@robtobey @jhonyy9 @AkashdeepDeb What is the value of B in this diagram? A.12 B.10.2 C.9 D.8
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i know its the law of Cosines
but im not sure how to substitute the values correctly \[a ^{2}=b^2+c^2-2bc*\cos(A)\]
@ganeshie8
You need to use \[b^2=a^2 + c^2 -2ac * \cos(B)\]
Since you want to find b
okay then what do i sub for \[b^2\]
Nothing, b is the value you are looking for. Once you figure out what equals b^2, take the square root of it to find b.
Use the law of sines to find angle B
so \[b^2=12^2+6.33^2-2(12)(6.33)*\cos(113.17)\]
No, you have to find angle B using the law of sines. You put angle A where angle B should be. Do you know the law of sines?
\[sinA/a=sinB/b=sinC/c\]
Yes. First use it to find angle C: \[12/(\sin113.17)=6.33/sinC\] \[(6.33\sin113.17)/12=sinC\] \[\sin^-1((6.33\sin113.17)/12)=C\] C=29.01°
Now find angle B using angles A and C: 180-(29.01+113.17) =37.82
Then plug that into your formula: \[b^2=12^2+6.33^2-2(12)(6.33)\cos37.82\]
8. But why would I use sines? Cosines is to find any angle or the unknown side length....which is what i need to find, isn't it?
Yes, you're right. Both law of sines and law of cosines are used for finding any angle or unknown side length.
8 is the correct answer.
We use law of sines or law of cosines depending on what information we are given and what we need to find.
ooohhh okay :) i still dont understand it fully, and I have 4 more questions about sines and cosines application
I'm sorry, I'm not very good at explaining things :)
its alright. I am horrible in math, i mean HORRIBLE!!
if i tag you in a few other questions would you mind helping me?
You're not horrible. Math is all about applying the use of formulas and theorems. You just need to keep practicing, learn to use what you're given to deduce a way to find your answer. If you keep working at it, you'll get better. And we here at Open Study will always be around when you're stuck on something :)
I can't, I'm sorry :( I'm just leaving. But open a new question and try tagging ganeshie again and a few others. I'm sure someone can help :) Good luck!
Thank you!!
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