In the picture we need to find out the angle, x<90, that D makes with a1/a2. We can use the cosine rule.
C^2=D^2+a2^2+2(a2)(D)cosx; B^2=D^2+a1^2-2(a1)(D)cosx.
[Note that if x=90 (Pythagorean assumption), C^2=D^2+a2^2 and B^2=D^2+a1^2 and C^2-B^2=a2^2-a1^2. (C-B)(C+B)=(a2+a1)(a2-a1).
77.6(2D+292.6)=40870(40870-2a1); 155.2D+22705.76=1670356900-81740a1.
155.2D=1670334194.24-81740a1. From this it's clear that D and a1 cannot be found uniquely. However, because the right-hand side must be positive a1<1670334194.24/81740=20434.72 approx. Any value in excess of this would make D negative. This means that a2>20435.28. a2>a1.]
If x≠90, C^2-B^2=a2^2-a1^2+2(a1+a2)Dcosx.
C-B=77.6, C=B+77.6; a1+a2=40870.
(C-B)(C+B)=40870(a2-a1)+2*40870Dcosx.
77.6(2D+292.6)=40870(40870-2a1)+81740Dcosx.
155.2D+22705.76=1670356900-81740a1+81740Dcosx.
D(155.2-81740cosx)=1670334194.24-81740a1. We have three variables, but none can be found uniquely. But we can confirm that a1<20434.72 and a2>20435.28, provided cosx<155.2/81740=0.0018987 and x>89.89 degrees. For x<89.89, the left-hand and right-hand sides must both be negative. If x=84 degrees, for example, (cosx is about 0.1), -8018.8D=1670334194.24-81740a1. So a1>20434.72 and a2<20435.28. So the magnitude of x affects the relationship between a1 and a2.
Perimeter = 40870+292.6+2D=41162.6+2D.
B+C>a1+a2 in order to produce a triangle. 2D+292.6>40870, D>20288.7, so B>20396.2 and C>20473.8.
EXAMPLE SOLUTION
- Let D=20400, and x=90, then a1=(1670334194.24-155.2*20400)/81740=20395.99 and a2=20474.01 (approx). B=20507.5, C=20585.1.
- Let D=20400, and cosx=0.00175 (x=89.90 approx), then a1=(1670334194.24-12.155)/81740=20434.72, a2=20435.28, B=20507.5, C=20585.1.
The examples show that 40870^2 is such a large number that it is barely affected when anything is subtracted from it so a1 and a2 are both close to 40870/2=20435.