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yes alright!

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

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 ??

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

why did multiply ??

you*

what about it ?

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

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

yes right

yes right

b^2 = 71.61
so b=what ?

8.4622

right

i'll try my best :-)

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

opposite over adjacent

yes now draw a right triangle

|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 ?

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 ?

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

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|

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

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

my pleasure