I need Calc help!!!?

Question

find the trigonometric value of Ó¨ with the given conditions:

Cos Ó¨ = -15/17 Sin Ó¨ > 0

How do i do this?

Answer 1

There can only be one solution to this problem, because the cosine is negative and the sine positive only in quadrant II.  However, there are actually several possible ways to find the find the angle, Θ.  The most direct way is to find arccos (-15/17), which can be done using a calculator.  That will give us Θ = arccos (-15/17) ≈ 151.9275°.  But it isn't much of a challenge, because it requires little thinking.  

Here's a second plan of attack.  With the given information, we can calculate sin Θ:

sin Θ = {√[(17)² - (-15)²]}/17

sin Θ = √(289 - 225)/17

sin Θ = (√64)/17

sin Θ = 8/17 (Notice that sin Θ > 0.)

Using the arcsin function, we see that Θ ≈ 28.0725°.  But since the required angle lies in quadrant II, then 90° < Θ < 180°, so this angle must be a reference angle.  Therefore, we subtract 28.0725° from 180° to obtain Θ ≈ 151.9275°.  Now we can take the cosine of 151.9275° and see if it closely approximates our given cosine value.  When we do that, we get a value of approximately -0.8824, which agrees with our given value down to the fourth decimal place.

Here is yet a third method for finding Θ.  We are given cos Θ, and we have calculated sin Θ = 8/17.  Now we can calculate tan Θ, and use the arctan function to find Θ.

tan Θ = sin Θ/cos Θ

tan Θ = 8/17/(-15/17)

tan Θ = 8/-15

tan Θ ≈ -0.5333

Now we can take arctan (-0.5333), which gets us Θ ≈ -28.0725.  For reasons we noted earlier, we know that Θ must lie in quadrant II, so this angle also must be a reference angle.  So we subtract its absolute value from 180°, which gets us the same angle as before, Θ ≈ 151.9275.

Answer 2

28'4" is your basic angle found. For Cos to be -ve it can only be possible at Quadrant II and Quadrant III. To find the angle at Q2 u take 180-28'4" u get 151'56". For Q3 180+28'4" = 208'4"  

So, Ó¨ = 151'56" , 208'4"    i used ' to denote degrees and " to denote minutes.

Answer 3

Cos Ó¨ = -15/17

Ó¨ = arccos(-15/17) = arccos(-0.88235)

Ó¨ = 2.65 radians or 151.93 degrees (using a calculator).