Two climbers are at points A and B on a vertical cliff face. To an observer C, 40 m from the foot of the cliff, on the level ground . A is at an elevation of `48^(@)` and B of `57^(@)` . What is the distance between the climbers ?

Answer

To find the distance between the climbers A and B on the vertical cliff face, we can follow these steps:

### Step 1: Understand the setup
We have an observer C who is 40 meters away from the foot of the cliff (point P). The angles of elevation to climbers A and B are given as 48 degrees and 57 degrees, respectively.

### Step 2: Draw the diagram
1. Draw a vertical line representing the cliff.
2. Mark point P at the bottom of the cliff (the foot).
3. Mark point C, which is 40 meters horizontally from point P.
4. Mark points A and B at different heights on the cliff.
5. Draw lines from point C to A and B, creating angles of elevation of 48 degrees and 57 degrees.

### Step 3: Calculate the height of climber B (BP)
Using triangle BPC:
- The tangent of angle 57 degrees is given by the formula:
\[ \tan(57^\circ) = \frac{BP}{PC} \]
- Here, \( PC = 40 \) meters.
- Rearranging gives:
\[ BP = PC \cdot \tan(57^\circ) \]
- Using the value of \( \tan(57^\circ) \approx 1.539 \):
\[ BP = 40 \cdot 1.539 \approx 61.57 \text{ meters} \]

### Step 4: Calculate the height of climber A (AP)
Using triangle APC:
- The tangent of angle 48 degrees is given by the formula:
\[ \tan(48^\circ) = \frac{AP}{PC} \]
- Again, \( PC = 40 \) meters.
- Rearranging gives:
\[ AP = PC \cdot \tan(48^\circ) \]
- Using the value of \( \tan(48^\circ) \approx 1.11 \):
\[ AP = 40 \cdot 1.11 \approx 44.4 \text{ meters} \]

### Step 5: Find the distance between climbers A and B
Now, we need to find the difference in height between climbers B and A:
\[ \text{Distance between climbers} = BP - AP \]
Substituting the values we calculated:
\[ \text{Distance} = 61.57 - 44.4 \approx 17.17 \text{ meters} \]

### Final Answer
The distance between the climbers A and B is approximately **17.17 meters**.

---