anonymous
  • anonymous
A ranch in the Australian Outback is shaped like triangle ACE, with m∠A = 42, m∠E = 103, and AC = 15 miles. Find the area of the ranch, to the nearest square mile. I know the answer is 44 but how do i get it?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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austinL
  • austinL
|dw:1377812200402:dw|
austinL
  • austinL
|dw:1377812269584:dw|
austinL
  • austinL
How would you solve for "h"?

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More answers

anonymous
  • anonymous
soh cah toa?
austinL
  • austinL
|dw:1377812372427:dw| Which one would we use?
anonymous
  • anonymous
sin(42)= h/15 ?
anonymous
  • anonymous
wait no
austinL
  • austinL
Yes :)
anonymous
  • anonymous
oh yay wait but when you solve that isnt it negative?
austinL
  • austinL
Yeah, that is odd...
austinL
  • austinL
@Luigi0210
anonymous
  • anonymous
ahh i hate this problem
Luigi0210
  • Luigi0210
You called the camel?
anonymous
  • anonymous
Umm i'm not sure can you help please
Luigi0210
  • Luigi0210
Sure, hold on. What is this?
anonymous
  • anonymous
Uh how would i find the area of triangle ACE with m
austinL
  • austinL
I know how it should work, but I arrive at a negative length for one of the sides of the triangle which is physically impossible.
Luigi0210
  • Luigi0210
Did you try: \[A= \frac{1}{2}a*b*sinC\]
anonymous
  • anonymous
how does that work
austinL
  • austinL
We would need to solve for another side.
austinL
  • austinL
|dw:1377813301432:dw|
Luigi0210
  • Luigi0210
Don't all triangles make 180?
anonymous
  • anonymous
so its 35 but how does that help me with side lengths
Luigi0210
  • Luigi0210
Oh, we cold use the law of sines to solve for a missing side
austinL
  • austinL
Yes, but for that we need two sides (Law of Sines). As well as the additional angle which is easy to solve for. \(\dfrac{a}{\sin(a)} = \dfrac{b}{\sin(b)} = \dfrac{c}{\sin(c)}\)
austinL
  • austinL
\(\dfrac{a}{\sin(42)} = \dfrac{15}{\sin(103)}\) We would then just solve for "a" Then we would need to calculate for the final angle.
Luigi0210
  • Luigi0210
^ That's what I was talking about doing
austinL
  • austinL
=D
Luigi0210
  • Luigi0210
I got 10.30097235
Luigi0210
  • Luigi0210
I haven't had to deal with triangles for a while.. so we put the calc in degree mode when solving right?
austinL
  • austinL
Yes, because the angles are measured in degrees.
anonymous
  • anonymous
yea
austinL
  • austinL
And yet we arrive back at the negative numbers.
Luigi0210
  • Luigi0210
Really? Is yours in radians? I got a negative answer in radians
anonymous
  • anonymous
yea mines in radians whoops
austinL
  • austinL
10.301.... Don't I feel silly now.
anonymous
  • anonymous
haha smallest mistake makes the biggest difference thanks
Luigi0210
  • Luigi0210
Yup, and now you guys are set :)
austinL
  • austinL
|dw:1377814037860:dw|
anonymous
  • anonymous
thank you :D
austinL
  • austinL
\(A=\dfrac{1}{2}a∗b∗\sin(C)\) a=15 b=10.301 C=?
anonymous
  • anonymous
um wait would you have to split it in the middle again to make the two right triangles
Luigi0210
  • Luigi0210
Nope, you don't have to do that
Luigi0210
  • Luigi0210
C is just referring to the angle measure
anonymous
  • anonymous
its 35
austinL
  • austinL
Correct. C=35
austinL
  • austinL
Plug and chug.
anonymous
  • anonymous
but don't we need to know the measure of AE?
Luigi0210
  • Luigi0210
Nope :)
Luigi0210
  • Luigi0210
Watch, you'll see the magic when you plug it in ;)
anonymous
  • anonymous
wait i'm confused- sorry
austinL
  • austinL
\(A=\dfrac{1}{2}a∗b∗\sin(C)\) a=15 b=10.301 C=35
austinL
  • austinL
There is your area formula. You know all the parts. Boom. :)
Luigi0210
  • Luigi0210
That sinC is not the general sinx=opp/hyp It's just sin(angle) If that's what you're confused on
anonymous
  • anonymous
i got like -33 or something
Luigi0210
  • Luigi0210
Did you do it all as one?
anonymous
  • anonymous
wait no i got it haha THANK YOU
austinL
  • austinL
\(A = \dfrac{1}{2}(15)\times10.301\times\sin(35)\) :D
Luigi0210
  • Luigi0210
Don't thank me, thank Austin, the bike riding donkey :P
anonymous
  • anonymous
haha thank y'all
austinL
  • austinL
You are very welcome!!

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