i suck at trigonometry.

So much for my dreams at being a fucking engineer.

Oh well, i could always be an english teacher. Requierments on puertorico aren't THAT high to be exact.

So any ways, i have bitch ass trigonometry homework.

And due to certain cases, my math skillz droped to that of a 7th grader.

so i'm here for help <3

i need the answers to well. all of them : ( this is part on, the shortest one.
I have afterwards 4 more parts made up of 20+ questions, so i'd find it a huge help if you'd help me with part 1 >.> the easiest one.

i'll be using * as my mark for degrees later on, sorry about that.

NOTE, i have 2 weeks to finish : ( it's worth a whole damn grade >.>


so anyways.

1. using exact values, verify that cos 60 (degrees) = 1 - 2 sin (root) 30 (degrees)

2. using exact values, find the value of:

A. 12 sin 30(degrees) - 6 tan 45 (degrees) + sec 45(degrees)/52

B. sin 2A+tan 3A/2 - cos A + sec (A+15 degrees), when A=30 ( degrees)

3. write the other functions of x if cos x= 3/5

4. In a right triangle sin A= 3/4 and C= 17. find A and B.

Be sure to draw a right triangle and letter it properly.

5. Find the following functions correct to five decimal places:

A. sin 22* 43'
B. cos 44* 56'
C. sin 49* 17'
D. tan 11* 37'
E. sin 79* 23' 30''
F. cot 19* 0' 25''
G. tan 64* 6' 45''
H. cos 29* 45' 48''
I. cot 78* 45' 48''
J. cos 15* 12' 51''

6. Find the acute angle A to the nearest second when.

A. sin A= 0.30985
B. cot A= 0.52613
C. cos A= 0.77273
D. tan A= 6.63831
E. sin A= 0.10655
F. cos A= 0.55570
G. sin A= 0.92112
H. tan A= 0.13476
I. cot A= 0.32323
J. cos A= 0.75273

7. The vertex angle of an isosceles triangle is 57* 24' and each o fit's equal sides is 375.5 ft long. Find the altitude of the triangle

8. Solve the follow right triangles:

A. a=117 ft, b= 16.35 ft
B. b=8* 29', a= 32.8 FT

9. A man standing on top of a building sees an automobile on a street below him. The angle of depression is 36* 22'. If his eye is 550 ft above the level of the automobile, how far is the automobile from him?

10. when the sun's angle of elevation is 30* the shadow of a post is 6 ft longer than when the angle is 45*. find the height of the post.

11. In a right triangle, given a=32.8 FT and A= 28* 30' 15''.
solve for B and C.

That's part 1. part 2-5 i'll answer my self ( i hope )

parts 2-5 include 20-45 questions per part.......oh lord.

So i'd be beyond grateful if you would help me with part 1 : (

They teach things a lot differently over the pond!

I'll have a look at them tomorrow - no promises, mind. It's been years since I did anything like that, so I may give up and do something else.

If I do manage to get any, then I won't just give you the answers. Rather, I'll show my working for a couple, and leave the rest to you - you need to learn.

Solution: the answer is trivial.

1. Eat spinach
2. Instantly grow ridiculously sized forearms
3. ?????
4. Profit!

All problems fixed, kyeruu. Now go get that engineering job tiger!

There is this really neat thing out there called a calculator. Get yourself one.



Just because I am bored though, I am going to work on the first one tomorrow.

Solution: the answer is trivial.

Heh-heh-heh-heh-heh.

Also,

So much for my dreams at being a fucking engineer. Oh well, i could always be an english teacher. Requierments on puertorico aren't THAT high to be exact. So any ways, i have bitch ass trigonometry homework. And due to certain cases, my math skillz droped to that of a 7th grader. so i'm here for help <3 i need the answers to well. all of them : ( this is part on, the shortest one. I have afterwards 4 more parts made up of 20+ questions, so i'd find it a huge help if you'd help me with part 1 >.> the easiest one. i'll be using * as my mark for degrees later on, sorry about that. NOTE, i have 2 weeks to finish : ( it's worth a whole damn grade >.> so anyways. 1. using exact values, verify that cos 60 (degrees) = 1 - 2 sin (root) 30 (degrees) 2. using exact values, find the value of: A. 12 sin 30(degrees) - 6 tan 45 (degrees) + sec 45(degrees)/52 B. sin 2A+tan 3A/2 - cos A + sec (A+15 degrees), when A=30 ( degrees) 3. write the other functions of x if cos x= 3/5 4. In a right triangle sin A= 3/4 and C= 17. find A and B. Be sure to draw a right triangle and letter it properly. 5. Find the following functions correct to five decimal places: A. sin 22* 43' B. cos 44* 56' C. sin 49* 17' D. tan 11* 37' E. sin 79* 23' 30'' F. cot 19* 0' 25'' G. tan 64* 6' 45'' H. cos 29* 45' 48'' I. cot 78* 45' 48'' J. cos 15* 12' 51'' 6. Find the acute angle A to the nearest second when. A. sin A= 0.30985 B. cot A= 0.52613 C. cos A= 0.77273 D. tan A= 6.63831 E. sin A= 0.10655 F. cos A= 0.55570 G. sin A= 0.92112 H. tan A= 0.13476 I. cot A= 0.32323 J. cos A= 0.75273 7. The vertex angle of an isosceles triangle is 57* 24' and each o fit's equal sides is 375.5 ft long. Find the altitude of the triangle 8. Solve the follow right triangles: A. a=117 ft, b= 16.35 ft B. b=8* 29', a= 32.8 FT 9. A man standing on top of a building sees an automobile on a street below him. The angle of depression is 36* 22'. If his eye is 550 ft above the level of the automobile, how far is the automobile from him? 10. when the sun's angle of elevation is 30* the shadow of a post is 6 ft longer than when the angle is 45*. find the height of the post. 11. In a right triangle, given a=32.8 FT and A= 28* 30' 15''. solve for B and C. That's part 1. part 2-5 i'll answer my self ( i hope ) parts 2-5 include 20-45 questions per part.......oh lord. So i'd be beyond grateful if you would help me with part 1 : (

