For a task of such quadrangle as a trapeze, not less than three of its parties have to be defined. Therefore, for an example, it is possible to consider a task in which condition lengths of diagonals of a trapeze and also one of side vectors are set.
Instruction
1. The figure from a statement of the problem is presented in figure 1. In this case it is necessary to assume that the considered trapeze is a quadrangle of ABCD in which lengths of diagonals of AC and BD and also AV side presented by a vector are set (ax, ay). The accepted basic data allow to find both bases of a trapeze (both top, and lower). The lower basis of AD at first will be found in a concrete example.
2. Consider ABD triangle. Length of its party of AV is equal to a vector module. Let |a| =sqrt ((ax) ^2+(ay) ^2) =a, then cosf =ax/sqrt (((ax) ^2+(ay) ^2), as the directing a cosine. Let the set diagonal of BD have p length, and required AD length x. Then, according to the theorem of cosines, P^2=a^2 + x^2-2axcosф. Or x^2-2axcosф + (a^2-p^2) =0.
3. Solutions of this square equations:X1=(2acosф+sqrt(4(a^2)((cosф)^2)-4(a^2-p^2)))/2=acosф+sqrt((a^2)((cosф)^2)-(a^2-p^2))==a*ax|sqrt(((ax)^2+(ay)^2)+sqrt((((a)^2)(ax^2))/(ax^2+ay^2))-a^2+ p^2) =AD.
4. For finding of the top basis of VS (its length by search of the decision is also designated x) the |=a module and also the second diagonal of BD=q and a cosine of the angle of AVS which, obviously, is equal (p-f) is used.
5. Further the triangle of AVS to which, as well as earlier, the theorem of cosines is applied is considered, and there is the following decision. Considering that cos (p-f)=-cosf, on the basis of the decision for AD, it is possible to write down the following formula, having replaced p with q: VS =-a*ax | sqrt (((ax) ^2+(ay) ^2) +sqrt ((((a) ^2) (ax^2)) / (ax^2+ay^2))-a^2+q^2).
6. This equation is quadratic and, respectively, has two roots. Thus, in this case it is necessary to choose only those roots which have positive value as length cannot be negative.
7. Primerpust in ABCD trapeze the side of AV is set by a vector (1, sqrt3), p=4, q=6. To find the trapeze bases. Decision. Using the algorithms received above it is possible to write down: |a| =a=2, cosf =1/2. AD=1/2+sqrt (4/4 - 4+16)=1/2 +sqrt (13)= (sqrt(13)+1)/2.BC=-1/2+sqrt(-3+36)= (sqrt(33)-1)/2.