Question

Need help with precalculus will give owl bucks

Answer 1

For each triangle , determine the indicated value

Answer 2

I know this is Cosine rule, but not sure how to start.

Answer 3

You are right, you need a cosine rule for this. Do you have an idea of what the rule says for cosine?

Answer 4

To solve a triangle is to find the lengths of each of its sides and all its angles. The sine rule
is used when we are given either a) two angles and one side, or b) two sides and a non-included
angle. The cosine rule is used when we are given either a) three sides or b) two sides and the
included angle.

Answer 5

I think its that one

Answer 6

Cosine rule states:
\(a^2=b^2+c^2-2bc*cos(A)\)
\(b^2=a^2+c^2-2ac*cos(B)\)
\(c^2=a^2+b^2-2ab*cos(C)\)

Answer 7

Keep in mind, you are solving for only C. Which is the correct equation to use?

Answer 8

O lol , c^2

Answer 9

Yes.

Answer 10

wait thats what i put

Answer 11

i got confused

Answer 12

You need to use this equation \(c^2=a^2+b^2-2ab*cos(C)\)

Answer 13

ok

Answer 14

Now, plug the a, c, and b and solve for C. Can you do that?

Answer 15

wait wait before we proceed can yo usee if i used the right rule for this one

Answer 16

You are right.

Answer 17

I used thee sine rule

Answer 18

but cosine rule says if you have two sides and inlcuded angle then use cosine rule but i have included angle and i used sine rule

Answer 19

Yes. Because you got the degree and the side for A and a side for b. All you have to do is:
\(\Large\frac{sin(A)}{a}=\frac{sin(B)}{b}\)
\(\Large\frac{sin(70°)}{7}=\frac{sin(B)}{4}\)
Cross multiply and solve for B, don't forget to use inverse sine to solve for B at the end.

Answer 20

Same thing for cosine rule equation you had, don't forget to use inverse cosine to solve for C.

Answer 21

o man im so confused i had asked a question earlier and he did it without the inverse cosine

Answer 22

Ok. I have a good idea. How about you show your work here or write it on paper and take a picture of it.

Answer 23

You take the inverse to both sides in order to solve for C.

Answer 24

Ok thank you i will do it right now. OOO. This one i had to solve for side a

Answer 25

the one that he did not do inveerse

Answer 26

http://openstudy.com/updates/5673466fe4b032ed60dbd48a

Answer 27

\(c^2=a^2+b^2-2ab*cos(C)\)

Plug these in: a=6, b=12, c=17

\(17^2=6^2+12^2-2(6)(12)*cos(C)\)

\(289=180-144*cos(C)\)

Answer 28

Is that what you had?

Answer 29

haha no i am writing them right now but yes.

Answer 30

I have a question for you.

Answer 31

Ok

Answer 32

When you knew that you had to use the cosine rule, did you plug in a,b, and c or you did not attempt this question at all?

Answer 33

No i did not attempt the question at all because i couldnt tell if i did it right, also i just learned this three open study questions ago

Answer 34

Okay, can you show your work here by typing it out?

Answer 35

im stuck

Answer 36

\(\Large-\frac{109}{144}=cos(C)\)

Answer 37

What happens if you take the inverse cos to both sides? Have you learned about this before?