Ask your own question, for FREE!
Mathematics 9 Online
OpenStudy (anonymous):

Find the m angle T answers: 40.5° 38.5° 139.5° 124.5°

OpenStudy (anonymous):

OpenStudy (anonymous):

your using the law of cosine right?

jimthompson5910 (jim_thompson5910):

you are correct

jimthompson5910 (jim_thompson5910):

so what do you get when you do?

OpenStudy (anonymous):

gotta work it out

jimthompson5910 (jim_thompson5910):

alright

OpenStudy (anonymous):

I don't know how I would put it

OpenStudy (anonymous):

t^2 = s^2 + r ^2 ?

jimthompson5910 (jim_thompson5910):

the law of cosines is c^2 = a^2 + b^2 - 2*a*b*cos(C) so in this case, we would say t^2 = r^2 + s^2 - 2*r*s*cos(T)

jimthompson5910 (jim_thompson5910):

then plug everything in to get t^2 = r^2 + s^2 - 2*r*s*cos(T) 16^2 = 24^2 + 22^2 - 2*24*22*cos(T) from here, you solve for T

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

so I don't do anything with the cos (T) ?

jimthompson5910 (jim_thompson5910):

think of it as its own variable you solve for cos(T) first, then you further solve for T once you have isolated it

jimthompson5910 (jim_thompson5910):

so if you really want to, you can replace cos(T) with x that would mean 16^2 = 24^2 + 22^2 - 2*24*22*cos(T) turns into 16^2 = 24^2 + 22^2 - 2*24*22*x

jimthompson5910 (jim_thompson5910):

then you would solve 16^2 = 24^2 + 22^2 - 2*24*22*x for x like normal after you have isolated x, you replace x with cos(T) finally, you would solve for T using the arccosine function

OpenStudy (anonymous):

this is what I have so far 256 = 576 + 484 - 1056

jimthompson5910 (jim_thompson5910):

so it would be 256 = 576 + 484 - 1056*cos(T) or 256 = 576 + 484 - 1056x if you let x = cos(T)

OpenStudy (anonymous):

okay then add 576 + 484 ?

jimthompson5910 (jim_thompson5910):

yep

OpenStudy (anonymous):

then do I subtract from 256 ?

jimthompson5910 (jim_thompson5910):

correct

OpenStudy (anonymous):

from the 1060 ?

OpenStudy (anonymous):

or the 1056 x ?

jimthompson5910 (jim_thompson5910):

256 = 576 + 484 - 1056*cos(T) 256 = 1060 - 1056*cos(T) 256 - 1060 = -1056*cos(T) -804 = -1056*cos(T)

jimthompson5910 (jim_thompson5910):

on line 3, we are subtracting 1060 from both sides because you can't combine it with the 1056*cos(T) term

OpenStudy (anonymous):

oh right okay okay

OpenStudy (anonymous):

then ill divide it

jimthompson5910 (jim_thompson5910):

both sides by what

OpenStudy (anonymous):

-804?

jimthompson5910 (jim_thompson5910):

what do you get when you do that

OpenStudy (anonymous):

.7613

jimthompson5910 (jim_thompson5910):

so you now have 0.7613 = cos(T)

jimthompson5910 (jim_thompson5910):

or cos(T) = 0.7613

OpenStudy (anonymous):

ohh then solve for the rest of it ?

jimthompson5910 (jim_thompson5910):

yep, so you use the arccosine to undo the cosine

jimthompson5910 (jim_thompson5910):

arccosine and cosine are like division and multiplication

OpenStudy (anonymous):

ohh okay I guess hah so I don't solve it like I did in the beginning?

jimthompson5910 (jim_thompson5910):

I'm not sure what you mean exactly

OpenStudy (anonymous):

like would I still solve it t^2 = r^2 + s^2 since I have to find the angle of T

jimthompson5910 (jim_thompson5910):

oh no that only applies if you wanted to find the side lengths (not angles) and if this was a right triangle

OpenStudy (anonymous):

oh okay! so then how would I solve this one?

jimthompson5910 (jim_thompson5910):

cos(T) = 0.7613 arccos( cos(T) ) = arccos(0.7613) ... apply the arccos function to both sides T = arccos(0.7613) T = ??

OpenStudy (anonymous):

803.9328 ?

jimthompson5910 (jim_thompson5910):

way too high

jimthompson5910 (jim_thompson5910):

do you have a TI calculator?

OpenStudy (anonymous):

yeah I do the scientific one right?

jimthompson5910 (jim_thompson5910):

how are you getting 803 are you hitting "2nd" then "cos" to get arccos?

OpenStudy (anonymous):

no.. I don't know what the arcos is ?

OpenStudy (anonymous):

i multiplied 1056

jimthompson5910 (jim_thompson5910):

what did you type in to get 803 then?

jimthompson5910 (jim_thompson5910):

oh i see

jimthompson5910 (jim_thompson5910):

well you can type "arccos(0.7613) in degrees" into google without quotes to get this https://www.google.com/search?hl=&q=arccos%280.7613%29+in+degrees&sourceid=navclient-ff&rlz=1B3GGLL_enUS420US420&ie=UTF-8

jimthompson5910 (jim_thompson5910):

google will act like a calculator and spit the result out

OpenStudy (anonymous):

ohhhh okay i got that

jimthompson5910 (jim_thompson5910):

btw arccos is often found by hitting the \(\large \cos^{-1}\) key

jimthompson5910 (jim_thompson5910):

oh you did? well that's your final answer that's the measure of angle T

OpenStudy (anonymous):

yeah i typed that and i got 40.4?

jimthompson5910 (jim_thompson5910):

you got it correct, so why did you multiply it by 1056?

OpenStudy (anonymous):

i don't know i thought i had to but i guess not lol

jimthompson5910 (jim_thompson5910):

nope, once you isolate T, you get T = 40.4210626 degrees roughly

OpenStudy (anonymous):

so then it would be 40.5 ?

jimthompson5910 (jim_thompson5910):

it's actually closer to 40.415439021453 so I would go with 40.4 if you're rounding to one decimal places

jimthompson5910 (jim_thompson5910):

place*

OpenStudy (anonymous):

aha ohh but it doesn't give me the choice of that so i figure it would be 40.5

jimthompson5910 (jim_thompson5910):

odd, well there is a bit of roundoff error I guess

jimthompson5910 (jim_thompson5910):

40.5 works if they're using approximations

OpenStudy (anonymous):

i think so. but thank you so much! :)

jimthompson5910 (jim_thompson5910):

sure thing

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!