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

- anonymous

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- SolomonZelman

This is a right triangle?

- anonymous

I think so.

- SolomonZelman

Ok, then you can use the Pythagorean theorem to find the side b.

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## More answers

- anonymous

I would do that, but it is a degree instead of a number.

- SolomonZelman

I am saying to use the `. a² + b² = c² . `

- SolomonZelman

24²+b²=26²

- anonymous

Okay. Thank you so much!

- SolomonZelman

we aren'tdone yet, are we?

- SolomonZelman

Did you find the side b?

- SolomonZelman

(It will be a precise, natural number, answer)

- anonymous

I'm working on it.

- SolomonZelman

Ok, sure, take your time:)

- anonymous

I got 10.

- SolomonZelman

Oh, very good. that is right!

- SolomonZelman

|dw:1441584583951:dw|

- SolomonZelman

So you can already tell which choice is closer to the area.

- anonymous

24 right?

- jdoe0001

|dw:1441585425017:dw|

- SolomonZelman

I wonder that they do not have the answer choice for that....

- anonymous

So, the test makers didn't put the correct choice on it?

- SolomonZelman

No, I think it is rather our mistake that we assumed that your triangle is a right triangle.

- jdoe0001

hmmm

- anonymous

Okay, so how would we solve this now?

- SolomonZelman

I think we need to apply the law of cosines.
\(b^2=a^2+c^2-2ac{\rm Cos}(B)\)

- jdoe0001

have you covered the law of cosines yet?

- anonymous

Not yet, but this is supposed to be a Algebra 1 study.

- SolomonZelman

https://www.mathsisfun.com/algebra/trig-cosine-law.html

- jdoe0001

hmmm so.. haemm
I think is safe to assume then, that is a right triangle

- SolomonZelman

This is a good link. mathisfun is always easy to read:)

- SolomonZelman

yes jdoe, I assumed that the first I encountered it, but there is no option of 120.

- 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

Thank you for the link, I will check it out.

- 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

\(b^2=c^2+a^2-2(a)(c){\rm Cos}(B)\)
\(b^2=26^2+24^2-2(24)(26){\rm Cos}(33)\)
\(b=\sqrt{26^2+24^2-2(24)(26){\rm Cos}(33)}=14.33\)

- SolomonZelman

|dw:1441585266641:dw|

- anonymous

I got 171.96 as the area, but that is not one of the answers either.

- SolomonZelman

I am trying to find every thing in the triangle, perhaps, before doing area.

- SolomonZelman

\(\sin(33)/14.33{~~}={~~}\sin(C)/26\)

- SolomonZelman

|dw:1441585636236:dw|

- 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

(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

Okay, so is this Algebra, it seems more like Geometry to me?

- anonymous

Thank you guys so much. Have a great day. :)

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