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11/20/2019 5:07:11 PM
Posted: 5/22/2005 2:13:15 PM EST
[Last Edit: 5/22/2005 2:16:56 PM EST by IchWarrior]
So your standing at the bottom of a water fall, you look up to the top of the falls and for some reason you take not your angle of elevation is 60 degrees.


You go hiking 1000 feet away from the water fall, remaining at the same elevation (Okay, better yet you are on a boat and you float 1000 feet down river.

You look up to the top of the falls and your angle of elevation is 57.3 degrees.


How tall is the falls?


Cant for the life of me figger it out.

ETA: My only hunches is that you find a ratio of your angles of elevation and multiply that by the 1000 foot difference...

Link Posted: 5/22/2005 3:42:57 PM EST
This is a simple trig problem. The tangent of your angle of elevation is equal to the unknown height of the falls over the distance you are away from it. That is...

tan(57.3)=x/1000 solve for x.
Link Posted: 5/22/2005 3:47:31 PM EST
I'm lazy anymore...ok, maybe I have always been lazy, but I end up just figuring things like this out in a CAD program.
Link Posted: 5/22/2005 3:50:24 PM EST

Originally Posted By Illinigunner21:
This is a simple trig problem. The tangent of your angle of elevation is equal to the unknown height of the falls over the distance you are away from it. That is...

tan(57.3)=x/1000 solve for x.



i didnt know "simple" and "trig" went in the same sentence.
Link Posted: 5/22/2005 3:53:08 PM EST
the tangent of an angle = opposite side/adjacent side.

Therefore,

tangent(57.3) = x/1000 where x=height of waterfall

solve for X
Link Posted: 5/22/2005 4:24:15 PM EST
[Last Edit: 5/22/2005 4:31:32 PM EST by drjarhead]
Actually, you have use trig and algebra.

h = xtan60 = (1000+x)tan57.3


Edited cause I don't know my cos from my tan. Rusty. It's been years, hell decades.
Link Posted: 5/22/2005 4:26:19 PM EST
My answer is "perty damn big"
Link Posted: 5/22/2005 4:29:55 PM EST
[Last Edit: 5/22/2005 4:33:55 PM EST by Aalmeron]
law of sine

a/sinA=b/sinB=c/sinC

a= unknown
A=57.3
b=1000
B=90

simple enough

a/sin57.3=1000/1
a=1000*sin57.3
you can finish it from there
Link Posted: 5/22/2005 4:34:34 PM EST

Originally Posted By Aalmeron:
law of sine

a/sinA=b/sinB=c/sinC

a= unknown
A=57.3
b=1000
B=90

simple enough

a/sin57.3=1000/1
a=1000*sin57.3
you can finish it from there



Problem is b = a + 1000
Link Posted: 5/22/2005 4:40:08 PM EST

Originally Posted By drjarhead:

Originally Posted By Aalmeron:
law of sine

a/sinA=b/sinB=c/sinC

a= unknown
A=57.3
b=1000
B=90

simple enough

a/sin57.3=1000/1
a=1000*sin57.3
you can finish it from there



Problem is b = a + 1000



dont think so.
but shit, i did make a mistake..

ok

A= 57.3 a=?
B=32.7 b=1000
C=90 c=?

but other than that, the length of b does not equal the length of a plus its own length...

a/sin57.3=1000/sin32.7
a=(1000*sin57.3)/sin32.7
Link Posted: 5/22/2005 4:43:25 PM EST

Originally Posted By drjarhead:
Actually, you have use trig and algebra.

h = xtan60 = (1000+x)tan57.3


Edited cause I don't know my cos from my tan. Rusty. It's been years, hell decades.


Yep, because if you were right under the falls the first angle would be 90 degrees.
Link Posted: 5/22/2005 4:48:30 PM EST
[Last Edit: 5/22/2005 4:53:08 PM EST by Aalmeron]

Originally Posted By dport:

Originally Posted By drjarhead:
Actually, you have use trig and algebra.

h = xtan60 = (1000+x)tan57.3


Edited cause I don't know my cos from my tan. Rusty. It's been years, hell decades.


Yep, because if you were right under the falls the first angle would be 90 degrees.



ok i dont know what formula that is, but i know the law of sine gives hte right answer.
Link Posted: 5/22/2005 4:50:41 PM EST
Aw, hell, you're in trig and can't do that? Don't make me get my calculator out...
Link Posted: 5/22/2005 4:51:57 PM EST

Originally Posted By Aalmeron:

Originally Posted By dport:

Originally Posted By drjarhead:
Actually, you have use trig and algebra.

h = xtan60 = (1000+x)tan57.3


Edited cause I don't know my cos from my tan. Rusty. It's been years, hell decades.


