A = 41°, B =61°, c = 19 solve using the law of sines? show steps please so I can actually learn how to do this! Thank you!:) also, please round to the nearest tenth if possible. Thanks again. :)

Question

johnweldon1993 · Answer

Alright so the law of sines

$$\large \frac{a}{\sin A}= \frac{b}{\sin B} =\frac{c}{\sin C}$$

where a , b  and c are the sides of your triangle
and A, B and C are the angles of your triangle

What would we do first? do you know?

anonymous · Answer

Nope. honestly i have no idea |dw:1384820963510:dw|

johnweldon1993 · Answer

Dont worry about it, we'll go through it step by step...

anonymous · Answer

wait, like 41/sinA 61/sinB and 19/sinC?

johnweldon1993 · Answer

So what you have KNOWN so far is

A = 41 degrees
B = 61 degrees
and
c = 19

if we substituted all this into that formula up there we would have

$$\large \frac{a}{\sin 41}= \frac{b}{\sin 61} =\frac{19}{\sin C}$$

What we want to do...is complete 1 of these ratios that we can compare the others to...

anonymous · Answer

oh, okay, thanks. so how do i do that? lol

johnweldon1993 · Answer

you were close...just remember the A, B and C are the angles...and go on the bottom...and the a, b and c are the side lengths and go on top...

johnweldon1993 · Answer

So now...you have 2 of the 3 angles in a triangle right? 

What do the angles of a triangle sum up to?

anonymous · Answer

180, so 41+61= 102. so angle 3 is 78?

johnweldon1993 · Answer

That is correct! so now we have

$$\large \frac{a}{\sin 41}= \frac{b}{\sin 61} =\frac{19}{\sin 78}$$

johnweldon1993 · Answer

Now that you have 1 completed ratio...you can use it to find the missing variables...for example...

johnweldon1993 · Answer

$$\large \frac{b}{\sin 61} =\frac{19}{\sin 78}$$

and solve for b...

anonymous · Answer

and how do I do that?

johnweldon1993 · Answer

Well you want to isolate b right? so multiply both sides of this equation by sin61

$$\large b = \frac{19\times \sin 61}{\sin 78}$$

throw this in your calculator and solve for b...what do you get?

anonymous · Answer

uhmm i got -12.etc etc so i think i did it wrong. i couldnt get the first part of it to go over the second. if that makes sense?

johnweldon1993 · Answer

make sure you calculator is in "degree" mode

anonymous · Answer

uhm my calculator is on my phone and doesnt have options like that.:/

johnweldon1993 · Answer

Even if you turn it to the side? well regardless you get... https://www.google.com/#q=(19*sin(61degrees))%2F(sin(78degrees))

johnweldon1993 · Answer

And remember to round to the nearest tenth so that comes to b = 17

anonymous · Answer

lol no it turns scientific if i do that though. & oh okay thanks. so now what?

johnweldon1993 · Answer

So now you have to solve for a...you have 'b' and sinB....so use the ratio you just solved and use it to solve for 'a' the same way we just did for 'b'

johnweldon1993 · Answer

$$\large \frac{a}{\sin 41} =\frac{17}{\sin 61}$$

and solve for 'a'

anonymous · Answer

so a/sin41*19/sin78?

johnweldon1993 · Answer

Yes you could also do that @PitbullsAreForever ...since every ratio is equal... but remember it is  =  not  *

anonymous · Answer

oh so then i will multiply 78 by 19? or??im lost i think :/

wolf1728 · Answer

It can all be done without using the law of sines:

sin (41°) = hgt / 19
hgt = sin (41°) * 19
hgt = 12.465

side AD² = 19² - 12.465²
side AD = 14.34

tan 78° = 12.465 / side DC
side DC = 12.465 / 4.70
side DC = 2.65

side b = 14.34 + 2.65 = 16.99

side a = 12.465² + 2.65²
side a = 12.74