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Trapezium

Can angles A and B be solved? Neither the area nor the perimeter was given. Thank you very much if you can help! :)

anakin skywalker
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  • If it is a trapezium, are the sides of length $5.9$ and $9$ parallel? –  Aug 03 '14 at 08:36
  • I see, sometimes it is confusing by what a trapezium mean between british and american definitions. But in the question, I'm referring to the american definition of a trapezium which there are no parallel sides. So, 5.9 and 9 are not parallel. Thank you for your response @Rahul :) – anakin skywalker Aug 03 '14 at 08:45
  • You can take a look at my answer (also others') in http://math.stackexchange.com/questions/713435/geometry-question-regarding-existence-of-a-quadrilateral/716657#716657 – Mick Aug 03 '14 at 11:15

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If I'm not mistaken, I think the followings tell us that the angles $A,B$ are not determined. Set $$A(0,0),B(9,0),C(9+7\cos(180^\circ-B),7\sin(180^\circ-B)),D(4\cos A,4\sin A).$$ Then, we have $$CD^2=5.9^2=\{9+7\cos(180^\circ-B)-4\cos A\}^2+\{7\sin(180^\circ-B)-4\sin A\}^2$$ $$\iff 5.9^2=9^2+7^2+4^2-126\cos B-72\cos A+56\cos A\cos B-56\sin B\sin A.$$ Here, even if we suppose that $0\le A,B\le 90^\circ,$ we have $$5.9^2=9^2+7^2+4^2-126y-72x+56xy-56\sqrt{(1-x^2)(1-y^2)}$$ where $x=\cos A, y=\cos B, \sin A=\sqrt{1-x^2}, \sin B=\sqrt{1-y^2}.$

For example, if $A=90^\circ$, then $\cos B=y=39.19/70\approx 0.559857.$

If $A=45^\circ$, then $\cos B=y\approx 0.254432.$

Since we have at least two concrete examples, it's obvious that there are infinitely many pairs $(A,B)$ because $C$ is on the circle whose center is $B$ with radius $7$ and $D$ is on the circle whose center is $A$ with radius $4$.

mathlove
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    More simply: You can take diagonal $AC$ to be some particular length, say $d$. Then you can build a $d$-$4$-$5.9$ triangle to one side of it, and an $d$-$9$-$7$ triangle on the other side. Note that $d$ has to be between $5.9-4$ and $5.9+4$ to make the first triangle viable; and it has to be between $9-7$ and $9+7$ to make the second triangle viable. Therefore, the range of $d$ is $2$ to $9.9$; and because there's a range, the trapezium is not uniquely determined by the lengths of its edges. – Blue Aug 03 '14 at 10:23