Yep, because if you were right under the falls the first angle would be 90 degrees.



ok i dont know what formula that is, but i know the law of sine gives hte right answer.



I can't even remember the law of sine. It's been too long, so you're WAAAY ahead of me.
Link Posted: 5/22/2005 4:54:21 PM EST

Originally Posted By Aalmeron:

Originally Posted By dport:

Originally Posted By drjarhead:
Actually, you have use trig and algebra.

h = xtan60 = (1000+x)tan57.3


Edited cause I don't know my cos from my tan. Rusty. It's been years, hell decades.


Yep, because if you were right under the falls the first angle would be 90 degrees.



ok i dont know what formula that is, but i know the law of sine gives hte right answer.



But:

sin R = height of falls/ hypotenuse ::: so I am not sure how that helps.

tan R = height of falls / distance
Link Posted: 5/22/2005 4:56:00 PM EST

Originally Posted By dport:

Originally Posted By Aalmeron:

Originally Posted By dport:

Originally Posted By drjarhead:
Actually, you have use trig and algebra.

h = xtan60 = (1000+x)tan57.3


Edited cause I don't know my cos from my tan. Rusty. It's been years, hell decades.


Yep, because if you were right under the falls the first angle would be 90 degrees.



ok i dont know what formula that is, but i know the law of sine gives hte right answer.



I can't even remember the law of sine. It's been too long, so you're WAAAY ahead of me.



law of sine gets 1557.67 feet high.

that tangent formula not only has 2 unknowns, but is set to equal the hypotonuse. why are you guys trying to figure out the hyp? he wants the hight of the falls, not the distance from where the person is, to the hight of the falls.

anyway, the hypotonuse is 1851.03 feet tall
Link Posted: 5/22/2005 4:58:42 PM EST

Originally Posted By Aalmeron:

Originally Posted By dport:

Originally Posted By Aalmeron:

Originally Posted By dport:

Originally Posted By drjarhead:
Actually, you have use trig and algebra.

h = xtan60 = (1000+x)tan57.3


Edited cause I don't know my cos from my tan. Rusty. It's been years, hell decades.


Yep, because if you were right under the falls the first angle would be 90 degrees.



ok i dont know what formula that is, but i know the law of sine gives hte right answer.



I can't even remember the law of sine. It's been too long, so you're WAAAY ahead of me.



law of sine gets 1557.67 feet high.

that tangent formula not only has 2 unknowns, but is set to equal the hypotonuse. why are you guys trying to figure out the hyp? he wants the hight of the falls, not the distance from where the person is, to the hight of the falls.

anyway, the hypotonuse is 1851.03 feet tall


With the hyp and the angle you can always get the opposite leg.
Link Posted: 5/22/2005 4:58:53 PM EST

Originally Posted By drjarhead:

Originally Posted By Aalmeron:

Originally Posted By dport:

Originally Posted By drjarhead:
Actually, you have use trig and algebra.

h = xtan60 = (1000+x)tan57.3


Edited cause I don't know my cos from my tan. Rusty. It's been years, hell decades.


Yep, because if you were right under the falls the first angle would be 90 degrees.



ok i dont know what formula that is, but i know the law of sine gives hte right answer.



But:

sin R = height of falls/ hypotenuse ::: so I am not sure how that helps.

tan R = height of falls / distance



so you're using soh cah toa so tangent=oposite over adjacent
tan57.3=x/1000
tan57.3*1000=x which is hte hight of the falls
x=1557.67

same answer
Link Posted: 5/22/2005 5:03:02 PM EST
[Last Edit: 5/22/2005 5:05:52 PM EST by drjarhead]

Originally Posted By Aalmeron:

Originally Posted By dport:

Originally Posted By Aalmeron:

Originally Posted By dport:

Originally Posted By drjarhead:
Actually, you have use trig and algebra.

h = xtan60 = (1000+x)tan57.3


Edited cause I don't know my cos from my tan. Rusty. It's been years, hell decades.


Yep, because if you were right under the falls the first angle would be 90 degrees.



ok i dont know what formula that is, but i know the law of sine gives hte right answer.



