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等腰三角形中某些元素未定时,常会产生多解,需分类考察,举例探析如下.例1 (2006年黑龙江)一条东西走向的高速公路上有两个加油站A、B,在A的北偏东45°方向还有一个加油站C,C到高速公路的最短距离是30千米,B、C间的距离是60千米.要想经过C修一条笔直的公路与高速公路相交,使两路交叉口P到B、C的距离相等,请求出交叉口P与加油站A的距离(结果可保留根号).
Some elements in an isosceles triangle are not timed, and often produce multiple solutions. They need to be investigated and classified. Examples are discussed below. Example 1 (Heilongjiang, 2006) There are two gas stations A and B on an east-west highway, north of A. There is also a gas station C in the direction of the east-east 45°, and the shortest distance from C to the highway is 30 km, and the distance between B and C is 60 km. In order to pass through C, a straight highway intersects with the highway, making The distance between the two intersections P to B and C is equal, and the distance between the intersection P and the gas station A is requested (the result can retain the root number).