Need help with Principle angles

• Jul 18th 2007, 08:20 PM
supersaiyan
Need help with Principle angles
Determine the priniciple angle

-187

410

-67

905

-282

-730

135

1249

I know the answers but i am not sure how to get those answers because sometime they add or subtract 360 and sometime they add 180

Can somebody plz explain me how to get principle angle

Thanks

and 2 more question

Evaluate y = cos (theta) for 0 degree less than or equal to (Theta) less than or equal to 540 degree when y =-7 (answer to nearest degree)

Evaluate y = sin (theta) for -90 degree less than or equal to (Theta) less than or equal to 540 degree when y =-0.3 (answer to nearest degree)
• Jul 19th 2007, 07:00 AM
topsquark
Quote:

Originally Posted by supersaiyan
Determine the priniciple angle

-187

410

-67

905

-282

-730

135

1249

I know the answers but i am not sure how to get those answers because sometime they add or subtract 360 and sometime they add 180

Can somebody plz explain me how to get principle angle

Thanks

and 2 more question

Evaluate y = cos (theta) for 0 degree less than or equal to (Theta) less than or equal to 540 degree when y =-7 (answer to nearest degree)

Evaluate y = sin (theta) for -90 degree less than or equal to (Theta) less than or equal to 540 degree when y =-0.3 (answer to nearest degree)

By "principle angle" I'm assuming you mean "reference angle?" (I've never heard the term before.)

Anyway, to get a reference angle you are looking for the magnitude of the smallest angle between the given angle and either the + or - x axis.

For example, if you look at the angle -187 degrees you will see that it falls in the second quadrant, so the reference angle will be measured from the -x axis. So the reference angle will be 7 degrees.

As to how you would actually go about finding the value of this angle, my first recommendation is to write the angle as a positive angle measured from the +x axis. If the angle is negative, simply add 360 degrees to it. In the case of -187 degrees we have that -187 + 360 = +173 degrees. Then to find the reference angle we simple subtract this from 180 degrees: 180 - 173 = 7 degrees.

Another example:
410
This is bigger than 360 degrees, so subtract 360 degrees:
410 - 360 = 50 degrees.
This is a first quadrant angle, so we are looking for the angle measured from the +x axis. This is just 50 degrees.

A final example:
-67
This is negative so add 360 degrees to it:
-67 + 360 = 293 degrees
This is a fourth quadrant angle, so we are looking for the angle measured from the +x axis. This is 360 - 293 = 67 degrees.

Quote:

Evaluate y = cos (theta) for 0 degree less than or equal to (Theta) less than or equal to 540 degree when y =-7 (answer to nearest degree)
y can't be -7. Cosine only goes from -1 to 1. There's a typo here somewhere.

Quote:

Evaluate $y = sin( \theta )$ for $-90^o \leq \theta \leq 540^o$ when y =-0.3 (answer to nearest degree)
Use the inverse button on your calculator to find the inverse sine of -0.3. I get
$\theta = -17.4576^o$

Now, we may add or subtract any multiple of 360 degrees to this. We can't subtract 360 degrees from this because that would be less than -90 degrees. But we can add 360 degrees to this because this is less than 540 degrees. So another solution would be:
$\theta = -17.4576^o + 360^o = 342.542^o$

As we can't add another 360 degrees to this without it being larger than 540 degrees, we are done. So the solution set is
$\theta = -17^o, 343^o$
to the nearest degree.

-Dan
• Jul 19th 2007, 11:23 AM
Soroban
Hello, supersaiyan!

What is the definition of a "principal angle" in your course?

I will assume that it is an angle between 0° and 360°.

For all problems, make a sketch of the angle or simply visualize it.

Quote:

Determine the prinicipal angle

$1)\;-187^o$

Starting at the positive x-axis, move clockwise (CW) 187°
. . We end up in Quadrant 2.

Now, what positive angle will get us to the same position?

Since there are 360° in a full circle, going 187° CW
. . is the equivalent to going: . $360^o - 187^o \:=\:173^o$ counterclockwise (CCW).

Therefore: . $-187^o \:\equiv \;173^o$

Quote:

$2)\;410^o$

Starting at the positive x-axis, move CCW 410°
. . We end up in Quadrant 1.

Since: . $410^o \:=\:360^o + 50^o$, we made a "full circle", plus 50°.

We can reach the same position by moving just 50°.

Therefore: . $410^o \:\equiv \:50^o$

Quote:

$3)\;-67^o$

We move 67° CW, down into Quadrant 4.

This is the same as moving CCW: $360^o - 67^o \:=\:293^o$ CCW.

Therefore: . $-67^o \:\equiv \:293^o$

Quote:

$4)\;905^o$

We move CCW 905°.

Since $905^o\:=\:360^o + 360^o + 185^o$,
. . we go "around the circle" twice, plus another 185°
We end up in Quadrant 3.

Therefore: . $905^o \:\equiv \:185^o$

Are you getting the idea?

• Jul 20th 2007, 05:00 AM
supersaiyan
Thanks alot both of you
i Think i got the idea