1250 in the original problem is decomposed into 49+576+625.
Then the original problem becomes A2+B2+C2-14a-48b-50c+49+576+625 = 0.
The deformation is (a-7)2+(b-24)2+(c-25)2=0.
Because the square must be greater than or equal to zero, all three are equal to zero, so a=7, b=24, c=25, and the Pythagorean theorem is satisfied, so the triangle is a right triangle! !
The second question:
The key to this problem is to find the side length of a triangle. Because drawing is not very convenient, you need some space to imagine, depending on your understanding, landlord.
First, make a garden with a center and a radius of 5, and record it as a circle; Make a circle with B as the center and 4 as the radius, and write it as circle B; Make a circle with C as the center and 3 as the radius, and write it as circle C. Then three circles have a focus in triangle ABC, which is P.
Circle a, circle b and circle c intersect, that is, the length of three sides of triangle ABC can be expressed as the sum of the radii of two intersecting circles, and then the intersection part is subtracted.
Suppose that the intersection of circle A and circle B is set to X, the intersection of circle A and circle C is set to Y, and the intersection of circle B and circle C is set to Z..
Then there is: 5+4-X=5+3-Y=4+3-Z is converted into 9-X=8-Y=7-Z XYZ, and the values are as follows:
(3,2, 1); (4,3,2); (5, 4, 3) ...(9, 8, 7) * * * There are seven possible groups and seven possible side lengths, which are 6, 5, 4, 3, 2, 1 0. According to the meaning of the question, if the point P is in the triangle and the tangent AP=5, then the side length of the triangle is at least greater than 5.
So the triangle is an equilateral triangle with a side length of 6.
In this way, the problem becomes the problem of triangular BPC. Where BC=6, BP=4 and CP=3, find the angle BPC.
Let p be a vertical line, intersect BC at point o, let BO=m, and let co = n.
Then m+n = 6; 4 square meters =3 square meters = n square meters
The solution is m = 43/12; n=29/ 12
Sin angle BPO = m/4 = (43/12)/4 = 43/48; So the angle BPO=arcSin43/48
Sin angle CPO = n/3 = (29/12)/3 = 29/36; So the angle CPO=arcSin29/36.
Angle BPC= Angle BPO+ Angle CPO=arcSin43/48+arcSin29/36.