Falco276
There are many ropes keeping a hot air balloon from floating away before a balloon race. One of these ropes is fixed to the ground at a 45° angle. Another is fixed to the ground at a 30° angle.
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Falco276
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If the hot air balloon is 21 feet off the ground, what is the distance between the ground directly underneath the balloon and the second rope rounded to the nearest hundredth of a foot? Type the numeric answer only in the box below.
Falco276
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Falco276
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@GCR92
GCR92
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ok let me see
Falco276
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I think I know.
The vertical angle measurement is equal to the horizontal measurement of the right angle.
Falco276
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21 feet by 21 feet.
I just have to find out the measurement to rope 2
GCR92
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they are asking you the distance between rope 2 and underneath the balloon right?
Falco276
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yes, exactly.
GCR92
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just to make sure i understood well, i remind you i'm not english!! hehehe
GCR92
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ok
GCR92
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lemme get a paper so i can explain better
Falco276
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|dw:1358712594988:dw|
GCR92
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the distance between rope 2 and 1 is x right? as the drawing says
Falco276
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yes, but I found the half of X which is 45.
Falco276
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look at the drawing, am I right?
Falco276
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I only have to fing the measurement that is the question mark.
Falco276
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*find
GCR92
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you have 2 angles if you notice
GCR92
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even though they dont give you a number, you can see that the angle underneath the balloon is 90, a right triangle
Falco276
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i noticed that.
GCR92
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apply pitagoras
GCR92
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ah well you need to find the sides first, nevermind
Falco276
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sides?
GCR92
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you gonna solve the problem like this
GCR92
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x: distance from rope 2 to underneath the balloon - distance from rope 1 to underneath the balloon
Falco276
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distance from rope 1 to underneath the balloon = 45.
Falco276
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Im trying to figure out rope 2
GCR92
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how do you know its 45? 45 is the given angle
Falco276
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no, not from the angle. On how I got 45 was that the vertical measurement leading to the balloon has the same measurement that leads to rope 1.
Falco276
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NO NO NO NO!!!!!
GCR92
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Im solving it, wait
Falco276
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I have it wrong!
GCR92
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it has a lot to do with sinus and cosinus
Falco276
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its 21!!!!!!
Its 21!!!!
GCR92
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well first you need to remember that sinus = op side / h
GCR92
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we are measuring from rope 1 to balloon now, so sin 30 right?
GCR92
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i tell you that sin 30 = 1/2
GCR92
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so you gonna put 1/2 = 21ft / h right?
GCR92
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solve for h and then you have the hypotenuse
GCR92
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what do you think you can do next
GCR92
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there are many ways to solve this problem
Falco276
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um, okay.
GCR92
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also remember that all the angles of a triangle = 180, if this doesn't happen, it's not a triangle
GCR92
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so if you have a 90º angle, a 30º angle...what's the other angle?
Falco276
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wait, so were trying to find a measurement form rope 1 to the balloon, am I right?
GCR92
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yep
GCR92
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i got it already, just waiting to see how you proceed
GCR92
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that problem is too difficult for you, i mean, they should teach you trigonommetry first, there's no way you can do that without using trigonommetry
Falco276
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I never took trigonometery before.
GCR92
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I know you haven't heard of the trigonometric functions
GCR92
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you need to do it 2 steps
Falco276
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no, never. I just take basic geometry.
GCR92
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ok look
GCR92
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|dw:1358713814089:dw|
GCR92
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do you remember what was the sinus and what was the cosinus?
Falco276
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(sigh)
I don't remember. (honest, i don't want to be mean to you.)
GCR92
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hahaha, dont worry about it, i'll explain again, but you should take note of it because you are going to need it in EVERY problem
Falco276
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okay, I'll take your note.
GCR92
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|dw:1358714056382:dw|
GCR92
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Now you need to MEMORIZE this scheme, mathematics are also a subject to study!
GCR92
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now ill draw you a triangle and you'll tell me which is the oppsite and which is the adjacent side ok?
Falco276
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ok.
GCR92
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|dw:1358714498016:dw|
Falco276
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I think the opposite side is left from angle A
the adj side is right from Angle A.
GCR92
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nope
GCR92
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just read its name
GCR92
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opposite = infront of the angle
GCR92
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adjacent = sidewards the angle
GCR92
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alongside of it
Falco276
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fine! Isn't the adjecent side the- How do I Explain it!
Isnt the adjecent side the sidewards of the angle?
GCR92
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draw it
Falco276
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|dw:1358715165594:dw|
GCR92
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no, the other one, that's the hypotenuse
Falco276
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|dw:1358715296994:dw|
Falco276
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@hba
GCR92
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yep
hba
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Yes @Falco276
Falco276
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@hba
please help me!
hba
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@GCR92 is helping you,Just follow him :)
GCR92
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I have to explain the whole trigonometry before you are able to solve that problem...
