Right Triangles?

Question

How can i find the length and the angle of a line that comes from the right angle of a triangle and divides the hypotenuse exactly in half?

Answer 1

the length of the line will be equal to half the length of the hypotenuse as the circumcise will have the center at half the hypotenuse and pass through the three vertices so making the given line also a radius
for angle you may use the sine law
a/sinA=b/sinB=c/sinC=2R

Answer 2

the line that bisect the hypotenuse from 90 degrees angle is equil to half the length of the hypotenuse.
To Prove this, for example: a 3-4-5 Triangle.
You have to draw a diagram so it is easy to see.
1) After draw a line from the 90 degrees angle to the midpoint of the hypotenuse. The length from the midpoint to the endpoint of the Hypotenuse is 2.5.
2) Now, draw a segment that is perpendicular to the leg that has a length of 3 from the midpoint of the hypotenuse. It will form 2 small right triangles.
3) Now, find the angle that is formed by the hypotenuse and the 3-lenth leg by using Sin^-1 (4/5) and the angle will be 53.13010235 degrees. (don't round anything)
4) Use sin(53.13010235)*2.5 to find the high of the small triangle on the right side, and the answer should be 2.
5) use Pythagorean theorem to solve for the leg of the small triangle. a^2+2^2=2.5^2 and the answer should be 1.5.
6) since the whole leg is 3 and the leg of the small triangle is 1.5 so the length of the left triangle is also 1.5
7)Use Pythagorean theorem angle to find the hypotenuse of the left triangle (which is also the segment from the 90 degrees angle to the midpoint of the Hyponenuse) of the left small tringle.
1.5^2+2^2=c^2 and the answer should be 2.5
There, i proved it. Hope this helps

Answer 3

If it divides the hypotenuse in half, I'm thinking that if you then drop a vertical line from the point of intersection to the base that it has also divided the base in half. On both sides of that vertical line, you will have mirror image right triangles. The vertical side will be half the original vertical, the hypotenuse will be half the original hypotenuse and the new angle coming out of the right angle will be the same as the angle originally on the other side of the base from the right angle.

Answer 4

Expand your triangle with the duplicate over the hypotenuse to make a rentangle. Then you see your hypotenuse is one diagonal of a rectangle while your line is the half of the other diagonal. Therefore, your line is half of the hypotenuse, and the angles are the same as two (not 90 degrees) angles of the triangle.

Answer 5

Let the right triangle b ABC where angle B is the right angle, and AC is the hypotenuse. Let D be the midpoint of the hypotenuse.

Then by the law of cosines,
BD^2 = AD^2 + AB^2 - 2*AD*AB cos( angle DAB), where angle DAB = arcsinBC/BA).

I assume you mean the angle DBA when you say "the angle of
the line". If so, use law of sines:

AD/sin (angle DBA) = BD/ sin (angle DAB)
Solve above for sin (angle DBA) then take arcsin to find the angle.

Answer 6

Remember the diagonals of a rectangle are bisectors of each other. So if you make your right triangle into a rectangle, you'll see that you have two equivalent diagonals. So the line from the right angle to the halfway point will be exactly half the length of the hypotenuse.

The picture below will help...

Answer 7

Please do not answer this question about math by 'vtb'. It is part of an ongoing competition called the USAMTS, which gives you a month to write proofs for five problems. All work has to be your own, and this is a clear violation of that policy. This is not the first time that 'vtb' has used Yahoo answer to solve USAMTS problems. So once again, you are just helping this user cheat on the competition if you answer this question.

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This question pertains to his previous question about trapazoids, found here:
http://answers.yahoo.com/question/index;_ylt=AuyLfz91pvvhZijNEVkpTk7sy6IX?qid=20061110150816AAxjGN8