In right ∆ABC, angle B = 90°, angle C = 40°, and angle BC = 10. What are the other two side lengths of the triangle? A) AC = 13.1, AB = 8.4 B) AC = 18.7, AB = 15.8 C) AC = 14.3, AB = 10.2 D) AC = 15.6, AB = 11.9

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In right ∆ABC, angle B = 90°, angle C = 40°, and angle BC = 10. What are the other two side lengths of the triangle? A) AC = 13.1, AB = 8.4 B) AC = 18.7, AB = 15.8 C) AC = 14.3, AB = 10.2 D) AC = 15.6, AB = 11.9

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Could really use the help, don't have to give me the answer outright just need to know how to solve this problem, can someone please guide me in the right direction? thank you.
|dw:1434457799463:dw| \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]
so you need one angle and one side to find the other side of right triangle after that you can apply Pythagorean theorem

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@Nnesha Ahh so I solve it using SOHCAHTOA which is what my teacher taught me awhile back, I haven't done this sort of stuff in forever, might need some more assistance haha
yes alright!
as you can there is one angle and 2 sides in SOHCAHTOA formula i posted above so you need one angle and one side to solve for 3rd variable cos = adjacent over hyp so we to use cos function \[\huge\rm \cos (40) = \frac{ 10 }{ h }\] h = hypotenuse solve for h
|dw:1434458317996:dw| adjacent side that touch the angle
@Nnesha I got D as my answer
how did you get D ?
\[\huge\rm \cos (40) = \frac{ 10 }{ h }\] solve for h and it's 40 degree
do you know how to solve for h ??
@Nnesha Yea, I ended up getting 7.66
what is cos(40) make sure mode is 'n degree
@Nnesha I got cos(40)=0.76604=10/h then I rearranged that, then multiplied 0.7660 x 10 and that was what I got, im trying to get an idea from this
why did multiply ??
you*
@Nnesha https://www.mathsisfun.com/algebra/trig-finding-side-right-triangle.html
what about it ?
\[\huge\rm .766=\frac{ 10 }{ h }\] you should multiply both sides by H not by 10
\[\huge\rm .766 \times h = \frac{ 10 }{\cancel h} \times \cancel{h}\]
@Nnesha Then divide both sides by .766? I ended up getting 13.054 or 13.1
yes right
right |dw:1434460279833:dw| now you can apply Pythagorean theorem to find 3rd side of right triangle \[\huge\rm a^2 +b^2 =c^2\] plug in values and solve for 3rd side remember c is the longest side of right triangle which is hypotenuse
alright I did 10^2+b^2=13.1^2 then got 100+b^2=171.61 then subtracted 100 from both sides got b^2=71.61, I squared both sides and got B=8.4622
yes right
b^2 = 71.61 so b=what ?
8.4622
right
so its A, thank you so much haha, idk if you would be willing to help me with one more problem considering how long it took awhile to complete this haha
i'll try my best :-)
@Nnesha haha ok thank you, I appreciate it In ∆ABC, angle B is a right angle and tan A = 13/16 . What is angle C ?
what's the definition of tan ?? look at my first comment tan equal what ?
opposite over adjacent
yes now draw a right triangle
@Nnesha Alright done
|dw:1434461030870:dw| which side is 13 and which one is 16 ?
opposite should be 13 and adjacent should be 16?
yes right u should write 13 and 16 ?
@Nnesha It looks funny to me, the adjacent side is longer than the opposite side?
that doesn't matter hypotenuse should be bigger than other 2 sides doesn't matter a is longer or b
which side is opposite of angle A ?
@Nnesha the side that's 13
yes so |dw:1434461354773:dw|where u should write 13 ? and where suppose to be 16 ?
between A and B would be the 13 and between B and C would be the 16
|dw:1434461430223:dw| yep right now tanC = opposite over adjacent substitute values solve for C
@Nnesha Im having difficulty understanding what to do >_< would I multiply both sides by one of the numbers or?
did u substitute values of opposite and adjacent side let me know u get after that
Ok yea I just push in Tan(C)=13/16
well wait we did it wrong
tan A = 13/16 not tan C
so it should be like this |dw:1434462085204:dw|
@Nnesha Ahhh ok, I knew something was off, so then Tan(C)=16/13
now tan = opposite over adjacent \[\huge\rm tan (C)= \frac{ 16 }{ 13 }\] yes that's right
so u need to find invs of tan \[\huge\rm C = \tan^{-1} \frac{ 16 }{ 13 }\] use calculator
Alright, I got 0.8884
make sure mode 'n degree
@Nnesha Yea, it keeps giving me the same answer for some reason C=0.88847979
bec mode of calculator is on rad so it should be on degree
hit rad and then type that
http://prntscr.com/7hm54j
I got this now C=50.906141113771
looks good
@Nnesha Thank you for everything, I really appreciate it
my pleasure

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