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

- anonymous

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- schrodinger

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- 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

|dw:1434457799463:dw|
\[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]

- 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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## More answers

- 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

yes alright!

- 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

|dw:1434458317996:dw|
adjacent side that touch the angle

- anonymous

@Nnesha I got D as my answer

- Nnesha

how did you get D ?

- Nnesha

\[\huge\rm \cos (40) = \frac{ 10 }{ h }\]
solve for h
and it's 40 degree

- Nnesha

do you know how to solve for h ??

- anonymous

@Nnesha Yea, I ended up getting 7.66

- Nnesha

what is cos(40) make sure mode is 'n degree

- 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

why did multiply ??

- Nnesha

you*

- anonymous

@Nnesha https://www.mathsisfun.com/algebra/trig-finding-side-right-triangle.html

- Nnesha

what about it ?

- Nnesha

\[\huge\rm .766=\frac{ 10 }{ h }\]
you should multiply both sides by H not by 10

- Nnesha

\[\huge\rm .766 \times h = \frac{ 10 }{\cancel h} \times \cancel{h}\]

- anonymous

@Nnesha Then divide both sides by .766? I ended up getting 13.054 or 13.1

- Nnesha

yes right

- 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

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

yes right

- Nnesha

b^2 = 71.61
so b=what ?

- anonymous

8.4622

- Nnesha

right

- 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

i'll try my best :-)

- 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

what's the definition of tan ?? look at my first comment tan equal what ?

- anonymous

opposite over adjacent

- Nnesha

yes now draw a right triangle

- anonymous

@Nnesha Alright done

- Nnesha

|dw:1434461030870:dw|
which side is 13 and which one is 16 ?

- anonymous

opposite should be 13 and adjacent should be 16?

- Nnesha

yes right u should write 13 and 16 ?

- anonymous

@Nnesha It looks funny to me, the adjacent side is longer than the opposite side?

- Nnesha

that doesn't matter
hypotenuse should be bigger than other 2 sides
doesn't matter a is longer or b

- Nnesha

which side is opposite of angle A ?

- anonymous

@Nnesha the side that's 13

- Nnesha

yes so |dw:1434461354773:dw|where u should write 13 ?
and where suppose to be 16 ?

- anonymous

between A and B would be the 13 and between B and C would be the 16

- Nnesha

|dw:1434461430223:dw|
yep right now
tanC = opposite over adjacent substitute values solve for C

- anonymous

@Nnesha Im having difficulty understanding what to do >_< would I multiply both sides by one of the numbers or?

- Nnesha

did u substitute values of opposite and adjacent side let me know u get after that

- anonymous

Ok yea I just push in Tan(C)=13/16

- Nnesha

well wait we did it wrong

- Nnesha

tan A = 13/16
not tan C

- Nnesha

so it should be like this |dw:1434462085204:dw|

- anonymous

@Nnesha Ahhh ok, I knew something was off, so then Tan(C)=16/13

- Nnesha

now tan = opposite over adjacent \[\huge\rm tan (C)= \frac{ 16 }{ 13 }\]
yes that's right

- Nnesha

so u need to find invs of tan \[\huge\rm C = \tan^{-1} \frac{ 16 }{ 13 }\]
use calculator

- anonymous

Alright, I got 0.8884

- Nnesha

make sure mode 'n degree

- anonymous

@Nnesha Yea, it keeps giving me the same answer for some reason C=0.88847979

- Nnesha

bec mode of calculator is on rad
so it should be on degree

- Nnesha

hit rad and then type that

- Nnesha

http://prntscr.com/7hm54j

- anonymous

I got this now C=50.906141113771

- anonymous

@Nnesha

- Nnesha

looks good

- anonymous

@Nnesha Thank you for everything, I really appreciate it

- Nnesha

my pleasure

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