yanni
  • yanni
i neeeed serious help.....
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
what do you need help with?
yanni
  • yanni
In the figure below, find the exact value of y. (Do not approximate your answer.)
yanni
  • yanni
i attached the problem

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More answers

anonymous
  • anonymous
find the side of the larger triange, using pyth theorem
anonymous
  • anonymous
Okay. It is a 45 45 90 triangle. So we know the hypotenuse is five. The hypotenuse in a 45 45 90 triangle is x(sqrt2). So sqrt2 is 1.41. 5/1.41 is 3.546. So x=3.546. Since it equals x.
mathstudent55
  • mathstudent55
The three triangles are similar, so the lengths of their sides are proportional. 6/5 = 5/y
mathstudent55
  • mathstudent55
You don't know it's a 45-45-90 triangle.
anonymous
  • anonymous
then the side that you found is the hypothenuse of 45-45-90 triangle. using that hypotheneuse is twice as large as legs, find the size of the legs
mathstudent55
  • mathstudent55
If the altitude is drawn to the hypotenuse of a right triangle, each leg of the right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to the leg.
anonymous
  • anonymous
simple - use 45-45-90 logic that y -y - ysqrt2 5 = ysqrt2 solve for y
mathstudent55
  • mathstudent55
@JuanitaM Why do you keep mentioning a 45-45-90 trianlge when we don't have one here?
anonymous
  • anonymous
yes an altitude was droppened to form 90 degree at base labeled 6
mathstudent55
  • mathstudent55
The large triangle has hypotenuse of length 6 and a leg of length 5. The other leg cannot possibly measure 5, so it's not a 45-45-90 triangle. Since all triangles are similar, none of them are 45-45-90 triangles.
anonymous
  • anonymous
Find the 3rd side of the triangle, label this x. \[x=\sqrt{6^2-5^2}=\sqrt{11}\] Then find the line going straight down the triangle, label this z. You can do this two ways (you'll need both ways). \[z=\sqrt{\sqrt{11}^2-(6-y)^2}=\sqrt{11-(6-y)^2}\] \[z=\sqrt{5^2-y^2}\] If you make them equal to each other, you'll be able to solve for y. \[\sqrt{11-(6-y)^2}=\sqrt{5^2-y^2}\] \[11-(6-y)^2=25-y^2\] \[11-(36-12y+y^2)=25-y^2\] \[2y^2-12y-50=0\] \[\therefore y=\sqrt{34}+3\]
anonymous
  • anonymous
Is this too confusing?
anonymous
  • anonymous
|dw:1360214471836:dw| Maybe that will make it more understandable :)
mathstudent55
  • mathstudent55
@Chelsea04 Can you try to calculate your answer for y as a number rounded off to the nearest tenth?
mathstudent55
  • mathstudent55
@yanni Did you understand anything about this problem or are you more confused than when you asked the question?
agent0smith
  • agent0smith
Chelsea, your answer cannot be correct. Your value of y is greater than the 6 at the base of the triangle!
agent0smith
  • agent0smith
Chelsea, this is your mistake: \[11−(36−12y+y^2)=25−y^2\]\[12y−50=0 \] mathstudent55 has the correct answer: The three triangles are similar, so the lengths of their sides are proportional. 6/5 = 5/y You can confirm it by using cosine.
mathstudent55
  • mathstudent55
@agent0smith Finally a voice of reason. Thanks!
agent0smith
  • agent0smith
haha no prob. I forgot to make the drawing:
agent0smith
  • agent0smith
I'll call that angle x |dw:1360216763172:dw| First, look at the bigger/outer triangle \[\cos x = \frac{ 5 }{ 6 }\] Now, look at the inner triangle: \[\cos x = \frac{ y }{ 5 }\] |dw:1360216959551:dw|
anonymous
  • anonymous
Right, oops!!! That seems like a much faster way!!!
agent0smith
  • agent0smith
once you get cosx = 5/6 and cos x = y/5, then you can equate 5/6 = y/5
anonymous
  • anonymous
Either way is correct, just mine was more algebraic (and a little wrong!) than trignometric.
agent0smith
  • agent0smith
Yes, your way works (other than your mistake) but will probably take longer.

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