I can't even remember the law of sine. It's been too long, so you're WAAAY ahead of me.



law of sine gets 1557.67 feet high.

that tangent formula not only has 2 unknowns, but is set to equal the hypotonuse. why are you guys trying to figure out the hyp? he wants the hight of the falls, not the distance from where the person is, to the hight of the falls.

anyway, the hypotonuse is 1851.03 feet tall



You are way off.

h is height of the falls or o for opposite side.

You don't use both unkowns. Solve for x using the right side of the equation:

xtan60 = (1000+x)tan57.3

Where x is the distance from the fall. x +1000 is the second distance.

then,

h = xtan60

solve for h.


The Law of sines BTW concerns any triangle, even oblique, with regards to the ratios btwn angles and opposite sides. IOW the ratio of the sin of one of the angles divided by the oppsite side is the same as all the others.

I don't see how it helps.

Gee guys, I haven't done this shit since 1983



Link Posted: 5/22/2005 5:10:20 PM EST
Ow my head hurts!
Link Posted: 5/22/2005 5:11:04 PM EST

Originally Posted By drjarhead:

Originally Posted By Aalmeron:

Originally Posted By dport:

Originally Posted By Aalmeron:

Originally Posted By dport:

Originally Posted By drjarhead:
Actually, you have use trig and algebra.

h = xtan60 = (1000+x)tan57.3


Edited cause I don't know my cos from my tan. Rusty. It's been years, hell decades.


Yep, because if you were right under the falls the first angle would be 90 degrees.



ok i dont know what formula that is, but i know the law of sine gives hte right answer.



I can't even remember the law of sine. It's been too long, so you're WAAAY ahead of me.



law of sine gets 1557.67 feet high.

that tangent formula not only has 2 unknowns, but is set to equal the hypotonuse. why are you guys trying to figure out the hyp? he wants the hight of the falls, not the distance from where the person is, to the hight of the falls.

anyway, the hypotonuse is 1851.03 feet tall



You are way off.

h is height of the falls or o for opposite side.

You don't use both unkowns. Solve for x using the right side of the equation:

xtan60 = (1000+x)tan57.3

Where x is the distance from the fall. x +1000 is the second distance.

then,

h = xtan60

solve for h.


The Law of sines BTW concerns any triangle, even oblique, with regards to the ratios btwn angles and opposite sides. IOW the ratio of the sin of one of the angles divided by the oppsite side is the same as all the others.

I don't see how it helps.

Gee guys, I haven't done this shit since 1983




h = xtan60 doesnt make sence.
where did you get 60?
and x and h are two un knowns
Link Posted: 5/22/2005 5:14:39 PM EST

Originally Posted By Aalmeron:

Originally Posted By drjarhead:

Originally Posted By Aalmeron:

Originally Posted By dport:

Originally Posted By Aalmeron:

Originally Posted By dport:

Originally Posted By drjarhead:
Actually, you have use trig and algebra.

h = xtan60 = (1000+x)tan57.3


Edited cause I don't know my cos from my tan. Rusty. It's been years, hell decades.


Yep, because if you were right under the falls the first angle would be 90 degrees.



ok i dont know what formula that is, but i know the law of sine gives hte right answer.



I can't even remember the law of sine. It's been too long, so you're WAAAY ahead of me.



law of sine gets 1557.67 feet high.

that tangent formula not only has 2 unknowns, but is set to equal the hypotonuse. why are you guys trying to figure out the hyp? he wants the hight of the falls, not the distance from where the person is, to the hight of the falls.

anyway, the hypotonuse is 1851.03 feet tall



You are way off.

h is height of the falls or o for opposite side.

You don't use both unkowns. Solve for x using the right side of the equation:

xtan60 = (1000+x)tan57.3

Where x is the distance from the fall. x +1000 is the second distance.

then,

h = xtan60

solve for h.


The Law of sines BTW concerns any triangle, even oblique, with regards to the ratios btwn angles and opposite sides. IOW the ratio of the sin of one of the angles divided by the oppsite side is the same as all the others.

I don't see how it helps.

Gee guys, I haven't done this shit since 1983




h = xtan60 doesnt make sence.
where did you get 60?
and x and h are two un knowns



The first angle, at distance x, is 60 deg. The second angle at distance 1000 + x is 57.3 deg.
Link Posted: 5/22/2005 5:18:10 PM EST

Originally Posted By IchWarrior:
So your standing at the bottom of a water fall, you look up to the top of the falls and for some reason you take not your angle of elevation is 60 degrees.


