Question

in triangle RST m 13 years ago

Answer 1

Well... this is trickier... maybe

Answer 2

This calls for the Law of Cosines, instead

Answer 3

Anytime you're ready @katherineregister :)

Answer 4

oh yeah sorry computer is actng up ughhh any way so what is the law of cosines

Answer 5

On second thought, never mind
this is still Law of Sines

Answer 6

its where you have to find the T to geet the S isnt it

Answer 7

Yes.

Answer 8

Similar principles...
But we do need angle T.

Once again, we have a determining pair... which is it?

Answer 9

oh god i hate these i realy cant do them i dont know where to even begin( besides the triangle)

Answer 10

r and
R

Answer 11

and relevant would be T and t

Answer 12

Yes :)
Relevant for now, okay?
In the end, what we're really looking for is angle S.

Since it's an angle we're looking for, we'll use the incarnation of the Law of Sines with the sines on top (as opposed to earlier, where the sides were on top)

As per the law of Sines, we have...
\[\huge \frac{\sin(R)}{r}=\frac{\sin(T)}{t}\]

Answer 13

Catch me so far?

Answer 14

yes

Answer 15

Well, plug away :)
and solve for sin(T)

Answer 16

Yup :)
You're doing great...

Answer 17

so i change it to

Answer 18

right?

Answer 19

No.
A bit off there...
don't remove important details...

Answer 20

No rush, Kat... take your time
But for now, we're solving for sin(T)
and not yet T
but don't worry, we'll get there...

Answer 21

thats what i meant lol

Answer 22

Your instructor probably won't see it that way.
Better be sure :)
so... in the end...
\[\Large \sin(T) = ?\]

Answer 23

.5908= sin T

Answer 24

That's good :D
Now use that trusty calculator again, use that \(\Large \sin^{-1}\) function

Answer 25

\[\Large 0.5908 = \sin(T)\]\[\Large \sin^{-1}(0.5908) = T\]

Answer 26

36.2137

Answer 27

Rounded off to the nearest degree? ;)

Answer 28

and then i use that and use the equation 180- ( R+T) correct

Answer 29

Oh, right
Yes :)

Answer 30

Glad you caught that, I got lost :3

I forgot it was S we were looking for


derrrp

Answer 31

lol i got 44degrees that is rounded

Answer 32

i gtg

Answer 33

Nicely done :)

Answer 34

Oh sure :)

But great work on the law of sines there :)

Answer 35

thank you i mught need your help with this tommorow and i think its law of cosines tommorow

Answer 36

Okay, you wanna heads-up?

Answer 37

The law of cosines is basically an extension of the Pythagorean theorem...
It's got a trickier formula, but it's a lot easier to use, you normally just plug in right away...
The Law of Cosines states...
\[\huge c^2 = a^2 + b^2 \color{red}{-2ab \ \cos(C)}\]
Or, if it's an angle you're looking for..
\[\huge \color{blue}{\cos(C) = \frac{a^2 + b^2 -c^2}{2ab}}\]