# Homework Help: Trig- Solve for the angle(s)

1. Mar 11, 2009

### Oscar Wilde

1. The problem statement, all variables and given/known data
Solve for x, 0 degress (is less than or equal to) x (is less than or equal to) 360 degrees. Give your answer to the nearest tenth of a degree. Use tables or a calculator when necessary.

1. tan (x+15 degrees)=1

2. sec (x-5 degrees)=2

2. Relevant equations
sec= 1/cos Isolating the variable...

3. The attempt at a solution

Ok so there's only one part thats tripping me up. But first let me show you my work

#1 tan x= 1- 15 degrees
tan^-1 (1)=45 degrees
45 degrees - 15 degrees= 30 degrees
Tangent is positive in the first and third quadrant, I have the first, so I obtain my third quadrant angle by 180 deg+30 deg=210
ANSWER: 30,210 (this wasn't the one I was having problems with, but I'd be elated if someone could verify my answer)

#2 sec x= 2+ 5 degrees
(1/cos) x= 2 + 5 deg
entire group to the negative first power, which gives me cosine. This is the part I am unsure of. Do I raise the degree portion (5 deg) to the negative first as well, or do I exempt it?
I chose to include, which yields
cos x= (1/2) + .2 deg
cos x^-1(1/2)=60 deg +.2 deg
60.2 deg in the first quadrant. Cosine is positive in the first and fourth quadrant. I obtain the fourth quadrant value by 360-60.2 deg=299.8 degrees

But anyway my question is highlighted in the bolded row. Help much appreciated, thanks.

2. Mar 11, 2009

### Mentallic

sorry I can't follow exactly what you've done.

The first step you took:

<== this is incorrect. But somehow you managed to get the right answer in the end

How about trying another approach for both questions. It might make it easier for you to understand.

$$tan(x+15^o)=1$$

now, let $$x+15^o=u$$

therefore, $$tan(u)=1$$

now solve for u: $$u=tan^{-1}1=45^o,225^o$$

and substituting u back to solve for x: $$x+15^o=45^o, 225^o$$

Hence, $$x=30^o,210^o$$

Try it for Q2

Remember, $$sec(x)=\frac{1}{cos(x)}$$

3. Mar 12, 2009

### Staff: Mentor

This technique works if the answer is in the back of the book and you write a bunch of stuff between the start of the problem and the answer. It doesn't matter what the "bunch of stuff" is, just so long as you start with the problem and end with the answer. This technique is not as useful when you don't have the answer.

First off, tan(x + y) is almost never equal to tan(x) + y, which is a major flaw in your second equation.

Next, if you start with tan(x) = 1 - 15 degrees (putting aside the question of how one might subtract 15 degrees from 1?), and take the arctangent of both sides, I get
x = tan-1(1 - 15 degrees). There's that pesky 1 - 15 degrees business again.

Just as tan(x + y) != tan(x) + y, tan-1(a + b) != tan-1(a) + b.

The fact that you happened to stumble onto the right answer without doing much of anything that's mathematically allowed is nothing short of a miracle. It would be another miracle if your teacher gave you any credit for this work.

4. Mar 12, 2009

### Oscar Wilde

Mentallic, your help is much appreciated. Here's #2

2. sec (x - 5 degrees)=2

(1/cos)^-1(x - 5 degrees)=2^-1

cos (x - 5 degrees)=1/2
cos^-1(1/2)= x - 5 degrees
60 degrees= x - 5 degrees
x= 65 degrees
Cosine is also positive in the 4th quadrant, I obtain the second value by: 360 degrees-65 degrees= 295 degrees

Once again, thank you Mentallic for your assistance, it really helped. Matt, I also appreciate your help. I am a little embarassed that I did not acknowledge the parentheses. Rather humorous. Thanks a bunch,

Oscar

5. Mar 12, 2009

### Chaos2009

So close but no cigar.

Your work is right in finding the first angle, but your second angle is a little wrong. You need to find your second value of cos^-1 before you add those 5 degrees.

cos^-1 (1/2) = 60 and the second value is 360 - 60 or 300 when you are working with degrees.

6. Mar 13, 2009

### Staff: Mentor

Yes, much better this time. You arrived at the right answer for one of the angles, and your work supports your result.

One thing to be aware of is the notation you used. cos^(-1) is most often used to mean the inverse cosine function, rather than the reciprocal of it, as I think you meant. No harm done, since we understood what you meant, but keep that in mind.

The steps you used were correct, but a little convoluted.
Here's one sequence of operations that's a little more direct (all angles in degrees):
sec(x - 5) = 2
==> 1/cos(x - 5) = 2
==> cos(x - 5) = 1/2 (taking the reciprocal of both sides)
==> x - 5 = cos-1(1/2) (meaning is inverse cosine, or arccos)
==> x = cos-1(1/2) + 5
==> x = 60 + 5 = 65 degrees

The other angle whose cosine is 1/2 is 300 degrees, so for that value, x + 5 = ?