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

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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

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This is a right triangle?
I think so.
Ok, then you can use the Pythagorean theorem to find the side b.

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

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