The angles of elevation of the top of a tower from two points which are at distances of 10 m and 5 m from the base of the tower and in the same straight line with it are complementary. The height of the tower is

Options

A. 5m
B. 15m
C. `5sqrt2`m
D. `75 m`

Answer

To solve the problem, we need to find the height of the tower using the information given about the angles of elevation from two points. Let's denote the height of the tower as \( h \).

### Step-by-Step Solution:

1. **Define the Angles of Elevation**:
Let the angle of elevation from the point 10 m away from the tower be \( \theta \) and from the point 5 m away be \( 90^\circ - \theta \) (since they are complementary).

2. **Set Up the Trigonometric Relationships**:
From the point 10 m away, we can use the tangent function:
\[ \tan(\theta) = \frac{h}{10} \]
From the point 5 m away, we have:
\[ \tan(90^\circ - \theta) = \cot(\theta) = \frac{h}{5} \]

3. **Express \( \cot(\theta) \) in Terms of \( \tan(\theta) \)**:
We know that:
\[ \cot(\theta) = \frac{1}{\tan(\theta)} \]
Therefore, we can rewrite the second equation as:
\[ \frac{h}{5} = \frac{1}{\tan(\theta)} \]

4. **Substitute \( \tan(\theta) \)**:
From the first equation, we can express \( h \) in terms of \( \tan(\theta) \):
\[ h = 10 \tan(\theta) \]
Substituting this into the second equation gives:
\[ \frac{10 \tan(\theta)}{5} = \frac{1}{\tan(\theta)} \]
Simplifying this, we have:
\[ 2 \tan^2(\theta) = 1 \]

5. **Solve for \( \tan(\theta) \)**:
Rearranging gives:
\[ \tan^2(\theta) = \frac{1}{2} \]
Taking the square root:
\[ \tan(\theta) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

6. **Find the Height \( h \)**:
Now substituting back to find \( h \):
\[ h = 10 \tan(\theta) = 10 \cdot \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \]

### Final Answer:
The height of the tower is \( 5\sqrt{2} \) meters.