## anonymous one year ago 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

1. SolomonZelman

This is a right triangle?

2. anonymous

I think so.

3. SolomonZelman

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

4. anonymous

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

5. SolomonZelman

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

6. SolomonZelman

24²+b²=26²

7. anonymous

Okay. Thank you so much!

8. SolomonZelman

we aren'tdone yet, are we?

9. SolomonZelman

Did you find the side b?

10. SolomonZelman

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

11. anonymous

I'm working on it.

12. SolomonZelman

Ok, sure, take your time:)

13. anonymous

I got 10.

14. SolomonZelman

Oh, very good. that is right!

15. SolomonZelman

|dw:1441584583951:dw|

16. SolomonZelman

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

17. anonymous

24 right?

18. jdoe0001

|dw:1441585425017:dw|

19. SolomonZelman

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

20. anonymous

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

21. SolomonZelman

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

22. jdoe0001

hmmm

23. anonymous

Okay, so how would we solve this now?

24. SolomonZelman

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

25. jdoe0001

have you covered the law of cosines yet?

26. anonymous

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

27. SolomonZelman
28. jdoe0001

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

29. SolomonZelman

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

30. SolomonZelman

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

31. 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

32. anonymous

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

33. 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 =)

34. SolomonZelman

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

35. SolomonZelman

|dw:1441585266641:dw|

36. anonymous

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

37. SolomonZelman

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

38. SolomonZelman

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

39. SolomonZelman

|dw:1441585636236:dw|

40. 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.

41. 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)

42. anonymous

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

43. anonymous

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