The task of finding of a corner of a polygon at the known several of its parameters is rather simple. In case of definition of a corner between a median of a triangle and one of the parties it is convenient to use vector method. To set a triangle, there are enough two vectors of its parties.
Instruction
1. In fig. 1 triangle is completed to the parallelogram corresponding to it. At the same time it is known that in a point of intersection of diagonals of a parallelogram they are halved. Therefore the joint-stock company is median triangle AVS lowered from And on the party of VS. From this it is possible to conclude that it is necessary to find a corner φ between the party the EXPERT of a triangle and a median of the joint-stock company. The same corner, according to fig. 1, is available between a vector and and d vector corresponding to AD parallelogram diagonal. By the rule of a parallelogram the vector of d is equal to the geometrical sum of vectors of an and b, d = to a + to b.
2. It is necessary to find a way of definition of a corner φ. For this purpose it is necessary to use a scalar product of vectors. It is the most convenient to define a scalar product on the basis of the same vectors of an and d which is determined by a formula (a, d) = |a| |d| cosφ. Here φ – a corner between vectors of an and d. As the scalar product of the vectors set by coordinates is defined by expression: (a (ax, ay), d (dx, dy)) =axdx+aydy, |a| ^2 = ax^2 + ay^2, |d| ^2 = dx^2 + dy^2, that cosφ=(axdx+aydy) / (sqrt (ax^2 + ay^2) sqrt (dx^2 + dy^2)). Besides, the sum of vectors in a coordinate form is defined by expression: d (dx, dy) = a (ax, ay) + b (bx, by) = {ax+bx, ay+by }, that is dx = ax+bx, dy=ay+by.
3. Example. The triangle of AVS is set by vectors of a (1.1) and b(2, 5) according to fig. 1. To find a corner φ between its median the joint-stock company and the party of a triangle the EXPERT. Decision. As it was already shown above, for this purpose it is enough to find a corner between vectors and and d. This corner is set by its cosine and sqrt (dx^2 + dy^2)).1 is calculated according to the following identity of cosφ=(axdx+aydy) / (sqrt (ax^2 + ay^2). d (dx, dy) = {1+2, 1+5} = d(3, 6).2. cosφ=(3+6) / (sqrt(1+1)sqrt(9+36))=9 / (3sqrt (10)) =3/sqrt (10). φ=arcos (3/sqrt (10)).