anonymous
  • anonymous
I really need help with this please. :) Here is a picture of the problem: https://cdn.ple.platoweb.com/EdAssets/2aa74bea51264416af45cb620a392471?ts=635545843843870000 If a = 24 mm, c = 26 mm, and m∠B = 33°, what is the approximate area of ABC? 1. 261.67 mm2 2. 54.46 mm2 3. 169.93 mm2 4. 339.85 mm2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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SolomonZelman
  • SolomonZelman
This is a right triangle?
anonymous
  • anonymous
I think so.
SolomonZelman
  • SolomonZelman
Ok, then you can use the Pythagorean theorem to find the side b.

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More answers

anonymous
  • anonymous
I would do that, but it is a degree instead of a number.
SolomonZelman
  • SolomonZelman
I am saying to use the `. a² + b² = c² . `
SolomonZelman
  • SolomonZelman
24²+b²=26²
anonymous
  • anonymous
Okay. Thank you so much!
SolomonZelman
  • SolomonZelman
we aren'tdone yet, are we?
SolomonZelman
  • SolomonZelman
Did you find the side b?
SolomonZelman
  • SolomonZelman
(It will be a precise, natural number, answer)
anonymous
  • anonymous
I'm working on it.
SolomonZelman
  • SolomonZelman
Ok, sure, take your time:)
anonymous
  • anonymous
I got 10.
SolomonZelman
  • SolomonZelman
Oh, very good. that is right!
SolomonZelman
  • SolomonZelman
|dw:1441584583951:dw|
SolomonZelman
  • SolomonZelman
So you can already tell which choice is closer to the area.
anonymous
  • anonymous
24 right?
jdoe0001
  • jdoe0001
|dw:1441585425017:dw|
SolomonZelman
  • SolomonZelman
I wonder that they do not have the answer choice for that....
anonymous
  • anonymous
So, the test makers didn't put the correct choice on it?
SolomonZelman
  • SolomonZelman
No, I think it is rather our mistake that we assumed that your triangle is a right triangle.
jdoe0001
  • jdoe0001
hmmm
anonymous
  • anonymous
Okay, so how would we solve this now?
SolomonZelman
  • SolomonZelman
I think we need to apply the law of cosines. \(b^2=a^2+c^2-2ac{\rm Cos}(B)\)
jdoe0001
  • jdoe0001
have you covered the law of cosines yet?
anonymous
  • anonymous
Not yet, but this is supposed to be a Algebra 1 study.
SolomonZelman
  • SolomonZelman
https://www.mathsisfun.com/algebra/trig-cosine-law.html
jdoe0001
  • jdoe0001
hmmm so.. haemm I think is safe to assume then, that is a right triangle
SolomonZelman
  • SolomonZelman
This is a good link. mathisfun is always easy to read:)
SolomonZelman
  • SolomonZelman
yes jdoe, I assumed that the first I encountered it, but there is no option of 120.
jdoe0001
  • jdoe0001
well, right... I'd use the law of cosines, but if your teacher hasn't covered, there's no sense in giving you this exericse yet
anonymous
  • anonymous
Thank you for the link, I will check it out.
jdoe0001
  • jdoe0001
law of cosines and Heron's formula to get the area then again, you may not have covered heron's formula either you might think the formula has a two long legs, but it doesn't =)
SolomonZelman
  • SolomonZelman
\(b^2=c^2+a^2-2(a)(c){\rm Cos}(B)\) \(b^2=26^2+24^2-2(24)(26){\rm Cos}(33)\) \(b=\sqrt{26^2+24^2-2(24)(26){\rm Cos}(33)}=14.33\)
SolomonZelman
  • SolomonZelman
|dw:1441585266641:dw|
anonymous
  • anonymous
I got 171.96 as the area, but that is not one of the answers either.
SolomonZelman
  • SolomonZelman
I am trying to find every thing in the triangle, perhaps, before doing area.
SolomonZelman
  • SolomonZelman
\(\sin(33)/14.33{~~}={~~}\sin(C)/26\)
SolomonZelman
  • SolomonZelman
|dw:1441585636236:dw|
SolomonZelman
  • SolomonZelman
So the height is less that 14, but not much less than 14.33. We can guess it is 12, then it is approaches choice A.
SolomonZelman
  • SolomonZelman
(That is if you want to just approximate the area. But we can try to find the height.... which is a lot of sweating)
anonymous
  • anonymous
Okay, so is this Algebra, it seems more like Geometry to me?
anonymous
  • anonymous
Thank you guys so much. Have a great day. :)

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