Posted by Crystal on Saturday, November 24, 2012 at 8:18pm.
SKETCH the problem
draw altitude h down from C to line AB at D
call AD x
then DB is 350 -x
Now look at right triangle ACD
tan 32 = h/x = .625
so
x = h/.625 = 1.6 h
Now triangle BCD
tan 54 = h/(350-x)
1.38 = h/(350 -1.6 h)
482 - 2.20 h = h
3.2 h = 482
h = 151 meters
If you follow that you can do the other two.
Sorry, kilometers, not meters
Look down at the bottom of the page for related problems solved. Look particularly at this one:
http://www.jiskha.com/display.cgi?id=1279657357
How did you get x = h/.625 = 1.6 h ? How did you arrive at 1.6?
1/.625 = 1.6
tan 32 = h/x
so
x tan 32 = h
so
x = h/tan 32
so
x = h/.625
but 1/.625 = 1.6
so
x = 1.6 h
Thanks! I don't know how I missed that.
okay so I think I'm still kind of hopeless... my solution to the second question is way off, but here's what I did:
right triangle ABC. (facing like this: >) so that AB is the hypotenuse, and I labelled AC as h for the height of the cliff. I drew a dotted line from point A to about the middle of line BC and called that point D.
Made angle ABD 31 degrees, angle ADC 33 degrees, line BD x+300, line DC x.
for triangle ADC:
tan33=h/x
x=h/tan33
x=h/0.65
for triangle ABC:
tan 31=h/x+300+x
0.6 = h all over h/0.65 +300 + h/0.65
0.6 =h x (1.3/2h) + 300
0.6 - 300 = 1.3/h
-299.4 = 1.3/h
h = 1.3/299.4
h = 0.004 .....
Yeah, I guess I really don't get this... *sigh*
It was midnight. I went to bed.
Made angle ABD 31 degrees, angle ADC 33 degrees, line BD x+300, line DC x.
for triangle ADC:
tan33=h/x
x=h/tan33
x=h/0.65
for triangle ABC:
tan 31 = h/(x+300)
.6 = h/(x+300)
.6 x + 180 = h
.6(h/.65) + 180 = h
.92 h + 180 = h
180 = .08 h
h = 2250
need to carry more significant figures to get a more accurate answer
Completely understandable, thank you so much for checking back though!
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