BloomLocke367
  • BloomLocke367
Find the area.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
what
anonymous
  • anonymous
whats the question
BloomLocke367
  • BloomLocke367
|dw:1433197250937:dw|

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More answers

BloomLocke367
  • BloomLocke367
@Nnesha
anonymous
  • anonymous
hold on
BloomLocke367
  • BloomLocke367
@perl
anonymous
  • anonymous
okay this link will help https://www.mathsisfun.com/algebra/trig-area-triangle-without-right-angle.html
BloomLocke367
  • BloomLocke367
@Nnesha
Nnesha
  • Nnesha
http://prntscr.com/7c1xmi you need to find 3rd side of triangle it's not a right triangle so law of cosine ?familiar with this method ?
BloomLocke367
  • BloomLocke367
kind of, I sort of know what it is, but I don't really know how to use it
BloomLocke367
  • BloomLocke367
I have to eat dinner, be back in a sec
Nnesha
  • Nnesha
alright lol
Nnesha
  • Nnesha
have fun!
BloomLocke367
  • BloomLocke367
I'm back, I wasn't that hungry
BloomLocke367
  • BloomLocke367
@Nnesha
Nnesha
  • Nnesha
ahh alright so did you eat candy ? :P
BloomLocke367
  • BloomLocke367
no, I had like a few bites.. I think it's my nerves that have me not hungry.. I'm very stressed.
Nnesha
  • Nnesha
\[\huge\rm c^2 = a^2 +b^2 -2abcos C\] |dw:1433199653612:dw| plug in value solve for c
BloomLocke367
  • BloomLocke367
wait, don't I need to use law of sines for this?
Nnesha
  • Nnesha
mhm nope bec given is two sides and one angle
BloomLocke367
  • BloomLocke367
yes, you use cosine if you have 3 sides, and law of sines if you have 2 sides and an angle.. I'm almost positive
Nnesha
  • Nnesha
http://mathforum.org/library/drmath/view/69918.html you can use law of cosines to find 3rd side
BloomLocke367
  • BloomLocke367
but... https://www.mathsisfun.com/algebra/trig-sine-law.html ...
Nnesha
  • Nnesha
https://www.mathsisfun.com/algebra/trig-cosine-law.html
BloomLocke367
  • BloomLocke367
oh, okay.. sorry for doubting you
Nnesha
  • Nnesha
nah it's fine! i like it :P
Nnesha
  • Nnesha
okay solve for c
BloomLocke367
  • BloomLocke367
c^2=64+324-2(8)(18)cos(25), so far, right?
BloomLocke367
  • BloomLocke367
c^2=388-288cos(25) c^2=100cos(25)
BloomLocke367
  • BloomLocke367
right?
Nnesha
  • Nnesha
hold on >:P
BloomLocke367
  • BloomLocke367
okay
Nnesha
  • Nnesha
yep right solve for c
BloomLocke367
  • BloomLocke367
I got c^2=90.63 so c=9.52
BloomLocke367
  • BloomLocke367
@Nnesha
BloomLocke367
  • BloomLocke367
@Nnesha
Nnesha
  • Nnesha
hmm i got 11.269
BloomLocke367
  • BloomLocke367
how? (sorry, I had to go to bed) @Nnesha
BloomLocke367
  • BloomLocke367
@pooja195
geerky42
  • geerky42
Why do we need to find value of c? |dw:1433254516249:dw| Obviously, \(\large A_{\text{triangle}} = \dfrac12bh\) We already have base \(\overline{\text{QR}} = 8\) So we need to find h: Using fact that \(\sin\theta = \dfrac{\text{Opp}}{\text{Hyp}}\), we have \(\sin Q = \dfrac{h}{\overline{\text{PQ}}} = \dfrac{h}{18}\) Yield \(18\sin(25^\text o) = h\) So we have \(A_{\text{triangle}} = \dfrac12~\overline{\text{QR}}~h = \dfrac{1}{2}(8)(18\sin(25^\text o)~) = \cdots~?\)
BloomLocke367
  • BloomLocke367
I got 30.43
geerky42
  • geerky42
Me too. So that's your area.
geerky42
  • geerky42
Do you understand what I wrote?
BloomLocke367
  • BloomLocke367
oh, okay. kinda, I don't understand how you got the height, but the rest I get.
geerky42
  • geerky42
well, we can set up height of triangle here:
geerky42
  • geerky42
sorry hold on...
BloomLocke367
  • BloomLocke367
okay. Can you help with more?
geerky42
  • geerky42
|dw:1433254921641:dw| Then we can treat it as "new triangle" |dw:1433254928278:dw| From here, we have hypotenuse and angle Q and height. So using sine, we can solve for h. Then we use that height to solve for area of original triangle.
geerky42
  • geerky42
I can try.
BloomLocke367
  • BloomLocke367
ohhhhhhh okay. and it's the same height for both triangles?
geerky42
  • geerky42
Same angle and hypotenuse, so yeah same height,
BloomLocke367
  • BloomLocke367
alright. I have some more questions, if that's okay. I'll post a new question and tag you
geerky42
  • geerky42
sure

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