請解出上述題目---**Textual Information:**
**Question Stem:**
如下圖, A, B 兩點分別位於一河口的兩岸邊。某人在通往 A 點的筆直公路上, 距離 A 點 50 公尺的 C 點與距離 A 點 200 公尺的 D 點, 分別測得 ∠ACB=60°, ∠ADB=30°, 則 A 與 B 的距離為幾公尺?
**(English Translation of Question Stem for clarity, not to be included in the output unless requested):**
As shown in the figure below, points A and B are located on opposite banks of a river mouth. A person on a straight road leading to point A measures the angles ∠ACB = 60° and ∠ADB = 30° at point C, which is 50 meters from point A, and at point D, which is 200 meters from point A. What is the distance between A and B in meters?
**Other Relevant Text:**
- Distances from point A:
- C is 50 meters from A.
- D is 200 meters from A.
- Measured angles:
- ∠ACB = 60°
- ∠ADB = 30°
**Chart/Diagram Description:**
* **Type:** Geometric diagram representing a surveying problem involving distances and angles. It shows a river and points on opposite banks and on a road.
* **Main Elements:**
* **Points:** Points labeled A, B, C, and D. A, C, and D are collinear on a horizontal line. B is located above the line containing A, C, and D.
* **Lines:**
* A thick solid line connects B to C and B to D, forming triangles ABC and ABD.
* A dashed line connects B to A, appearing to be perpendicular to the line containing A, C, and D. This dashed line represents the distance AB.
* A dashed line connects B to C.
* A horizontal line segment is labeled with points A, C, and D in that order from left to right.
* **Shapes:** Triangle ABC and Triangle ABD are formed. Triangle ABD appears to be a right-angled triangle at A based on the dashed line from B to A, although this is not explicitly stated as a right angle symbol. The dashed line BA suggests that AB is the perpendicular distance from B to the line AD.
* **Coordinate Axes:** None present.
* **Data Points/Series:** Not applicable.
* **Angles:**
* Angle ∠ACB is marked and labeled as 60°. It is located at point C in triangle ABC.
* Angle ∠ADB is marked and labeled as 30°. It is located at point D in triangle ABD.
* **Labels and Annotations:** Points A, B, C, D are labeled. Angle values 60° and 30° are labeled.
* **Relative Position and Direction:** Points A, C, D are on a straight line. B is on the opposite side of the river from the line AD. The area above the line AD is shaded with wavy lines, representing the river. Points A, C, and D are on one bank (or a road leading to it), and B is on the opposite bank. Point C is between A and D on the line AD.
* **Legend:** None present.
* **River Representation:** The shaded area with wavy lines represents the river. Point B is on the river bank, and the line AD is on the other side, representing the road.
**Other Diagrams:**
There is another geometric figure to the right of the river diagram, which is a large equilateral triangle with smaller dashed equilateral triangles inside, resembling a fractal pattern (Sierpinski triangle). However, this figure is separate and not related to the question about the river and points A and B. Only the river diagram is relevant to the question asked.
视频信息
答案文本
视频字幕
We have a surveying problem where points A and B are on opposite banks of a river.
From point C, which is 50 meters from A, we measure angle ACB as 60 degrees.
From point D, which is 200 meters from A, we measure angle ADB as 30 degrees.
We need to find the distance between A and B.
Let's analyze the geometry step by step. Since points A, C, and D are collinear on the road,
we can find that CD equals 200 minus 50, which is 150 meters.
Now we focus on triangle BCD. Angle BDC is the same as angle ADB, which is 30 degrees.
Since angle ACB is 60 degrees, angle BCD is supplementary to it, so it's 120 degrees.
Using the triangle angle sum, angle CBD is 30 degrees.
Since two angles are equal, triangle BCD is isosceles with BC equal to CD.
Since triangle BCD has two equal angles of 30 degrees, it's isosceles,
so BC equals CD, which is 150 meters.
Now we can focus on triangle ABC where we know AC is 50 meters, BC is 150 meters,
and angle ACB is 60 degrees.
We can apply the Law of Cosines to find AB.
AB squared equals AC squared plus BC squared minus 2 times AC times BC times cosine of 60 degrees.
Substituting the values: AB squared equals 2500 plus 22500 minus 7500, which equals 17500.
Now let's complete the calculation. We found that AB squared equals 17500.
Taking the square root, AB equals the square root of 17500.
We can factor this as the square root of 2500 times 7.
Since the square root of 2500 is 50, we get AB equals 50 times the square root of 7.
Therefore, the distance between points A and B is 50 square root 7 meters.
This is our final answer to the surveying problem.
Let's summarize our complete solution. We started by finding that CD equals 150 meters.
Then we analyzed triangle BCD and discovered it's isosceles with BC equal to 150 meters.
Finally, we applied the Law of Cosines to triangle ABC to find that AB equals 50 square root 7 meters.
This surveying problem demonstrates how trigonometry and geometry can be used to measure distances
across obstacles like rivers. The final answer is 50 square root 7 meters.