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I'm working on some homework. With a simple yes or no if you have a right triangle ABC with B being the 90 degree angle and not a 45-45-90 triangle and you have the value of the hypotenuse and the right angle can you find...

  1. the lengths of both AB and BC?
  2. the angles of A and C?

using sine, cosine and tangent?

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  • $\begingroup$ Try drawing it. Start with two lines at right angles and see how many different ways you can fit a line of a given length to connect to the two lines. $\endgroup$ – Tom Collinge Feb 25 '14 at 11:37
  • $\begingroup$ @TomCollinge: I have deleted my answer. It seems I am the one guilty of misreading the problem. Apologies for my mistake. $\endgroup$ – MPW Feb 25 '14 at 12:22
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This is not really different from previous answers, but slightly more general. If you know the length of one side of a triangle and the size of the opposite angle, you know that the third angle lies on an arc of a circle. With a right-angle you know this is the full circle.

If $R$ is the circumradius, $a$ is the length of the side and $A$ the opposite angle, you have $\cfrac a{\sin A}=2R$

This relates also to the fact that in a fixed circle, the angle subtended by a chord is constant on each arc cut off by the chord.

Your problem is equivalent to "the angle in a semicircle is a right angle".

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Hint: How many ways can you lean a straight ladder against a wall?

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  • $\begingroup$ I don't completely understand. You can lean a ladder against the wall one way so the bottom of the ladder is on the ground and the top of the ladder is leaning against the wall but the ladder could be at different degrees. $\endgroup$ – Jessica M. Feb 25 '14 at 13:20
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    $\begingroup$ That's precisely the idea; you can have different degrees of the angles that aren't the right angle, and these correspond to different side lengths (a la law of sines). This ultimately means that you can have infinitely many triangles for a given hypotenuse. Another way you can think of it is that we need the triangle to satisfy the Pythagorean Theorem: $\overline{AB}^2+\overline{BC}^2=\overline{AC}^2$; we're given $\overline{AC}$, but if I choose any value less than $\overline{AC}$ for $\overline{AB}$, then I can solve for $\overline{BC}$, showing that there are infinitely-many solutions. $\endgroup$ – Hayden Feb 25 '14 at 18:29
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This is nothing more than an elaboration of Hayden's answer, but I'll post it anyway since I spent 10 minutes typing it on my phone while standing in a crowded train.

You can't tell anything about the angles of a right triangle if you only know its hypotenuse.

Imagine that the hypotenuse is a (fixed length) ladder leaning against a vertical wall, so that a right triangle is formed by the ladder, the wall, and the floor. You can slide the foot of the ladder back and forth, and this will change the angles of the triangle.

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  • $\begingroup$ i appreciate the effort. $\endgroup$ – Jessica M. Feb 25 '14 at 13:49

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