anonymous
  • anonymous
is anyone good with trigometry??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
trigo wat
anonymous
  • anonymous
trigonometric
anonymous
  • anonymous
maybe.

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amistre64
  • amistre64
euclid maybe; but hes dead :/
anonymous
  • anonymous
post wht you have
anonymous
  • anonymous
anonymous
  • anonymous
you can use the pythagorean theorem to find b and then use that to find a
anonymous
  • anonymous
this triangle is not possible... your hypotenuse will always be longer than each leg... this is not the case in this picture
amistre64
  • amistre64
b^2 + 13^2 = (12+a)^2 5^2 + a^2 = b^2
anonymous
  • anonymous
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?
amistre64
  • amistre64
(12+a)^2 - 13^2 = 5^2 + a^2
anonymous
  • anonymous
looks like similar triangles to me.
anonymous
  • anonymous
I'm trying to study for finals...
amistre64
  • amistre64
144 + a^2 + 24a - 196 = 25 + a^2 144 - 196 - 25 = -24a 144 - 196 - 25 = a -------------- -24
anonymous
  • anonymous
@amistre64 which one is that for?
amistre64
  • amistre64
a = 77/24..... the first one you posted
anonymous
  • anonymous
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?
amistre64
  • amistre64
4.1 sin(28)
anonymous
  • anonymous
thats all you do for that?
amistre64
  • amistre64
yes; unless you wanna put it into a calculator to get an approx distance
amistre64
  • amistre64
yeah, that first triangle set up could have been ration
anonymous
  • anonymous
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
anonymous
  • anonymous
PEOPLE! it is much easier to solve with similar triangles! 5/13 = a/5 --- > a = 25/13
amistre64
  • amistre64
dhat; yeah, i know :) just thought id give it a shot the other way lol
anonymous
  • anonymous
oh okay. go on. :)
amistre64
  • amistre64
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
amistre64
  • amistre64
given angles and a side; law of sines might be the best optio
anonymous
  • anonymous
how excatly would i do that.?
amistre64
  • amistre64
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
amistre64
  • amistre64
angB = 40
amistre64
  • amistre64
7 AC ----- = ------- tho sin 50 sin(40)
amistre64
  • amistre64
ac = abt 5.874 maybe
amistre64
  • amistre64
7 ab ----- = ----- = abt 9.12 sin 50 sin90
anonymous
  • anonymous
my two options are; A)
amistre64
  • amistre64
the trick is to dbl check that and see how close it is :)
amistre64
  • amistre64
a then
anonymous
  • anonymous
so how'd you get the 7 over sin 50 and the ab over sin90?
amistre64
  • amistre64
thats the law of sines; side a side b side c ----- = ------ = ------ sin(a) sin(b) sin(c)
amistre64
  • amistre64
7 ----- is a given; so the others just fall from that sin(50)
anonymous
  • anonymous
ohhh okay. My teacher sucked at teaching the law of sines so i can never get it.
anonymous
  • anonymous
amistre64
  • amistre64
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)
amistre64
  • amistre64
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
anonymous
  • anonymous
How about this.? Thank you for all your help too. If you need anything in return I'd be glad to
anonymous
  • anonymous
x = 23 sqrt(2) y = 23 + 46cos 30
anonymous
  • anonymous
hey you wrote this: tan(g) = sqrt(45-a^2) ------------ a^2 But how would i finish it.?
anonymous
  • anonymous
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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