[High School - Math] Trigonometry

[Cosmic M]

Ok, so I am studying for midterms, and i am good at math (got an A both quarters) but i cant remember everything. I dont quite remember how to do the application problems in the review packet. But, like all math, if i see it done once for me, ill remember it and be able to use it for any of them. So since you cant give me answers, i will change the numbers of one of the word problems and you can tell me how to solve it, and then i can solve the real problems on my own. Lets begin.

A building 180 feet tall casts a 70 foot long shadow. If a person looks down from the top of the building, what is the measure of the angle between the end of the shadow and the vertical side of the building (round to the nearest degree). (Assume the person's eyes are level with the top of the building).

So once again, if you can tell me how to arrive at that answer, i can then apply the method to the others.

[trumaestro]

http://i83.photobucket.com/albums/j314/trumaestro/situation.jpg

That's what's going on, right?

If it is, then you have a right triangle. You can use one of sine cosine or tangent ratios.
Remember to use the inverse of them to solve for angles.

[Cosmic M]

right, so there is a guy on the top, and we want the angle between the end on the shadow and the vertical side of the building, so does that mean we need another line from the top cutting the 70 in half?

[Sh4d0wD3v1L]

Trig like this is easy. The solution to most problems is draw a picture with a right triangle in it, and solve the missing side.

So, the building will be the leg (going bottom to top) and the shadow will be the base.


|(angle?)
|(building) 180ft (adjacent to angle)
|
|
|__________________
(shadow) 70ft (opposite of angle)

Now, you use the SOH-CAH-TOA to find which one to use...
(Sin=opposite/hypotenuse Cos=adjacent/hypotenuse Tan=opposite/adjacent)
In this case, the angle you are looking at, and the numbers you are given, you have the OPPOSITE (70ft shadow) and the ADJACENT (180ft bldg) sides to the angle.

Now, get out your calculator.
Make sure, first of all, your calculator is in DEGREE mode (hit the MODE key and look for it)
Hit the 2nd key, then tan. This the inverse tangent. This will give you the angle measure given 2 sides. Put 70/180 and close parentheses.

tan-1(70/180) = angle
tan-1(70/180) = ~21.25 degrees

Hope this helped, and feel free to ask if you need anything else.

[trumaestro]

The top is the vertical side of the building.

EDIT: I'm not entirely sure which angle you're to solve for, as the question isn't worded very clearly. It could be either angle that isn't 90 degrees.

[Cosmic M]

alright shadow, that makes sense, my math midterm is on wednesday, plenty of time to iron this stuff out :)

[Sh4d0wD3v1L]

Good luck to you then :)

In the mean time, I have my own trig to work on. Identities ftl.

[Cosmic M]

That was the last thing we learned before midterms. That topic is fresh in my mind.

[rushyrulz]

This specific problem is easy because there is so much given. All you would have to do is find a trig function that deals with the 180 and the 70. The only one there is is Tangent. So you would simply plug in the numbers from TOA. You take your opposite (70) and divide it by the Adjacent (180) You then multiply to get a decimal. You would then multiply the decimal by 100 to get the angle.

Since this is a study guide and not homework, I'll tell you the answer is 38.8° rounded up to 39°.

While we're at it, it would be good to go over finding the other side length. All you would have to do here is use Pythagorean's Theorem (prolly butchered the spelling).
So you use the theorem (a2+b2=c2)

a=70 b=180 c=?

C is always the hypotenuse.

so:
a2=4,900.
b2=32,400.
Add these together and get 37,300. Simply take the square root of this and that is your side length. The answer here is 193.1u.

[Cosmic M]

Yes, but you mixed up oppisite and adjacent. 180 is the oppisite, and 70 is adjacent, like shadow explained. You found the last angle, also it would be 21. 70/180 = .388 repeat and tan-1 of that is about 21, not 39.

[Sh4d0wD3v1L]

Not to bump this, but if 180 is the opp side, it's not finding the angle w/ the building. The way it is now (see: edit) is the angle with the building. The way I described it originally is the angle of the shadow relevant to the top of the bldg.....Which one are you looking for?

[masamursha]

you had it right the first time lol

[dore]

Not to bump this, but if 180 is the opp side, it's not finding the angle w/ the building. The way it is now (see: edit) is the angle with the building. The way I described it originally is the angle of the shadow relevant to the top of the bldg.....Which one are you looking for?

You did it right the first time. 70 is opposite, 180 is adjacent. So the answer is arctan(70/180).

[Cosmic M]

oh, so we are in fact looking for the 21 degree angle?

[rushyrulz]

Yes, but you mixed up oppisite and adjacent.

no

You did it right the first time. 70 is opposite, 180 is adjacent. So the answer is arctan(70/180).

yes

[Cosmic M]

well, the test is tommorow, wish me luck.

[180digi]

I'd like to pose a question involving trigonometry without creating a new topic if that is quite acceptable. A study question in my textbook says,

"...find the point (x,y) on the unit circle that corresponds to the real number t.

43.) t = 2(pi)/3

I know that the point is (-1/2, radical3/2), but is there any way to find this out without looking at a unit circle? We're being quizzed on this soon, and I'm not sure if we are going to be provided a unit circle for reference. Thanks.

[masamursha]

the unit circle is just showing you that the hypotenuse of that triangle will always be 1. You don't have to draw the circle, you could just draw the triangle where the hypotenuse is 1 on the coordinate plane. Drawing it on the coordinate plane will just help you make sure that you are in the right quadrant.

[GuidoHunter]

2pi = 360 degrees
2pi/3 = 120 degrees

That's in the second quadrant.

sin(120 degrees) = root(3)/2
cos(120 degrees) = -1/2

Or, since 120 degrees makes a 60 degree angle with the horizontal, you could just do sin(60) and cos(60), then just remember to make the x coordinate negative, what with cosines being negative in the second quadrant.

Now, if you don't know the sines and cosines of common angle measures, you're in trouble for any trig problem, so commit the sines, cosines, and tangents of 0, pi/6, pi/4, pi/3, and pi/2 to memory ASAP.

--Guido

http://andy.mikee385.com

[180digi]

Let me just make sure I understand this. I'll do a problem.

5.) t=pi/4

pi/4 x 180º/pi = 45º. I drew this triangle.

http://i28.photobucket.com/albums/c222/180digi/untitled-1.jpg

SOA-CAH-TOA, so it's (1/rad2, 1/rad2).

1/rad2 x rad2/rad2 = rad2/2

the point on the unit circle is (rad2/2, rad2/2)

[dore]

Just memorize the unit circle. You'll reference it for the rest of your math career so you ought to learn it now.

[masamursha]

yup you got it 180digi