Cosines

[Zfuss12]

I know your not supposed to post to many in a day but any help would be appreciated once again...I have come across a question while doing a review...
In triangle xyz x=8, y=6, and cosZ=.5. The length of z is closest to
(1) 12.2 (2) 10 (3) 148 (4)2

 

[daon]

Are you trying to find a side Z, given the cosine of an angle z? If you are just trying to find z given its cosine:

\(\displaystyle \L cosZ = \frac{1}{2} \Rightarrow Z = cos^{-1}(\frac{1}{2})\\ \Rightarrow Z = 60^o\) (one of those 'special' values you should know.)

With radian measure, Z = 1.0467

 

[Zfuss12]

Mr/Ms/Mrs Daon

Youve been a great help and i really apreciate it!

 

[daon]

Mr/Ms/Mrs Daon

Youve been a great help and i really apreciate it!

I edited my post, please make sure your question was asked correctly.

 

[Zfuss12]

I just have one more question.....how do you get the length of z from there?

 

[Zfuss12]

Yeh im trying to find the length of z from there Im sorry if my question was confusing

 

[daon]

Yeh im trying to find the length of z from there Im sorry if my question was confusing

Okay that changes the question.

You are given a triangle with sides xyz, side x = 8, side y = 6, side z = ?. But the angle across from Z we just found out: its 60^o.

The cosine law says that \(\displaystyle \L z^2 = x^2 + y^2 - 2(xy) cos(Z)\)

(come to think of it, we didn't need the angle )

 

[Zfuss12]

yeh thanks alot law of cosines i am familiar with

 

[daon]

yeh thanks alot law of cosines i am familiar with

So do you understand it from there?

 

[Zfuss12]

yes i got the side to be closest to 12.2