Falco276
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(sigh)
ok.
GCR92
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I explain as simple as it is, but not simpler
Falco276
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ok
GCR92
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ok i put the turbo now
GCR92
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we are going to solve 2 triangles
GCR92
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|dw:1358715724015:dw|
Falco276
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wait. I think I have it.
GCR92
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we are going to use the sinus to solve this one, why sinus? because the problem has given us the opposite side, wich is 21 ft
Falco276
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For the 45° triangle
GCR92
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and sinus = op side / h
GCR92
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and we can make a relationship
GCR92
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an equation
Falco276
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tan Ө = opposite / adjacent
tan Ө = height of the balloon off the ground / x
tan 45° = 15 / x
x = 15 / tan 45°
x = 15 / 1
x = 15
Falco276
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is this correct?
GCR92
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where did you get 15?
Falco276
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oh, no i meant 21
(facepalm)
GCR92
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You could've done it if you had the value of the x side
GCR92
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but that's what we are looking for
GCR92
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to use the tangent you need to have the value of 2 sides
GCR92
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I'd like to continue now
Falco276
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|dw:1358716138389:dw|
GCR92
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why?
Falco276
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|dw:1358716271703:dw|
Falco276
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Falco276
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now we take the other triangle.
GCR92
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the adjacent side is not 21, 21 is the only data they give us, you can't invent it
GCR92
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listen i dont want to be mean either but i want to help you and i want to finish, if you think you don't need to do what i tell you to solve it, go ahead
GCR92
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sin 30 = 1/2, we do that in order to calculate H
GCR92
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1/2 = 21 / h
GCR92
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h= 42
GCR92
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now we take the cosinus in order to calculate X
GCR92
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cos 30 = sqrt 3 / 2
GCR92
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sqrt 3 / 2 = 42 / x
GCR92
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x = 36.373
Falco276
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|dw:1358716402610:dw|
Falco276
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|dw:1358717381175:dw|
Falco276
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|dw:1358717481086:dw|
GCR92
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you can never write tan and let it alone
GCR92
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it is always tan OF SOMETHING
GCR92
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tan of x, tan of alfa, tan of 20
Falco276
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so the answer is 21 sq. root 3
GCR92
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ill solve it
GCR92
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2 triangles
GCR92
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take the 30º triangle and then, sin 30 = 1/2
GCR92
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1/2 = 21 / h, h = 42
GCR92
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now we have the h of the first triangle
GCR92
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now we can calculate the distance from the rope to underneath the balloon, that distance will be given by the cosinus
GCR92
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cos 30 = sqrt 3 / 2
GCR92
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sqrt 3 /2 = x/42
GCR92
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x = 36,373
GCR92
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now we go to the second triangle and apply exactly the same steps but with 45º
Falco276
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maybe my way of solving this problem was wrong.
sorry.
Falco276
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i tried my best.
GCR92
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I know but mathematics doesn't magically happen, you need to study a lot( and suffer lol) dont worry
GCR92
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if you do the same with the second triangle
GCR92
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sin 45 = sqrt 2 / 2
GCR92
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sqrt 2 / 2 = 21 / h
GCR92
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h = 29.698
GCR92
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now we calculate the cosinus
GCR92
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cos 45 = sqrt 2 / 2
GCR92
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sqrt 2 / 2 = x / 28,698
GCR92
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x = 20,292
GCR92
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distance from underneath the ballon to the second rope = 36, 373 + 20,292 = 56,665 feet
GCR92
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and it's solved
Falco276
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wait, but earlier you said that x is 36- never mind :P
GCR92
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that's the distance between underneath the balloon and rope 2
Falco276
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ah, i see.
GCR92
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and 20,292 ft is the distance between underneath the balloon and rope 1
GCR92
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sorry
Falco276
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got it.
GCR92
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36ft is the distance between rope 2 and 1
GCR92
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then you calculate from rope 1 to underneath the ballon and it gives you 20 ft
GCR92
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you add it and you obtain the final result
Falco276
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okay.
GCR92
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what i did is totally useless, i mean you can't learn in 20 min, i always believed that a teacher is need in order to learn something, anyway, wish you luck with this!
GCR92
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needed*
GCR92
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you did well! at least you had ideas! you only need to learn the theory well
GCR92
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i also think your teachers are crazy though
Falco276
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XD
Thanks Gonzalo! I promise to keep the theory at heart.
Yes, I finally understood the answer and also the work.
I'll tell you if need help on anything else.
GCR92
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You are welcome, sure, see you! :D
Falco276
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thanks! (thumbs up)
GCR92
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I also can be wrong, i mean if your teachers yells at you it is my fault, just don't take that in account, it's been ages since i dont take trigonometry :p
Falco276
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oh, i understand, dont worry! my teachers are really nice.