where's the hypotenuse for an obtuse angle?

[andrewleckie]

Where's the hypotenuse for an obtuse angle? For example, sin 100 degrees. I know it equals the ratio between the side opposite the angle and the hypotenuse, but the hypotenuse is unique to right-angled triangles (as I understand it, but maybe not). And ... right-angled triangles cannot have an obtuse angle!

Say I drew an obtuse angle floating in space. Where do I draw the hypotenuse?

 

[lookagain]

Where's the hypotenuse for an obtuse angle? For example, sin 100 degrees.
I know it equals the ratio between the side opposite the angle and the hypotenuse, but the hypotenuse is
unique to right-angled triangles (as I understand it, but maybe not). And ... right-angled triangles cannot have an obtuse angle!

Say I drew an obtuse angle floating in space. Where do I draw the hypotenuse?

Don't "draw an obtuse angle in space."

The terminal side of the 100 degree angle falls in Quadrant II. 10 degrees of the 100 degrees falls in that quadrant.
Locate a point on that terminal side and drop a perpendicular.

The perpendicular line segment from the point on the terminal side of the 100 degree angle to the horizontal axis
is the opposite side of the complement of the angle 10 degrees. That is, this opposite side is opposite the
80 degree angle. This 80 degree angle spans from the originally mentioned terminal side down to the horizontal axis.

The hypotenuse is the line segment that lies on the terminal side of the 100 degree angle. Its length spans from a point
chosen on the terminal side of the 100 degree angle that is in Quadrant II down to the center of the Unit Circle.

 

[wjm11]

Where's the hypotenuse for an obtuse angle? For example, sin 100 degrees. I know it equals the ratio between the side opposite the angle and the hypotenuse, but the hypotenuse is unique to right-angled triangles (as I understand it, but maybe not). And ... right-angled triangles cannot have an obtuse angle!

Say I drew an obtuse angle floating in space. Where do I draw the hypotenuse?

The picture supplied here:

http://www.purplemath.com/modules/unitcirc.htm

will help you understand Lookagain's explanation. As was stated, do not draw your angle "floating in space." Draw it on an x-y graph, with one side (ray) of the angle pointing to the right (positive x direction). Measure the angle counterclockwise to draw the second ray. Drop a perpendicular line TO THE X-AXIS to form your right triangle.

If your right triangle is in the first quadrant, where x and y are both positive, all six trig functions will be positive. As your triangle is moved to different quadrants, the signs on the trig functions will start changing (as x and/or y become negative).

Hope that helps.

 

[HallsofIvy]

Only right triangles have a "hypotenuse", defined as "the side opposite the right angle" and only right triangles have right angles.

What people are telling you is that "sine", "cosine", etc. do NOT apply only to right triangles. I can't know, of course, what specific definitions sine and cosine your textbook gives, but one common one is the "circle" definition: construct the unit circle on an xy-coordinate system (center at (0, 0), radius 1). For any non-negative number, t, start at (1, 0) and go around the circumference of the unit circle, counter-clockwise, at distance t (for t negative go clockwise). We define cos(t) and sin(t) to be the x and y components, respectively, of the ending point.

For an angle between 0 and 90 degrees, that gives the same thing as the "triangle" definition: drawing the line from (0, 0) to the point (x, y), and dropping a perpendicular to the x-axis, we have a right triangle with "near side" of length x, "opposite side" of length y, and hypotenuse of length 1: cos(t)= "near side over hypotenuse"= x/1= x, and sin(t)= "opposite side over hypotenuse"= y/1= y. But this allows us to define sin(x) and cos(x) for x any real number.

Notice, by the way, that "t" here is NOT an angle, but a distance along the circumference of the circle, and so is NOT measured in 'degrees' or 'radians'. However, the distance along the circumference, t, is proportional to the angle $\displaystyle \theta$: $\displaystyle t= r\theta$ IF $\displaystyle \theta$ is measured in radians.

To solve triangle problems involving triangles that are NOT right triangles, use the "sine law" ($\displaystyle \frac{sin(A)}{a}= \frac{sin(B)}{b}= \frac{sin(C)}{c}$, where A, B, and C are the angles and a, b, c are the lengths of the sides opposite to A, B, and C, respectively) and the "cosine law" ($\displaystyle c^2= a^2+ b^2- 2ab cos(C)$ where again a, b, c are the lengths of the sides opposite angles A, B C, respectively).