How Do I Solve sec(θ-150°)=4 for θ Within Specific Degree Limits?

[Needhelp2]

After a long summer, I finding my new C3 homework a bit tricky, so any help would be great!

Here is the question: sec(θ-150 degrees)=4

(solving for theta is greater than or equal to -180, but less than or equal to 180)

So I know that sec is the reciprocal of cos so I changed the equation to cos(θ-150)=1/4
from there I did the inverse of cos to get θ-150 = 75.52, then I added 150 to get θ= 225.5
I then drew a cos graph and from this would assume that the answers are 134.5 and -134.5... but the answer booklet for my textbook says they are -134.5 and 74.5!

This is probably so simple but I have no clue what I'm doing wrong!
(Sweating)

 

[topsquark]

After a long summer, I finding my new C3 homework a bit tricky, so any help would be great!

Here is the question: sec(θ-150 degrees)=4

(solving for theta is greater than or equal to -180, but less than or equal to 180)

So I know that sec is the reciprocal of cos so I changed the equation to cos(θ-150)=1/4
from there I did the inverse of cos to get θ-150 = 75.52, then I added 150 to get θ= 225.5
I then drew a cos graph and from this would assume that the answers are 134.5 and -134.5... but the answer booklet for my textbook says they are -134.5 and 74.5!

This is probably so simple but I have no clue what I'm doing wrong!
(Sweating)

The angle theta - 150 is a reference angle. What does that mean for your solutions?

-Dan

 

[Needhelp2]

The angle theta - 150 is a reference angle. What does that mean for your solutions?

-Dan

What is a reference angle? Sorry for such a silly question!

 

[Jameson]

After a long summer, I finding my new C3 homework a bit tricky, so any help would be great!

Here is the question: sec(θ-150 degrees)=4

(solving for theta is greater than or equal to -180, but less than or equal to 180)

So I know that sec is the reciprocal of cos so I changed the equation to cos(θ-150)=1/4
from there I did the inverse of cos to get θ-150 = 75.52, then I added 150 to get θ= 225.5
I then drew a cos graph and from this would assume that the answers are 134.5 and -134.5... but the answer booklet for my textbook says they are -134.5 and 74.5!

This is probably so simple but I have no clue what I'm doing wrong!
(Sweating)

It looks like you've got one answer but the second answer of 74.5 is confusing you. Where are $\cos(x)$ or $\sec(x)$ positive? In the first and fourth quadrants. Since your solution is positive then these need to be positive as well.

The way to set up your solution normally would be:

[math]\theta - 150 = 75.5[/math] and [math]\theta - 150 = -75.5[/math]

Due to restrictions though in directions we must rewrite the first angle as a negative angle so we get [math]\theta - 150 = -284.5[/math]

Make sense?

 

[Needhelp2]

Not really...(Sweating) I don't understand where you got the value of -284 from, and it isn't within the degree specification? (at least if it goes from -180 up to 180)
Sorry for being so dim, could you possibly take it step by step to get the second value of 74.5 and explain why 134.5 isn't a valid answer?

Im just so confused

Thanks though for all your help!

 

[Jameson]

Not really...(Sweating) I don't understand where you got the value of -284 from, and it isn't within the degree specification? (at least if it goes from -180 up to 180)
Sorry for being so dim, could you possibly take it step by step to get the second value of 74.5 and explain why 134.5 isn't a valid answer?

Im just so confused

Thanks though for all your help!

Sure :)

You're right that -284 isn't within the degree specification but that's not what $\theta$ is. We're looking at $\theta - 150$. $\theta$ must be from -180 to 180 but not $\theta - 150$. The -284.5 comes from the fact that if you look at the angle 75.5 degrees you can consider this as going counter-clockwise 75.5 degrees or go the other way a whole -284.5 degrees (clockwise because it's a negative angle) and you'll reach the same place. Notice that 284.5+75.5=360. Try drawing them both.

This problem has a restriction on $\theta$ which makes it tricky. You almost solved it yourself. Without the restrictions you could just solve:

(1) $\displaystyle \theta - 150 = 75.5$
(2) $\displaystyle \theta - 150 = -75.5$

as I said before. The reason you would choose (1) and (2) is because we need $\cos( \theta-150)$ to be positive and cosine is only positive in the first and fourth quadrant.

However due to the restrictions on $\theta$ if we solve both of these then the answer for (1) doesn't fall within the specified domain, so we must rewrite 75.5 as -284.5.