anonymous
  • anonymous
In triangle DEF, angle D = 44 degrees, angle E = 61 degrees, and segment EF = 20 inches. What is the length of segment DE?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
So you can add the two angles together then subtract by 180 first
anonymous
  • anonymous
angle f = 75
anonymous
  • anonymous
Then do law of sines.

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anonymous
  • anonymous
Laws of sines. You have the angle D (opposite to EF), just discover the angle F (opposite to DE). 44 + 61 + angle F = 180 angle F= 180-44-61 = 75 Now, laws of sines: EF/sin44º = ED/sin75º
anonymous
  • anonymous
so sin44/20=sin75/DE
anonymous
  • anonymous
sin44/20 = . 035, sin75/de = .035
anonymous
  • anonymous
Just solve for DE.
anonymous
  • anonymous
DE = sin75 x sin44/20?
anonymous
  • anonymous
No, DE = 20*sin75/sin44
anonymous
  • anonymous
27.8
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
A cruise ship travels 300 miles east, then turns 30 degrees north, and travels another 175 miles. What is the total distance from its starting point?
anonymous
  • anonymous
Laws of cosines. You have 2 sides and the angle between this sides. Just use the formula that I said to you yesterday: c² = a² + b² - 2abcos theta c is the side that you want to discover opposite to the angle theta.
anonymous
  • anonymous
172 miles
anonymous
  • anonymous
yes yes
anonymous
  • anonymous
In triangle XYZ, XY = 15, YZ = 21, and XZ = 27. What is the measure of angle Z to the nearest degree?
anonymous
  • anonymous
xy is opposite of angle Z
anonymous
  • anonymous
XZ is hypotenuse, YZ is adjacent
anonymous
  • anonymous
the triangle is right ?
anonymous
  • anonymous
i don't know, there's no picture or diagram
anonymous
  • anonymous
So you can't that there's a hypotenuse rs.
anonymous
  • anonymous
cant say*
anonymous
  • anonymous
oh, okay. i just assumed the hypotenuse was XZ because it's the longest segment
anonymous
  • anonymous
you have to use the other laws of cosine formula. Cos theta = (a² + b² - c²) / ( 2*a*b)
anonymous
  • anonymous
c is the side opposite to the angle that you want.
anonymous
  • anonymous
so XY is my c
anonymous
  • anonymous
yep
anonymous
  • anonymous
i got 89.9 but that's not an option
anonymous
  • anonymous
are you sure? cos theta = (27² + 21² - 15² )/2*27*21 = 0.83
anonymous
  • anonymous
do arccos with 0.83
anonymous
  • anonymous
33.9 which rounds up to 34
anonymous
  • anonymous
there's an option?
anonymous
  • anonymous
yep! 34 is an option
anonymous
  • anonymous
:)
anonymous
  • anonymous
verify the identity: http://puu.sh/3e9qD.png
anonymous
  • anonymous
It's to say that it's true or false?
anonymous
  • anonymous
Or re-do the identity ?
anonymous
  • anonymous
i have to show the steps of verifying that it is true or false
anonymous
  • anonymous
Ok, okay,
anonymous
  • anonymous
a minute
anonymous
  • anonymous
I got it, i will write in latex.
anonymous
  • anonymous
\[\frac{ \sec x }{ \csc x - \cot x } - \frac{ \sec x }{ \csc x + \cot x } = \frac{ \sec x (\csc x + \cot x) - \sec x (\csc x - \cot x) }{ \csc²x - \cot²x }\] \[\csc² x - \cot ²x = 1\] So, we will have, \[\sec x(\csc x + \cot x) - \sec x (\csc x - \cot x) =\] \[\sec x \csc x + \sec x \cot x - \sec x \csc x + \sec x \cot x\]\[2\sec x \cot x = 2 \frac{ 1 }{ \cos x } * \frac{ 1 }{ \tan x } = 2 \frac{ 1 }{ \cos x } * \frac{ \cos x }{ sen x } = \]\[2*\frac{ 1 }{ sen x } = 2\csc x\]
anonymous
  • anonymous
If you don't get it, i will try to explain better.
anonymous
  • anonymous
i get it. thank you so much.
anonymous
  • anonymous
Just to make it complete, I said that csc²x - cot²x = 1 because: \[\csc² x = \frac{ 1 }{ sen² x } \iff \cot² x = \frac{ \cos² x }{ sen² x }\]\[\csc² x - \cot² x = \frac{ 1 }{ sen² x } - \frac{ \cos² x }{ sen² x } = \frac{ 1- \cos² x }{ sen x }\]But you now that \[sen² x + \cos² x = 1\]So,\[sen²x = 1 - \cos²x\]To conclude,\[\csc²x - \cot²x = \frac{ 1 - \cos²x }{ sen²x } = \frac{ sen²x }{ sen²x } = 1\]

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