is anyone good with trigometry??

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is anyone good with trigometry??

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trigo wat
trigonometric
maybe.

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Other answers:

euclid maybe; but hes dead :/
post wht you have
you can use the pythagorean theorem to find b and then use that to find a
this triangle is not possible... your hypotenuse will always be longer than each leg... this is not the case in this picture
b^2 + 13^2 = (12+a)^2 5^2 + a^2 = b^2
what about this? How long is a string reaching from the top of a 8ft pole to a point 5 ft from the base of the pole?
(12+a)^2 - 13^2 = 5^2 + a^2
looks like similar triangles to me.
I'm trying to study for finals...
144 + a^2 + 24a - 196 = 25 + a^2 144 - 196 - 25 = -24a 144 - 196 - 25 = a -------------- -24
@amistre64 which one is that for?
a = 77/24..... the first one you posted
how about... A slide 4.1 m long makes an angle of 28 degrees with the ground. How high is the top of the slide above the ground?
4.1 sin(28)
thats all you do for that?
yes; unless you wanna put it into a calculator to get an approx distance
yeah, that first triangle set up could have been ration
what about find the missing abgle and side measure of triangle ABC, given that angle A=50 degrees, Angle C= 90 degrees, and CB=7
PEOPLE! it is much easier to solve with similar triangles! 5/13 = a/5 --- > a = 25/13
dhat; yeah, i know :) just thought id give it a shot the other way lol
oh okay. go on. :)
given 2 degrees and a side you can do law of cosines to get another side and then law of sine it with some careful planning
given angles and a side; law of sines might be the best optio
how excatly would i do that.?
never mind that; this is a 90 degree to begin with ABC, given that angle A=50 degrees, Angle C= 90 degrees, and CB=7
angB = 40
7 AC ----- = ------- tho sin 50 sin(40)
ac = abt 5.874 maybe
7 ab ----- = ----- = abt 9.12 sin 50 sin90
my two options are; A)
the trick is to dbl check that and see how close it is :)
a then
so how'd you get the 7 over sin 50 and the ab over sin90?
thats the law of sines; side a side b side c ----- = ------ = ------ sin(a) sin(b) sin(c)
7 ----- is a given; so the others just fall from that sin(50)
ohhh okay. My teacher sucked at teaching the law of sines so i can never get it.
cos(g) = a/sqrt(45) sin^2 + cos ^2 = 1 sin^2 + a^2/45 = 1 sin^2 = 1 - a^2/45 sin = sqrt(1-a^2/45) tan(g) = sqrt(1-(a^2/45)) --------------- a^2/sqrt(45)
tan(g) = sqrt(1-(a^2/45)) (sqrt(45)) ----------------------- a^2 tan(g) = sqrt(1-(a^2/45)(45)) -------------------- a^2 tan(g) = sqrt(45-a^2) ------------ a^2
How about this.? Thank you for all your help too. If you need anything in return I'd be glad to
x = 23 sqrt(2) y = 23 + 46cos 30
hey you wrote this: tan(g) = sqrt(45-a^2) ------------ a^2 But how would i finish it.?
With regard to 20110518.13.41.36.jpg: \[a=5 \text{Cos}\left[\frac{\pi }{2}-\text{ArcTan}\left[\frac{5}{12}\right]\right]=\frac{25}{13} \]\[b=5 \text{Sin}\left[\frac{\pi }{2}-\text{ArcTan}\left[\frac{5}{12}\right]\right]=\frac{60}{13} \]\[\sqrt{\left(\frac{25}{13}\right)^2+\left(\frac{60}{13}\right)^2}=5 \]

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