The angles of elevation of the top of a tower from two points on the ground at distances 32 m and 18 m from its base and in the same straight line with it are complementary. The height(in m) of the tower is .....

Options

A. 24
B. 20
C. 28
D. 16

Answer

To solve the problem step by step, we will denote the height of the tower as \( h \) meters. The distances from the base of the tower to the two points on the ground are 32 m and 18 m, respectively. The angles of elevation from these two points are complementary, meaning they add up to 90 degrees.

### Step 1: Define the Angles
Let the angle of elevation from the point 32 m away be \( \theta \). Consequently, the angle of elevation from the point 18 m away will be \( 90^\circ - \theta \).

### Step 2: Set Up the Tangent Ratios
Using the definition of tangent for the first point (32 m away):
\[ \tan(\theta) = \frac{h}{32} \]
For the second point (18 m away), we can use the cotangent since the angle is \( 90^\circ - \theta \):
\[ \tan(90^\circ - \theta) = \cot(\theta) = \frac{h}{18} \]

### Step 3: Relate the Two Equations
From the first equation, we can express \( h \) in terms of \( \theta \):
\[ h = 32 \tan(\theta) \]
From the second equation, we express \( h \) as well:
\[ h = 18 \cot(\theta) \]
Since both expressions equal \( h \), we can set them equal to each other:
\[ 32 \tan(\theta) = 18 \cot(\theta) \]

### Step 4: Use the Identity for Cotangent
Recall that \( \cot(\theta) = \frac{1}{\tan(\theta)} \). Substituting this into the equation gives:
\[ 32 \tan(\theta) = 18 \cdot \frac{1}{\tan(\theta)} \]

### Step 5: Multiply Both Sides by \( \tan(\theta) \)
To eliminate the fraction, multiply both sides by \( \tan(\theta) \):
\[ 32 \tan^2(\theta) = 18 \]

### Step 6: Rearrange the Equation
Rearranging gives:
\[ \tan^2(\theta) = \frac{18}{32} \]
This simplifies to:
\[ \tan^2(\theta) = \frac{9}{16} \]

### Step 7: Take the Square Root
Taking the square root of both sides, we find:
\[ \tan(\theta) = \frac{3}{4} \]

### Step 8: Substitute Back to Find \( h \)
Now, substitute \( \tan(\theta) \) back into one of the equations for \( h \):
\[ h = 32 \tan(\theta) = 32 \cdot \frac{3}{4} = 24 \]

### Final Answer
Thus, the height of the tower is:
\[ \boxed{24 \text{ m}} \]