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Why is is $x>90^\circ$?

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I thought that since we do not have the $5,12,13$ sides of a right triangle, the only way that's possible is the height of the triangle is leaning right, so $x<90$, but the answer key says that $x>90$.

The only way I can imagine how that's possible is the angle between the side with length five and $13$ is smaller than the angle between the side with length $5$ and $13$ in a $5,12,13$ triangle while the angle between the side with length $11$ and $13$ compared to the angle in the right triangle with sides $12$ and $13$ is unchanged.

Blue
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    $5^2+11^2=146\lt 169=13^2$. Since that side won't fit, we must open the angle to be more than $90^{\circ}$ to accommodate the side. If we closed the angle, we would need an even smaller side. – John Douma Jan 29 '23 at 03:17
  • Why can't you decrease the other two angles to make the side fit? @JohnDouma –  Jan 29 '23 at 03:18
  • I believe the side lengths are fixed. To decrease the other angles, we would have to lengthen at least one of the other sides. – John Douma Jan 29 '23 at 03:19
  • When are the angles fixed by side-lengths changeable rather than fixed? @JohnDouma? –  Jan 29 '23 at 03:25
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    I don't understand your last question. You suggested decreasing one or both of the non-right angles to fit the hypotenuse. To do that you must increase one of the lengths. I suggest you get a ruler and some craft sticks and cut them to ratios of those sides, i.e. 5:11:13 and put the triangle together. That may sound simplistic but you will gain the insight to see what I am talking about above. – John Douma Jan 29 '23 at 03:29

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Purely Computational solution:
Using Cosine Law: $\displaystyle c^{2} =a^{2} +b^{2} -2abcos( C)$. Substituting a = 5,b=11c=13 we get that
$\displaystyle cos( C) \ =\ ( 121+25-169) /( 2*11*5) \ =-23/110\ \ ( < 0)$
So angle (C) must be obtuse (>90).
Geometrically:
Lets say you are keeping the sides (c=13,a=5) as a constant. Then as you rotate the angle in anticlockwise(increasing angle C) the side (b) of the triangle should decrease in length to keep the other two sides constant.

mrtechtroid
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