From a boat on the lake, the angle of elevation to the top of a cliff is 12°50'. If the base of the cliff is 1366 feet from the boat, how high is the cliff (to the nearest foot)?

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From a boat on the lake, the angle of elevation to the top of a cliff is 12°50'. If the base of the cliff is 1366 feet from the boat, how high is the cliff (to the nearest foot)?

Mathematics
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1366(tan(12'50'')) '=degree and ''=minute for this one :)
okay :)
(79/6) degrees ??

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319.554 is what the calculator spits out :) that is if 79/6 is a good number. 12 and 50/60 = 12 5/6 = (72+5)/6 = 77/6 gotta recalc it :)
:)
311.183 lets try that one on fer size :)
i got 311.09
close enough....311 is prolly the answer then
:) thank you can you help me with my next one
maybe :) is it an easy one?
its kind of the same as this
lets here it
From a boat on the river below a dam, the angle of elevation to the top of the dam is 30°25'. If the dam is 1065 feet above the level of the river, how far is the boat from the base of the dam (to the nearest foot)?
these are basically tan problems
oh ok
[30(60)+5]/60 is going to be the number for my degrees.. tan(n) = 1065/x x = 1065/tan(n)...
1838.453 double check that...
1838.453 its what I get :)
1813 ft?
nope... messed up someplace... my number for degrees specifically..
[30(60)+ 25] /60
recalculating....
1814.03 is what i get with the right numbers :)
i did it that way and got 1838.7
wow lol
I added a "5" it aint 5/60 minutes... its 25/60 minutes so I had to get the right number for degrees :)
oh ok
im sure its 1814, or close to that :)
thank you :)

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