solve for angle B. b=9, a=6, c=4
any diagram?
the formula for the law of consines in this question is \[\cos(B) = \frac{a^2 +c^2 - b^2}{2ac}\] sorry about the mistake before
i got \[Cos(B)=a ^{2}+7^{2}+9^{2}/2(a)(9)\]
it should be \[\cos(B) = \frac{6^2 + 4^2 - 9^2}{2 \times 6 \times4}\] and looking at the diagram, it will be the largest angle, since its opposite the longest side
a is not 6, you don't know a from what i'm seeing
so when you get the answer... the next step is to find the angle... so it would be \[B = \cos^{-1}(answer)..... or.......B = \arccos(answer)\] depending on what you use to find an angle
opposite side of angle A is 6, a=6
i was looking at the wrong problem, yes a is 6 but i'm still confused haha i don't get this at all
campbell showed you the law of cosine, you just have to replace the variables
well the law of cosines is a standard formula... the question is just about identifying the information and substituting it into the correct place,
i got -29/48=cos(B)
you'll need to use brackets my method for input into a calculator is \[(6^2 + 4^2 - 9^2) \div (2 \times 4 \times 6)\] then you'll get a negative fraction answer...
great so on your calculator press shift or 2nd then cos then ANS close the brackets press equals and you should get the answer.
then ANS or answer keay just means you don't need to input the fraction again.
i don't think i'm right.. i got 0.82298
not quite so what sort of calculator do you have...?
i forgot my calculator at school so at the moment i'm using mathway.com
ok... hold on this online calculator will work. http://web2.0calc.com/ press 2nd then acos then +- 29/48 close brackets... and lastly the equals
127.168899655999
then to round that to the nearest 10th it would be 127.2 right?
yep thats great... well done
thank you so much! these take home tests are so hard for me
well remember if it take home... work with the formulas next to you... good luck
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