sarver1995
in a 45 45 90 triangle i need to find what x is
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sarver1995
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|dw:1342917395925:dw|
Romero
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|dw:1342917641735:dw|
sarver1995
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would it be 6 square root of 2
Romero
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Use Pythagoras theorem
\[x^2+x^2=12^2\]
Romero
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yes
Australopithecus
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you cant solve this with pythagoras theorem you have to use trig I'm pretty sure
Romero
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\[2x^2=12^2\]\[2x^2=144\]\[x^2=\frac{144}{2}\]\[x^2=72\]\[x= \sqrt{72}\]\[x= \sqrt{36*2}\]\[x=6 \sqrt{2}\]
Romero
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Yes you can you have two sides that are equal
Calcmathlete
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Hey @sarver1995 Been a while :)
45-45-90 triangles:
|dw:1342917784578:dw|
Australopithecus
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oh yeah just noticed :)
Calcmathlete
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\[x\sqrt{2} = 12\]\[x = \frac{12}{\sqrt{2}}\]\[x = \frac{12}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\]\[x = \frac{12\sqrt{2}}{2}\]\[x = 6\sqrt{2}\]There you go :)
Australopithecus
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SOH CAH TOA
Sin = opposite/hypotenuse
cos = adjacent/hypotenuse
tan = opposite/adjacent
cos(45) = x/12
cos(45)(12) = x
Romero
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from this I noticed that you can simply get one of the sides by multiplying the hypotenuse by
\[\frac{1}{\sqrt{2}}\]
Romero
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Only if the triangle is a 45 45 90
sarver1995
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|dw:1342917788335:dw|
Calcmathlete
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Same thing...use the ratio above...
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Romero
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same thing like I said
\[\frac{1}{\sqrt{2}}*x=10\]
Romero
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solve for x
sarver1995
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to calcmathlete yeah i am now reviewing and oh course still confused
i only have a few more lessons and i am finished with 11th grade. its been really rough
sarver1995
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|dw:1342918127585:dw| would x be 10 square root of 2
Calcmathlete
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Yes :)
\[x = 10\sqrt{2}\]
Romero
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Good job mate!
sarver1995
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|dw:1342918257888:dw| okay not sure how to do this one
sarver1995
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it cant be 1/2 square root of 2
Calcmathlete
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Does it tell you if it's a right triangle?
sarver1995
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45 45 90
Calcmathlete
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It tells you that?
sarver1995
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yes
Calcmathlete
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Ok.
The ratio again.
x, x, x√2
x = 1.2
x√2 = ?
sarver1995
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i will try to draw it again maybe i ddin't do it right
Calcmathlete
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By the way, if the x's get confusing because they are using x as a variable, I'll just use y.
sarver1995
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|dw:1342918455394:dw|
Calcmathlete
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Ok. Same situation.
The legs are always congruent in a 45-45-90 triangle.
The hypotenuse is always in this ratio:
\[Leg \times \sqrt{2}\]Can you figure it out since you have the leg?
sarver1995
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1.2 x square root of 2
sarver1995
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not sure on this one
Calcmathlete
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Yup. which can be rewritten as \(1.2\sqrt{2}\)
sarver1995
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okay i have one more like these can i try it with us
Calcmathlete
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Alright shoot away :)
sarver1995
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|dw:1342918731847:dw|
Calcmathlete
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\[x = \frac{\sqrt{24}}{\sqrt{2}} = \sqrt{\frac{24}{2}} = \sqrt{12} = ?\]
sarver1995
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2 square root of 3
Calcmathlete
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yup :)
sarver1995
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okay i have a different one i just need u to help me understand the problem it will take a little while to draw
Calcmathlete
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ok.
sarver1995
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|dw:1342918993022:dw| Refer to the figure to complete the following item.
Given:
If m arc vu =80° and marc st =40°, then 1 =
Calcmathlete
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Oh gosh...a proof?
sarver1995
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yeah lol
Calcmathlete
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Do you have to like fill in a blank or is it like write an entire proof?
sarver1995
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its just asking what angle 1 would equal
sarver1995
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just an answer
Calcmathlete
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I think that it's 50...
sarver1995
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can you tell me how you got that because i have three more problems like this one
Calcmathlete
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\[m\angle1 = \frac 12(arcVT - arcST)\]
sarver1995
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If arc vu m = 70° and m arc st = 30°, then 2 = would it be the same formula for this one
Calcmathlete
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I kind of have to go...sorry :/ If it maintains the same shape, it should be the same sort of formula...
sarver1995
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thanks for your help its really appreciated and God bless you