I just gave up my dream at being anything better then a Mc Donalds worker.

I wouldn't throw my dream away, and I'm sure you haven't but I'll just let you know I'm in college now and I'm a second year ME student, and stuff like that you don't do by hand. I'm not saying to slack off, because learning how to do it now by hand builds fundamentals, I'm just saying I wouldn't sweat it too much. At this level, we aren't expected to remember trig identities, or even some of the most basic integrals - we have tables for all of that. As far as exact angles and such - calculators.

I would say the most important thing you can take away from trigonometry for wanting to be an engineer is the basic concept of it, because you will be working with a lot of complex geometry and trigonometry later on down the road.

I wouldn't throw my dream away, and I'm sure you haven't but I'll just let you know I'm in college now and I'm a second year ME student, and stuff like that you don't do by hand. I'm not saying to slack off, because learning how to do it now by hand builds fundamentals, I'm just saying I wouldn't sweat it too much. At this level, we aren't expected to remember trig identities, or even some of the most basic integrals - we have tables for all of that. As far as exact angles and such - calculators. I would say the most important thing you can take away from trigonometry for wanting to be an engineer is the basic concept of it, because you will be working with a lot of complex geometry and trigonometry later on down the road.

Exact angle? That's easy, dude, 90 degrees.

funny thing, teacher told me i need to write down process, so i can't calculator my way to correct answers anymore, fuckkkkkk

funny thing, teacher told me i need to write down process, so i can't calculator my way to correct answers anymore, fuckkkkkk

I wasn't talking about this homework set but rather I was trying to give you some comfort for the future. What you've got is pretty easy anyways... all you have to do is memorize the unit circle and you've got it.

cos(60) = 1 - 2sin^2(30) // I'm assuming you meant sin^2 here

Well the cos(60) from the unit circle is 1/2, the sin(30) is also 1/2
1/2 = 1 - 2*(1/2)^2
1/2 = 1/2

or the one with cos(x) = 3/5
Since cosine is equal to adjacent/hypotenuse
adjacent = 3
hyp = 5
Applying pythagoream's theorem to a right triangle you get opposite to be 4
So the sin(x) = opposite/hyp = 4/5
tan(x) = opposite/adjacent = 4/3

The whole adj, hyp, and opp things you will just have to remember and later on down the road it becomes second nature

Cosine(theta) = adj/hyp = x/hyp
Sine(theta) = opp/hyp = y/hyp
Tangent(theta) = opp/adj = y/x

Since the sec, csc, and cot functions are just the inverse of their counterparts(and trust me rarely used except in computations), then you just invert the above fractions for their values... but really the easier way to think about it is sec(theta) = 1/cos(theta), and just let the algebra work it out.

So if you want the sec(60), then just take 1/cos(60) = 1/0.5 = 2

You've really progressed since the Army days eh?

You've really progressed since the Army days eh?

Lol I knew all that stuff in the army days, the army's just not interested in it. But I've learned, and I quote, a 'shitload' in my two years of college.

What school? Is the army paying?

Trig's pretty easy if you're in the class, but I can't remember a lot of it. I haven't had it in like 3 years at least.

What school? Is the army paying?

VA Tech. Yes.

Bump! If they actually want you to find exact trigonometric values to n decimal places of "unit circle unfriendly" angles without a calculator, look into taylor series and fourier series. They converge fairly quickly, but it will still probably take, say, 5-10 minutes of number crunching for 5 decimal places even if you're fairly fast.

Or use WolframAlpha to do all your homework.

http://www34.wolframalpha.com/input/...45(degrees)/52

VA Tech. Yes.

Cool.