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## anonymous one year ago In triangle ABC , m angle A=41 degrees, m angle B=32 degrees and AC= 9 inches what is AB to the nearest tenth of an inch a. 13.1 in b. 13.6 in c. 16.2 in d. 16.9 in

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1. anonymous

|dw:1432846650149:dw| I'm assuming it looks like this?

2. anonymous

i guess? it didnt have a picture on this one

3. johnweldon1993

So this can just be solved with the law of sines You have your triangle *drawn above* and you have the length of 1 other side So we can use $\large \frac{sin(32)}{9} = \frac{sin(107)}{AB}$ And solve for AB

4. anonymous

ok and i solve for AB how? @johnweldon1993

5. johnweldon1993

Wait, first, have you seen this before? The law of sines?

6. anonymous

can you walk me through this? or would that be asking for to much? @johnweldon1993 and no i changed schools and skiped like 3 units so im trying to get all of this

7. johnweldon1993

No never asking too much :) but first, what class is the for? I just dont want to introduce something you wouldnt even be learning :)

8. anonymous

Algebra 2

9. johnweldon1993

Okay then yeah this applies :) Okay so...in general You have a triangle |dw:1432852105182:dw|

10. johnweldon1993

So first we need to solve for angle C We know that the interior angles of a triangle always add to 180 So since we have 41...and 32...we have 73 of those 180 degrees So angle C must be the remaining 107 degrees right?

11. anonymous

ok...

12. johnweldon1993

does that make sense? Gotta make sure you're staying with me :)

13. anonymous

yes im with you

14. johnweldon1993

So now, we apply the law of sines What THAT states...is: $\large \frac{sin(A)}{a} = \frac{sin(B)}{b} = \frac{sin(C)}{c}$ Where A,B and C are the angle measurements we have and a,b and c are the side lengths we have

15. johnweldon1993

Now the way we use it....is if we have angle A, then side "A" would be the side the angle faces |dw:1432852557818:dw|

16. anonymous

that's familiar

17. johnweldon1993

So...since you have side B right...here it would be AC |dw:1432852698148:dw| And you want to solve for side C which here would be AB |dw:1432852734405:dw| we can write $\large \frac{sin(\text{Angle of B)}}{\text{Length of side B = 9}} = \frac{sin(\text{angle of C)}}{\text{length of side C = ???}}$ $\large \frac{sin(32)}{9} = \frac{sin(107)}{AB}$

18. johnweldon1993

And then we just solve for AB Now to do that...we just cross multiply $\large ABsin(32) = 9sin(107)$ And divide both sides by the sin(32) so $\large AB = \frac{9sin(107)}{sin(32)}$

19. anonymous

ok.. um sorry its taking a while for me to catch on

20. johnweldon1993

No that's fine :) just tell me if you want me to explain a part over :)

21. anonymous

it was pretty explanatory just takes a lot of practice thank you so much!

22. johnweldon1993

Of course :)

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