Finding the length of the hypotenouse without Pythagorous

[ecorocks]

Hi! This is my first post, so I don't know the formalities--I guess I should introduce myself? I'm a senior in college studying Evolutionary Biology, with two minors in Philosophy and Biomolecular Technology. My major demands alot of applied math, and all seniors have to take Physics, so I'll be here alot making sure I know my basic trig skills for the class. I'll help other users if I can too, I've taken Calculus and Statistics already!

My Physics professor is a nice guy, but he's foriegn, and I really don't understand what he is saying.

I have a triangle: I only know two lengths--it is 8m and 215m; niether are the hypotenuse. The angle facing the hypotenuse is 135 degrees. How do I find the other angles, and the length of the hypotenuse?

 

[stapel]

Re: Finding the length of the hypotenouse without Pythagorou

The angle facing the hypotenuse is 135 degrees.

The "hypotenuse" is the long side of a right triangle. By definition, the facing angle measures ninety degrees. If the facing angle measures 135°, then the triangle is not right, and the side in question cannot be the hypotenuse.

How much trigonometry and/or geometry do you know? For instance, are you familiar with the Law of Sines?

Thank you.

Eliz.

 

[ecorocks]

It's been 5 years since I studied Trig, I'm trying to get it all back. Law of sines sounds familiar--I couldnt tell you verbatim what it is though.

 

[Denis]

Nothing really wrong with your "kind of question", except pretty difficult
to answer here, due to all the typing necessary to explain;
you can get answers quickly and already typed out(!) by using
http://www.google.com ; like this site for right triangles:
http://id.mind.net/~zona/mmts/trigonome ... angle.html

 

[soroban]

Re: Finding the length of the hypotenouse without Pythagorou

Hello, ecorocks!

Welcome aboard!

You refer to a "hypotenuse", which exists only in a right triangle.
. . Your triangle is obviously not a right triangle.
I'll reword the problem . . .

I have a triangle.
Two of the sides have lengths: 8 m and 215 m.
The angle between them is 135°.
Solve the triangle.

    B
      *
       *    *
        *         *
         *              *   a
          *                   *
     c = 8 *                        *
            *                             *
             * 135°                             *
              *   *   *   *   *   *   *   *   *   *   *
            A               b = 215                     C

Use the Law of Cosines to find side \(\displaystyle a.\)

\(\displaystyle a^2\;=\;b^2\,+\,c^2\,-\,2bc\cos A \;=\;215^2\,+\,8^2\,-\,2(215)(8)\cos135^o \;=\;48721.44733\)

. . Hence: \(\displaystyle \L\,a\:\approx\:220.73\)


Then: \(\displaystyle \,\cos B\;=\;\L\frac{a^2\,+\,c^2\,-\,b^2}{2ac}\)\(\displaystyle \;=\;\L\frac{220.73^2\,+\,8^2\,-\,215^2}{2(220.73)(8)}\)\(\displaystyle \;=\; 0.725075007\)

. . Hence: \(\displaystyle \L\,B\:\approx\:43.52^o\)


Then: \(\displaystyle \,C\:=\:180^o\,-\,135^o\,-\,43.52^o\;\;\Rightarrow\;\;\L C\:=\:1.48^o\)