Question

 

Which is a possible value for x?#

 18

 15

 1

 17

Answer 1

Let y=AB=CD (y represents the lengths of AB and CD); let a=BD

Cosine Rule:

13²=a²+y²-2aycos(3x-4),

16²=a²+y²-2aycos(47).

Subtract these two equations:

16²-13²=87=2ay(cos(3x-4)-cos(47)).

cos(3x-4)=87/(2ay)+cos(47).

Since the RHS must be greater than cos(47), cos(3x-4)>cos(47). This means 3x-4<47°, 3x<51°, x<17° because, for example, if x=15°, 3x-4=41° and cos(41)>cos(47)

This reduces the possible values of x to 1 or 15. We have just shown that x=15° is a possibility:

cos(41)-cos(47)=87/(2ay)=0.0727 approx. So 2ay=87/0.0727=1196.514 approx, so ay=598.257.

When x=1°, 3x-4=-1°, which is a negative angle which cannot apply to an angle in a triangle. Therefore, the solution to this question is x=15°, even though we can't determine the length of BD; or the equal lengths AB and CD.

Note that x=15° is a possible solution, but not an actual solution. The actual solutions differ for different values of a and y (different lengths for BD and AB, CD). In fact, x=15° produces no solutions for the lengths of BD, AB, CD. Below is a specific example of the quadrilateral in which a=y=AB=BD=CD in length (triangles ABD and BDC are isosceles). (In this scaled figure x=13.936° approx.)