Which quadrilaterals always have diagonals that are perpendicular?
2 answers:
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These are a huge pain. First set up your initial triangle with A and B as your base angles and C as your vertex angle. Now drop an altitude and call it h. You need to solve for h. Use sin 56 = h/13 to get that h = 10.8. The rule is that if the side length of a is greater than the height but less than the side length of b, you have 2 triangles. h<a<b --> 10.8<12<13. Those are true statements so we have 2 triangles. Side a is the side that swings, this is the one we "move", forming the second triangle. First we have to solve the first triangle using the Law of Sines, then we can solve the second. 
to get that angle B is 64 degrees. Now find C: 180-56-64=60. And now for side c: 
and c=12.5. That's your first triangle. In the second triangle, side a is the swinging side and that length doesn't change. Neither does the angle measure. Angle B has a supplement of 180-64 which is 116. So the new angle B in the second triangle is 116, but the length of b doesn't change, either. I'll show you how you know you're right about that in just a sec. The only angle AND side that both change are C and c. If our new triangle has angles 56 and 116, then C has to be 8 degrees. Using the Law of Sines again, we can solve for c:
and c = 2.0. We can look at this new triangle and determine the side measures are correct because the longest side will always be across from the largest angle, and the shortest side will always be across from the smallest angle. The new angle B is 116, which is across from the longest side of 13. These are hard. Ugh.
The Answer Is 3a^2 3b^2 3c^2
Since all the variables equal each other, you may pick one variable to copy in the equation.
And since they're the same numbers, squaring a will have no effect in the equation.