The angles of elevation of the top of a tree 220 meters high from two points lie on the same plane are 30° and 45°. What is the distance (in metres) between the two points?

Options

A. 193.22
B. 144.04
C. 176.12
D. 161.05

Answer

To solve the problem, we need to find the distance between two points from which the angles of elevation to the top of a tree are given. The tree is 220 meters high, and the angles of elevation from the two points are 30° and 45°.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a tree of height 220 meters. We denote the top of the tree as point T, the base of the tree as point B, and the two points from which the angles of elevation are measured as points A and C.

2. **Setting Up the Right Triangles**:
- From point A (where the angle of elevation is 30°), we can form a right triangle ABT.
- From point C (where the angle of elevation is 45°), we can form another right triangle CBT.

3. **Using Trigonometry**:
- For triangle ABT (angle of elevation = 30°):
\[ \tan(30°) = \frac{BT}{AB} \]
Here, \( BT = 220 \) meters (height of the tree). Therefore:
\[ \tan(30°) = \frac{220}{AB} \]
We know that \( \tan(30°) = \frac{1}{\sqrt{3}} \). Thus:
\[ \frac{1}{\sqrt{3}} = \frac{220}{AB} \]
Rearranging gives:
\[ AB = 220\sqrt{3} \]

- For triangle CBT (angle of elevation = 45°):
\[ \tan(45°) = \frac{BT}{BC} \]
Again, \( BT = 220 \) meters. Therefore:
\[ \tan(45°) = \frac{220}{BC} \]
Since \( \tan(45°) = 1 \):
\[ 1 = \frac{220}{BC} \]
Rearranging gives:
\[ BC = 220 \]

4. **Finding the Distance Between Points A and C**:
The distance between points A and C is the sum of distances AB and BC:
\[ AC = AB + BC = 220\sqrt{3} + 220 \]
Factoring out 220 gives:
\[ AC = 220(\sqrt{3} + 1) \]

5. **Calculating the Numerical Value**:
Using the approximate value of \( \sqrt{3} \approx 1.732 \):
\[ AC \approx 220(1.732 + 1) = 220(2.732) \approx 600.84 \]

### Final Answer:
The distance between the two points A and C is approximately 600.84 meters.