Find area of the triangle?
http://puu.sh/3dKTt.jpg

- anonymous

Find area of the triangle?
http://puu.sh/3dKTt.jpg

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- anonymous

It's easy! You use 1/2 * SideAB * SideAC * sen79º

- primeralph

|dw:1371003691620:dw|

- anonymous

@zairhenrique so it's 601.25?

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## More answers

- anonymous

|dw:1371003684934:dw|

- anonymous

Yes! 601.25

- anonymous

mm²

- anonymous

got it! easy. can you help with another? I have to find the side measurement:
http://puu.sh/3dL0o.png
still laws of cosine

- anonymous

laws of cosine? i think it's laws of sin.

- anonymous

i'm learning laws of cosine now. it's probably laws of sin, this is on a review for the entire unit

- anonymous

It's laws of sin. See it:

- anonymous

|dw:1371003900445:dw|

- anonymous

31/sin38º = b/sin74º
b = 31*sin74º/sin38º

- anonymous

b = 48.4
48.4/74 = 50.35

- anonymous

no, no... b is the side.

- anonymous

well, 48.4/sin74

- anonymous

you want b.

- anonymous

just stop kk , its 48.4 rs

- anonymous

lol thank you.

- anonymous

http://puu.sh/3dLcu.png

- anonymous

You have to find the angle c, and after that, you will find the angle b.
So, 13/sin40º = 19/sinc
sinc = 19*sin40º/13

- anonymous

When you discover the angle c, just do 180 - 40º - c, and you will have the angle b :)

- anonymous

angle c is 69.96 right? i did your last equation 19 x sin40/13 then did arcsin with that answer and got 69.96

- anonymous

Yep, c is rounded to 70º.

- anonymous

got it :)

- anonymous

the answer is 70º. 180 - 40º - 70º = 70º

- anonymous

b = c

- anonymous

supposed to use laws of cosine for this one:
http://puu.sh/3dLmF.png

- anonymous

yep!

- anonymous

Do you know how to do that?

- anonymous

the only law of cosine that was included in this lesson was
cos C = a^2 + b^2 + c^2/2ab

- anonymous

Use that one:
\[c²=a²+b²-2ab \cos \Theta \]
|dw:1371004510363:dw|

- anonymous

In the picture that you posted, you have a, b, and theta, you just have to use the formula to find c :]

- anonymous

c = 169.26?

- anonymous

yep

- anonymous

rounded up to 169.3

- anonymous

how would i use laws of cosine to find an angle

- anonymous

Use my last draw and the equation will be:
\[\cos \theta = \frac{ a² + b² - c² }{ 2ab }\]

- anonymous

and when you have the value of cos theta, just use arccos

- anonymous

so for this
http://puu.sh/3dLNW.png
i'd do 17^2 + 22^2 - 30^2, then divide by 2 x 17 x 22, then do arccos on that number

- anonymous

in this case, the 'c' of my formula is the side BC, of 17^2

- anonymous

so, the correct is 30^2+22^2-17^2 then divide by 2*30*22

- anonymous

okay gotcha

- anonymous

the 'c' in my formula is the side opposite to the angle that you want.

- anonymous

okay makes sense

- anonymous

are you on the highschool ?

- anonymous

yep!
for tat answer i got 33.9

- anonymous

perfect

- anonymous

next one: http://puu.sh/3dLYk.png
i'll do a^2 + b^2 - c^2, but it'll be more like this:
11^2 + 17^2 - 12^2, because b is across from the angle i'm solving for, right?

- anonymous

yes rs

- anonymous

dont forget to divide by 2*11*17

- anonymous

gotcha

- anonymous

my answer is 44.77 for that one

- anonymous

thats it

- anonymous

next one is: http://puu.sh/3dM9b.png
j will be my c, so
7^2 + 6.58^2 - 10^2 / 2 x 7 x 6.58, then that answer for arccos

- anonymous

yep

- anonymous

i feel like this is much easier than laws of sine hahah

- anonymous

it's really much easier because there's a formula ready to give you the cos value rs

- anonymous

exactly! for that one i got 94.59 which rounds up to 95

- anonymous

thats right

- anonymous

for this one: http://puu.sh/3dMl7.png
use the formula you gave me earlier: c^2 = a^2 + b^2 - 2ab cos theta

- anonymous

yep

- anonymous

9.5

- anonymous

9.5

- anonymous

;)

- anonymous

you're the most helpful person i've worked with on openstudy. thank you so much.

- anonymous

you're welcome :] i really like math so, it's fun for me being here rs

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