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

In summary: You can rewrite -75.5 as 284.5 but that is outside the given range. So we get:(1) $\displaystyle \theta - 150 = 75.5$(2) $\displaystyle \theta - 150 = -284.5$Adding 150 to each equation and we get:(1) $\displaystyle \theta = 225.5$(2) $\displaystyle \theta = -134.5$So $\theta$ is either 225.5 or -134.5. The reason why 134.5 isn't valid is because it makes $\theta - 150 = -15.5$ and $\cos(-15.5)$ is negative. We need
  • #1
Needhelp2
17
0
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)

 
Mathematics news on Phys.org
  • #2
Needhelp said:
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
 
  • #3
topsquark said:
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!
 
  • #4
Needhelp said:
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:

\(\displaystyle \theta - 150 = 75.5\) and \(\displaystyle \theta - 150 = -75.5\)

Due to restrictions though in directions we must rewrite the first angle as a negative angle so we get \(\displaystyle \theta - 150 = -284.5\)

Make sense?
 
  • #5
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 :confused:

Thanks though for all your help!
 
  • #6
Needhelp said:
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 :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.
 
Last edited:

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

What are the reciprocal trig functions?

The reciprocal trig functions, also known as the inverse trig functions, are the inverse operations of the basic trigonometric functions (sine, cosine, tangent, cotangent, secant, and cosecant). They are used to find the angle measure given the ratio of sides in a right triangle.

What are the six reciprocal trig functions?

The six reciprocal trig functions are: arcsine (sin⁻¹), arccosine (cos⁻¹), arctangent (tan⁻¹), arccotangent (cot⁻¹), arcsecant (sec⁻¹), and arccosecant (csc⁻¹).

How are the reciprocal trig functions related to the basic trigonometric functions?

The reciprocal trig functions are the inverse operations of the basic trigonometric functions. For example, arcsine is the inverse of sine, arccosine is the inverse of cosine, and so on. This means that they undo the effects of the basic trigonometric functions.

What is the domain and range of the reciprocal trig functions?

The domain of the reciprocal trig functions is the set of all real numbers, except for certain values that result in undefined outputs (such as dividing by zero). The range of the reciprocal trig functions is also the set of all real numbers, but there are restrictions on certain values to ensure that the outputs are defined.

How are the reciprocal trig functions used in real-life applications?

The reciprocal trig functions are commonly used in fields such as engineering, physics, and surveying to solve problems involving angles and sides of triangles. They are also used in navigation, astronomy, and architecture to calculate distances, heights, and angles. Additionally, they are important in calculus and other advanced mathematical concepts.

Back
Top