Posted: 5/22/2005 3:13:15 PM EDT
|
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... |
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 |
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. |
But: sin R = height of falls/ hypotenuse ::: so I am not sure how that helps. tan R = height of falls / distance |
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. |
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 |
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. |
SOHCAHTOA...  |
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 |
looks good to me... |
