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Mathematics 21 Online
OpenStudy (anonymous):

Alphonse (point A) is over a 2500-meter landing strip in a hot-air balloon. At one end of the strip, Beatrice (point B) sees Alphonse with an angle of elevation measuring 39°. At the other end of the strip, Collette (point C) sees Alphonse with an angle of elevation measuring 62°. a. What is the distance between Alphonse and Beatrice? b. What is the distance between Alphonse and Collette? c. How high up is Alphonse?

OpenStudy (anonymous):

OpenStudy (anonymous):

You can find angle A. 180 degrees minus (angle B + angle C). Once you have that angle, use the same technique as previous problems.

OpenStudy (anonymous):

wait what that support to mean...

OpenStudy (anonymous):

how to set the problem with 82

OpenStudy (anonymous):

82?

OpenStudy (anonymous):

Let D be the point on the earth' s surface directly below the ballon. Let x be AD, the height of the balloon and y be the distance between B and D. Then the two equations below can be derived from the given information:\[\left\{\frac{x}{y}=\text{Tan}[39/180 \pi ],\frac{x}{2500-y}=\text{Tan}[62/180 \pi ]\right\} \]There are two solutions, x and y. Only x is required: x = 1415.14 and is the answer to question C. Question A:\[\text{AB}=1415.14\text{ Csc}\left[\frac{13 \pi }{60}\right]=2248.68 \]Question B answer:\[\text{AC}\text{=}1415.14\text{ Csc}\left[\frac{31 \pi }{90}\right]=1602.75 \]

OpenStudy (anonymous):

so that mean tan-1(39/180pi)?

OpenStudy (anonymous):

what Csc?

OpenStudy (anonymous):

OK, Robotey was showing you another way, but I remember you are still in high school. So, as I was saying you don't have angle A. But all triangles are 180, so 180-39-62 would give you A.

OpenStudy (anonymous):

you just wonder why 82 because i add 39 and 62 which 101 the subract it then got 79......oop my bad it 79

OpenStudy (anonymous):

OK, you know, distance Alphonse and Beatrice is line c, (Sin 62)/c=(Sin 79)/2500

OpenStudy (anonymous):

can i crosss it or not

OpenStudy (anonymous):

Any way you like, just get c=....

OpenStudy (anonymous):

i got 2248.62

OpenStudy (anonymous):

Good, same as Robotey, different method.

OpenStudy (anonymous):

that would be easy for me

myininaya (myininaya):

omg this problem took me forever alot of mistakes lol i finally got it to work this is probably the long way

myininaya (myininaya):

if you want to make fun of me chaguanas,you can

OpenStudy (anonymous):

I would, only there is nothing coming up on your attachment. How would you handle the height of Alphonse?

myininaya (myininaya):

the attachment isn't working? :(

OpenStudy (anonymous):

nada

myininaya (myininaya):

OpenStudy (anonymous):

that too hard to rerember this

myininaya (myininaya):

try again if it doesn't work i will try to explain

OpenStudy (anonymous):

Still nothing. Hikari, you got b and c?

myininaya (myininaya):

the height is the first thing i found...

OpenStudy (anonymous):

not really

OpenStudy (anonymous):

Distance between Alphonse and collette is line b. (Sin A)/a=(Sin B)/b

OpenStudy (anonymous):

so how i am able to find the angle

OpenStudy (anonymous):

You have everything, except line b. So put in all info and solve for b.

myininaya (myininaya):

so i used the pytagrean thm twice i called the point that she began her elevation G k? (this means BG+GC=2500 so we have AG^2+BG^2=AB^2 AG^2+GC^2=AC^2 solved both of the for AG^2 and set them equal AB^2-BG^2=AC^2-GC^2 AB^2-AC^2=BG^2-GC^2 AB^2-AC^2=(BG+GC)(BG-GC) AB^2-AC^2=(2500)(BG-GC) then i went to the side to get every term in terms of AG AB=AG/sin39 so AB^2=AG^2/sin^2(39) and AC=AG/sin62 so AC^2=AG^2/sin^2(62) also BG=AG/tan39 and GC=AG/tan62 then i got \[AG^2(\frac{1}{\sin^239}-\frac{1}{\sin^2(62)})-AG(2500)(\frac{1}{\tan39}-\frac{1}{\tan62})=0\]

myininaya (myininaya):

So AG cannot be zero so AG=1415.142464

OpenStudy (anonymous):

so i am still using 39 and 62

myininaya (myininaya):

the rest easy AC=1415.142464/sin62 AB=1415.142464/sin39

OpenStudy (anonymous):

Yes, Hikari. Wow, Myininaya, a lot of work.

myininaya (myininaya):

yes u know thats why i was hoping the attachment would work lol

myininaya (myininaya):

i have three other attemps like this but they are in the garbage because i kept making an error then finally PERFECTO! lol

OpenStudy (anonymous):

Great. Now if we can find the elementary method to show hikari.

OpenStudy (anonymous):

now i know c is 2248.7 so i can use that

myininaya (myininaya):

i think my way is elementary but long However I think the problem could have been solved with less steps

OpenStudy (anonymous):

Yes, Hikari, you can use c if you like.

OpenStudy (anonymous):

so sin62/B=sin39/2248.7

OpenStudy (anonymous):

Mix up (Sin 39)/b=(Sin 62)/2248.7

myininaya (myininaya):

and then with that fix you will have the same answer as rob and i

myininaya (myininaya):

i think you have one more leg remaining that you need to find the length do you know law of cosines?

OpenStudy (anonymous):

i got 1602.79 so can i round it or not

myininaya (myininaya):

i assume rounding is okay i believe it is impossible to have the exact value anyways

OpenStudy (anonymous):

so now the c question

OpenStudy (anonymous):

Ok, if you draw a line from A to bottom, it makes a right angle. So you can use sin. Sin B=Alphonseheight/2248.7. Solve for Alphonse height.

OpenStudy (anonymous):

sinB=A/2248.7

OpenStudy (anonymous):

sinB=A/2248.7

OpenStudy (anonymous):

Yeah, but call it something else don't call it A. Let's call it p. Sin 39=p/2248.7

OpenStudy (anonymous):

i got 1415.10

OpenStudy (anonymous):

Good! Same thing Myinaya have. Good job.

myininaya (myininaya):

gj hirkari

OpenStudy (anonymous):

thank for helping me and esay for me to remember this thank you

OpenStudy (anonymous):

:)

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