You go hiking 1000 feet away from the water fall, remaining at the same elevation (Okay, better yet you are on a boat and you float 1000 feet down river.

You look up to the top of the falls and your angle of elevation is 57.3 degrees.


How tall is the falls?


Cant for the life of me figger it out.

ETA: My only hunches is that you find a ratio of your angles of elevation and multiply that by the 1000 foot difference...




SOHCAHTOA...
Link Posted: 5/22/2005 5:25:17 PM EST
tanR = height / distance

tan60 = h / x
xtan60 = h

tan57.3 = h / (1000 + x)
(1000 + x)tan57.3 = h

therefore:

xtan60 = (1000 + x)tan57.3

tan60/tan57.3 = (1000 + x) / x

1.11x = 1000 + x

0.11x = 1000

x = 1000/0.11= 9091 ft.

Therefore:

9091 tan60 = 15746ft high falls.


Link Posted: 5/22/2005 5:48:25 PM EST
[Last Edit: 5/22/2005 5:54:52 PM EST by Aalmeron]

Originally Posted By drjarhead:

Originally Posted By Aalmeron:

Originally Posted By drjarhead:

Originally Posted By Aalmeron:

Originally Posted By dport:

Originally Posted By Aalmeron:

Originally Posted By dport:

Originally Posted By drjarhead:
Actually, you have use trig and algebra.

h = xtan60 = (1000+x)tan57.3


Edited cause I don't know my cos from my tan. Rusty. It's been years, hell decades.


Yep, because if you were right under the falls the first angle would be 90 degrees.



ok i dont know what formula that is, but i know the law of sine gives hte right answer.



I can't even remember the law of sine. It's been too long, so you're WAAAY ahead of me.



law of sine gets 1557.67 feet high.

that tangent formula not only has 2 unknowns, but is set to equal the hypotonuse. why are you guys trying to figure out the hyp? he wants the hight of the falls, not the distance from where the person is, to the hight of the falls.

anyway, the hypotonuse is 1851.03 feet tall



You are way off.

h is height of the falls or o for opposite side.

You don't use both unkowns. Solve for x using the right side of the equation:

xtan60 = (1000+x)tan57.3

Where x is the distance from the fall. x +1000 is the second distance.

then,

h = xtan60

solve for h.


The Law of sines BTW concerns any triangle, even oblique, with regards to the ratios btwn angles and opposite sides. IOW the ratio of the sin of one of the angles divided by the oppsite side is the same as all the others.

I don't see how it helps.

Gee guys, I haven't done this shit since 1983




h = xtan60 doesnt make sence.
where did you get 60?
and x and h are two un knowns



The first angle, at distance x, is 60 deg. The second angle at distance 1000 + x is 57.3 deg.



oh crap, you're right.

lets do it this way.

C=90
c=who cares

A1=60
a1=what we want (w)

B1=30
b1=x

A2=57.3
a2=w

B2=32.7
b2=X+1000

since using hte law of sine on both of these gives...

w/sin60=x/sin30
w*sin30/sin60=x


w/sin57.3=x/sin32.7
w+1000*sin32.7/sin57.3=x

w*sin30/sin60=w+1000*sin32.7/sin57.3
.577w=.477(w+1000)
1.2087w=w+1000
.2087w=1000
w=4790.87feet

which doesnt look right
Link Posted: 5/22/2005 5:50:06 PM EST

Originally Posted By Doctor_Chicago:
Ow my head hurts!



+1,000,000,000
Link Posted: 5/22/2005 5:57:58 PM EST

Originally Posted By drjarhead:
tanR = height / distance

tan60 = h / x
xtan60 = h

tan57.3 = h / (1000 + x)
(1000 + x)tan57.3 = h

therefore:

xtan60 = (1000 + x)tan57.3

tan60/tan57.3 = (1000 + x) / x

1.11x = 1000 + x

0.11x = 1000

x = 1000/0.11= 9091 ft.

Therefore:

9091 tan60 = 15746ft high falls.





looks good to me...
Link Posted: 5/22/2005 6:00:26 PM EST
I get 15,470.7 feet.
Link Posted: 5/22/2005 6:10:45 PM EST

Originally Posted By danno-in-michigan:
I get 15,470.7 feet.



That would be more correct. I dropped a few digits during my calcs.
Link Posted: 5/24/2005 5:06:43 PM EST

Originally Posted By danno-in-michigan:
I get 15,470.7 feet.



I confirm with this number
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