The angle of elevation of a tower at point A due north of it is `30^(@)` and at another point due East of A is `18^(@)`. If AB=a, then the height of the tower is_____.

Answer

To solve the problem, we need to find the height of the tower using the angles of elevation from two different points. Let's denote the height of the tower as \( h \) and the distance from the base of the tower to point A as \( AB = a \).

### Step 1: Set up the scenario
Let:
- \( T \) be the top of the tower,
- \( B \) be the base of the tower,
- \( A \) be the point due north of the tower,
- \( C \) be the point due east of \( A \).

From point \( A \), the angle of elevation to the top of the tower \( T \) is \( 30^\circ \). From point \( C \), the angle of elevation to the top of the tower is \( 18^\circ \).

### Step 2: Use tangent for point A
From point \( A \):
\[ \tan(30^\circ) = \frac{h}{AB} \]
Substituting \( AB = a \):
\[ \tan(30^\circ) = \frac{h}{a} \]
We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so:
\[ \frac{1}{\sqrt{3}} = \frac{h}{a} \]
This gives us:
\[ h = \frac{a}{\sqrt{3}} \quad \text{(Equation 1)} \]

### Step 3: Use tangent for point C
Now, we need to express the height \( h \) in terms of the distance from point \( C \) to the base of the tower \( B \). The distance from \( A \) to \( C \) is \( AC \), which is equal to \( AB + BC = a + BC \).

From point \( C \):
\[ \tan(18^\circ) = \frac{h}{BC} \]
Let \( BC = x \):
\[ \tan(18^\circ) = \frac{h}{x} \]
This gives us:
\[ h = x \cdot \tan(18^\circ) \quad \text{(Equation 2)} \]

### Step 4: Relate distances
From the right triangle formed by points \( A \), \( B \), and \( C \):
Using the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Substituting \( AC = a + x \) and \( AB = a \):
\[ (a + x)^2 = a^2 + x^2 \]
Expanding the left side:
\[ a^2 + 2ax + x^2 = a^2 + x^2 \]
Cancelling \( a^2 + x^2 \) from both sides:
\[ 2ax = 0 \]
This implies:
\[ x = 0 \quad \text{(which is not possible)} \]

### Step 5: Solve for height
We can express \( x \) in terms of \( a \) using the tangent relationships. From Equation 1 and Equation 2:
\[ \frac{a}{\sqrt{3}} = x \cdot \tan(18^\circ) \]
Rearranging gives:
\[ x = \frac{a}{\sqrt{3} \cdot \tan(18^\circ)} \]
Substituting \( x \) back into Equation 2:
\[ h = \frac{a}{\sqrt{3} \cdot \tan(18^\circ)} \cdot \tan(18^\circ) \]
Thus:
\[ h = \frac{a}{\sqrt{3}} \]

### Final Result
The height of the tower is:
\[ h = \frac{a}{\sqrt{3}} \]