anonymous
  • anonymous
right triangle ABC with right angle c and CD perpendicular to AB at D if AB is 8 and AC is 4 find AD
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1332363734112:dw|
jagatuba
  • jagatuba
Do you know your trig formulas for solving right triangles?
anonymous
  • anonymous
Seg1/leg=leg/hyp.

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anonymous
  • anonymous
?
anonymous
  • anonymous
idk if thats correct
jagatuba
  • jagatuba
Well you are going to have to go through a few steps to solve this problem. Any triangle can be solved as long as you know the length of one side and two other values, either side lengths or angles. This one is tricky because we only have 2 known values 90 degree angle (at D) and a length of 4 on the hypotenuse of triangle ACD; and one only partially known length, AD < 8. Now it's been a while since I had trig so I might have to enlist some help from @myininaya or @AccessDenied they are good at this stuff, but if they aren't online you might have to wait for my slow brain to figure out the problem before I can explain it.
mysesshou
  • mysesshou
will help in a bit. i solved it on paper. i'm finishing up another person. just hang on a sec :)
anonymous
  • anonymous
ok
mysesshou
  • mysesshou
I'm sure there are many ways of solving it, but here is how i did it ;)
mysesshou
  • mysesshou
|dw:1332368012219:dw|
mysesshou
  • mysesshou
|dw:1332368139181:dw|
mysesshou
  • mysesshou
solve fore theta: using SOH CAH TOA, choose CAH: \[cos(\theta) = \frac{4}{8}\] \[\cos^{-1}(4/8) = \theta\] \[\theta = 60 \] degrees
mysesshou
  • mysesshou
|dw:1332368424307:dw| solve for x
mysesshou
  • mysesshou
\[cos(\theta) = \frac{x}{4}\] \[4*cos(60) = x\] x=2
mysesshou
  • mysesshou
How does this look? check my numbers with calculator :)
mysesshou
  • mysesshou
@jagatuba , I think this is one method :)
mysesshou
  • mysesshou
Hopefully someone can double check :)
mysesshou
  • mysesshou
Since it has been a long time since I've done trig .... ok, bye.
mysesshou
  • mysesshou
I think it also works since 4 is to 8 as the 2 is to 4, since they are of similar shape.
AccessDenied
  • AccessDenied
There is a special case in Geometry for the situation of a right triangle with an altitude to the hypotenuse. I could show you the proof for the theorem if it would supplement this explanation to you, but I will omit it for now to keep it short. Just ask if you would like me to go over it, or you may be able to find it online / in a textbook! :) Basically, this theorem creates a proportion that relates the leg, its adjacent 'hypotenuse segment', and the entire hypotenuse. The leg is essentially the geometric mean of the hypotenuse and the segment adjacent to the leg. |dw:1332437959857:dw| So, in this particular problem, we have the hypotenuse segment AD which we want to find, the given leg AC, and the entire hypotenuse AB. We'd just set up the proportion: \[ \frac{AD}{AC} = \frac{AC}{AB};~~solve~for~{AD}, ~~AC=4,~AB=8. \] (That's not necessarily the only way to write it, but that's one way)
mysesshou
  • mysesshou
Ah, @AccessDenied, That makes sense. I have forgotten so many of the rules and proofs. I suppose I should reread my books. Your method is simpler (and faster for exams). :) Thanks ! :)
mysesshou
  • mysesshou
I had geometry back sometime around 2001-2002 I think O_O
AccessDenied
  • AccessDenied
No problem! I actually went through this recently in Geometry (I'm in Geometry and Algebra 2 currently, tho I do plenty of studying myself), so it's still fresh for me. :P
mysesshou
  • mysesshou
I so wish there was a site like this when I was in algebra through DiffEq. I loved math and it was so much fun, but I have since forgotten most of it. :) I like helping people and challenges, so I would have been on here for hours.

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