anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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.
Nnesha
  • Nnesha
|dw:1434457799463:dw| \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]
Nnesha
  • Nnesha
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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anonymous
  • anonymous
@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
Nnesha
  • Nnesha
yes alright!
Nnesha
  • Nnesha
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
Nnesha
  • Nnesha
|dw:1434458317996:dw| adjacent side that touch the angle
anonymous
  • anonymous
@Nnesha I got D as my answer
Nnesha
  • Nnesha
how did you get D ?
Nnesha
  • Nnesha
\[\huge\rm \cos (40) = \frac{ 10 }{ h }\] solve for h and it's 40 degree
Nnesha
  • Nnesha
do you know how to solve for h ??
anonymous
  • anonymous
@Nnesha Yea, I ended up getting 7.66
Nnesha
  • Nnesha
what is cos(40) make sure mode is 'n degree
anonymous
  • anonymous
@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
Nnesha
  • Nnesha
why did multiply ??
Nnesha
  • Nnesha
you*
anonymous
  • anonymous
@Nnesha https://www.mathsisfun.com/algebra/trig-finding-side-right-triangle.html
Nnesha
  • Nnesha
what about it ?
Nnesha
  • Nnesha
\[\huge\rm .766=\frac{ 10 }{ h }\] you should multiply both sides by H not by 10
Nnesha
  • Nnesha
\[\huge\rm .766 \times h = \frac{ 10 }{\cancel h} \times \cancel{h}\]
anonymous
  • anonymous
@Nnesha Then divide both sides by .766? I ended up getting 13.054 or 13.1
Nnesha
  • Nnesha
yes right
Nnesha
  • Nnesha
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
anonymous
  • anonymous
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
Nnesha
  • Nnesha
yes right
Nnesha
  • Nnesha
b^2 = 71.61 so b=what ?
anonymous
  • anonymous
8.4622
Nnesha
  • Nnesha
right
anonymous
  • anonymous
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
Nnesha
  • Nnesha
i'll try my best :-)
anonymous
  • anonymous
@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 ?
Nnesha
  • Nnesha
what's the definition of tan ?? look at my first comment tan equal what ?
anonymous
  • anonymous
opposite over adjacent
Nnesha
  • Nnesha
yes now draw a right triangle
anonymous
  • anonymous
@Nnesha Alright done
Nnesha
  • Nnesha
|dw:1434461030870:dw| which side is 13 and which one is 16 ?
anonymous
  • anonymous
opposite should be 13 and adjacent should be 16?
Nnesha
  • Nnesha
yes right u should write 13 and 16 ?
anonymous
  • anonymous
@Nnesha It looks funny to me, the adjacent side is longer than the opposite side?
Nnesha
  • Nnesha
that doesn't matter hypotenuse should be bigger than other 2 sides doesn't matter a is longer or b
Nnesha
  • Nnesha
which side is opposite of angle A ?
anonymous
  • anonymous
@Nnesha the side that's 13
Nnesha
  • Nnesha
yes so |dw:1434461354773:dw|where u should write 13 ? and where suppose to be 16 ?
anonymous
  • anonymous
between A and B would be the 13 and between B and C would be the 16
Nnesha
  • Nnesha
|dw:1434461430223:dw| yep right now tanC = opposite over adjacent substitute values solve for C
anonymous
  • anonymous
@Nnesha Im having difficulty understanding what to do >_< would I multiply both sides by one of the numbers or?
Nnesha
  • Nnesha
did u substitute values of opposite and adjacent side let me know u get after that
anonymous
  • anonymous
Ok yea I just push in Tan(C)=13/16
Nnesha
  • Nnesha
well wait we did it wrong
Nnesha
  • Nnesha
tan A = 13/16 not tan C
Nnesha
  • Nnesha
so it should be like this |dw:1434462085204:dw|
anonymous
  • anonymous
@Nnesha Ahhh ok, I knew something was off, so then Tan(C)=16/13
Nnesha
  • Nnesha
now tan = opposite over adjacent \[\huge\rm tan (C)= \frac{ 16 }{ 13 }\] yes that's right
Nnesha
  • Nnesha
so u need to find invs of tan \[\huge\rm C = \tan^{-1} \frac{ 16 }{ 13 }\] use calculator
anonymous
  • anonymous
Alright, I got 0.8884
Nnesha
  • Nnesha
make sure mode 'n degree
anonymous
  • anonymous
@Nnesha Yea, it keeps giving me the same answer for some reason C=0.88847979
Nnesha
  • Nnesha
bec mode of calculator is on rad so it should be on degree
Nnesha
  • Nnesha
hit rad and then type that
Nnesha
  • Nnesha
http://prntscr.com/7hm54j
anonymous
  • anonymous
I got this now C=50.906141113771
anonymous
  • anonymous
@Nnesha
Nnesha
  • Nnesha
looks good
anonymous
  • anonymous
@Nnesha Thank you for everything, I really appreciate it
Nnesha
  • Nnesha
my pleasure

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