt{3}} = \frac{h}{AQ} \implies AQ = h \sqrt{3} \]

### Step 4: Set Up the Right Triangle from Point B
In triangle PBQ, we can again use the tangent function:
\[ \tan(18^\circ) = \frac{h}{BQ} \]
Using the known value of \( \tan(18^\circ) \):
\[ \tan(18^\circ) = \frac{\sqrt{5} - 1}{2} \]
Thus, we can write:
\[ \frac{\sqrt{5} - 1}{2} = \frac{h}{BQ} \implies BQ = \frac{2h}{\sqrt{5} - 1} \]

### Step 5: Relate the Distances
Since point B is due east of point A, we can express \( AB \) as:
\[ AB = AQ + BQ \]
Substituting the values we found:
\[ AB = h\sqrt{3} + \frac{2h}{\sqrt{5} - 1} \]

### Step 6: Use Pythagoras Theorem
Using the Pythagorean theorem in triangle ABQ:
\[ BQ^2 = AB^2 + AQ^2 \]
Substituting the values:
\[ \left(\frac{2h}{\sqrt{5} - 1}\right)^2 = a^2 + (h\sqrt{3})^2 \]
Expanding both sides:
\[ \frac{4h^2}{(\sqrt{5} - 1)^2} = a^2 + 3h^2 \]

### Step 7: Solve for \( h^2 \)
Rearranging gives:
\[ \frac{4h^2}{(\sqrt{5} - 1)^2} - 3h^2 = a^2 \]
Factoring out \( h^2 \):
\[ h^2 \left(\frac{4}{(\sqrt{5} - 1)^2} - 3\right) = a^2 \]
Thus,
\[ h^2 = \frac{a^2}{\frac{4}{(\sqrt{5} - 1)^2} - 3} \]

### Step 8: Final Expression for \( h \)
Taking the square root gives:
\[ h = \frac{a}{\sqrt{\frac{4}{(\sqrt{5} - 1)^2} - 3}} \]

### Conclusion
The height of the tower \( h \) can be expressed as:
\[ h = \frac{a}{\sqrt{2 + 2\sqrt{5}}} \]