What is the value of cscθ?

Question

The terminal ray of angle θ drawn in standard position passes through (1,−3).

What is the value of cscθ?

Enter your answer in simplest form.

cscθ = 

 

Here's my work:

1) I plotted (1, -3) on a graph and then drew a point from the orgin.

2) Then drew from the point (1, -3) straight up to the x axis, so it forms a right triangle.

3) From there, I calculated the length of each leg of the triangle by using the pythagorean theorem. 

 - the shortest leg 1 unit

 - the one longer than it is 3 units

 - the hypotenuse is 3.16227766 ( I rounded it to 3.16)

4) To get the value of cscθ, I used my knowledge that cscθ is hypotenuse over opposite.

This is where I got confused. I'm havin a hard time determining which angle it's actually asking for. Is cscθ the angle near the orgin or the one from the point given?

 

I also don't really know what it means by simplest form. Does that mean fraction or decimal? Or something else perhaps?

 

If someone could help me, I'd appreciate that a lot:) Thank you!

Answer 1

Hi Ariana. Use the angle near the origin. To find the hypotenuse, you use the Pythagorean Theorem. So it’s 1 squared + -3 squared = c squared. You will get the square root of 10 squared for the hypotenuse.

So for csc, the hyp./ opp. = the square root of 10 / -3. This is the answer, since csc is negative in Quadrant 4. This can’t be simplified unless you are converting a square root to a decimaled number.

Answer 2

nicely done...

the angle is AOB where A is at (1,0), B is at (1,-3), and O is the origin

so opposite is (-3) and adjacent = 1

As your pythagorean says: hypotenuse = sqrt( 1^2 + (-3)^2)

= sqrt( 1 + 9)

= sqrt(10) <--- just leave it in "simplest form";

leave it just like that...

cosecant = hypotenuse / opposite

= sqrt(10)/ -3

= - sqrt(10)/3

Also remember: Add Sugar to Coffee

A: ALL trig functions are positive in quadrant 1

S: Sine function is positive in quadrant 2

T: Tangent function is positive in quadrant 3

C: cosine function is positive in quadrant 4

THe angle is in quadrant 4, so the sine is negative; therefore the cosecant is also negative, which it is

Answer 3

If the beam starts at 0,0.

This angle is 290° (3rd quad)

csc290=1/sin290=-1.064.....

To get 290; a=1, b=3, c=10^.5=3.162...

We want α.

The formula for a rt triangle is: a/sinα=b/sinβ=c/sinγ (In this case=c because the sign of a rt angle=1).

To get c; a^2+b^2=c^2.

See above.

In this case α=θ

So: a/sinα=c=1/sinα=3.162...

1/3.162...=sinα

= .316....

α=18.434...

Third quad=18.434+270=288.434...°