anonymous
  • anonymous
yep
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Nnesha
  • Nnesha
http://prntscr.com/7dmxnk what quadrant ?
anonymous
  • anonymous

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anonymous
  • anonymous
does this help
Nnesha
  • Nnesha
yes you forgot 3 there
anonymous
  • anonymous
my bad
Nnesha
  • Nnesha
|dw:1433549520753:dw| draw a right triangle in 3rd quadrant sin = opposite over hyp now apply Pythagorean theorem to find 3rd side of right triangle a^2 + b^2 = c^2
anonymous
  • anonymous
i don't get it sorry
Nnesha
  • Nnesha
given is sin = -2/3 sin = opposite over hypotenuse|dw:1433549857252:dw| that's how we have to draw a right triangle in 3rd quadrant \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]
Nnesha
  • Nnesha
-2 is opposite of angle a and longest side is hyp so that's how \[\huge\rm sin a = -\frac{ 2 }{ 3 }=\frac{ opposite }{ hypotenuse }\]
anonymous
  • anonymous
so cos =-?/3 and tan =-2/?
Nnesha
  • Nnesha
that's why first you should find 3rd side of right triangle a^2 +b^2 =c^2 substitute value solve for 3rd side
anonymous
  • anonymous
So -2^2+-3^2=13^2
Nnesha
  • Nnesha
hypotenuse is the longest side of right triangle which is c in the formula
Nnesha
  • Nnesha
c=3 a=-2 solve for b
anonymous
  • anonymous
b=5
Nnesha
  • Nnesha
\[\huge\rm (-2)^2 + b^2 = 3^2 \] \[\huge\rm b^2 = 9 -4\] b^2 is 5 what's b ?
anonymous
  • anonymous
(-2)^2+5^=3^
anonymous
  • anonymous
3^2
Nnesha
  • Nnesha
what did you substitute 5 ? b square = 5 not b b^2 = 5 how would you solve for b ? how would u cancel square ?
anonymous
  • anonymous
so sin=-2/3 cos -5/3 tan -2/5
Nnesha
  • Nnesha
\[\huge\rm (-2)^2 + b^2 = 3^2 \] \[\huge\rm b^2 = 9 -4\] \[\huge\rm b^2 = 5\] b square is 5 not b take square root both sides to cancel out square
anonymous
  • anonymous
so 20??
Nnesha
  • Nnesha
reread
Nnesha
  • Nnesha
square rO_Ot not square
anonymous
  • anonymous
i don't get sorry
Nnesha
  • Nnesha
\[\huge\rm x^2 = 4\] solve for x
anonymous
  • anonymous
-2
Nnesha
  • Nnesha
how did you get -2 ?
anonymous
  • anonymous
x^2-4=2
Nnesha
  • Nnesha
aww @Loser66 is here :-) can you help i have to go 4 prayer :3
Nnesha
  • Nnesha
you should take square root both sides \[\huge\rm \sqrt{ x^2} =\sqrt{ 4}\]
Nnesha
  • Nnesha
\[\huge\rm (-2)^2 + b^2 = 3^2 \] \[\huge\rm b^2 = 9 -4\] \[\huge\rm \sqrt{ b^2} = \sqrt{5}\] same here take square root both sides
Nnesha
  • Nnesha
alright cya and remember if you get radical sign at the bottom then multiply both top and bottom by it's conjugate :3
anonymous
  • anonymous
thanks then what
anonymous
  • anonymous
Nnesha
  • Nnesha
so cos =-?/3 and tan =-2/? so-cos-sqrt{5}/3 it's in 3rd quadrant so cos and sin values are negative tan = opposite over adjacent -2/sqrt{5} multiply top and bottom by sqrt{5} then solve
anonymous
  • anonymous
-10/25
anonymous
  • anonymous
Nnesha
  • Nnesha
you can't multiply square root with constant term
anonymous
  • anonymous
what
Nnesha
  • Nnesha
\[\frac{ -2 }{ \sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = ???\]
Nnesha
  • Nnesha
-2 times sqrt{2} is not equal to 10
anonymous
  • anonymous
-2.828
Nnesha
  • Nnesha
exact form not a decimal
anonymous
  • anonymous
-2 sqrt 2
Nnesha
  • Nnesha
-2 times sqrt{5} =-2sqrt{5} solve denominator
Nnesha
  • Nnesha
how did you get sqrt{2} ??
anonymous
  • anonymous
-2 √2
Nnesha
  • Nnesha
how did you get that ?
anonymous
  • anonymous
-2 sqrt 2= -2 √2
Nnesha
  • Nnesha
my question is how did you get sqrt{2} ?
anonymous
  • anonymous
u said it
anonymous
  • anonymous
can we move one to the question please
Nnesha
  • Nnesha
http://prntscr.com/7dnps2
anonymous
  • anonymous
25
Nnesha
  • Nnesha
-2 times sqrt{5} = -2sqrt{5} sqrt{5} times sqrt{5} = ??
Nnesha
  • Nnesha
sqrt{5} times sqrt{5} not equal to 25
anonymous
  • anonymous
5
Nnesha
  • Nnesha
yes right
anonymous
  • anonymous
then what
Nnesha
  • Nnesha
that's tan
Nnesha
  • Nnesha
csc = ? identity
anonymous
  • anonymous
5/3?
Nnesha
  • Nnesha
remember we are in3rd quadrant where tan is positive
Nnesha
  • Nnesha
tan =2sqrt{5}/5
Nnesha
  • Nnesha
csc =1/sin replace sin by its value multiply top with the reciprocal of the bottom
anonymous
  • anonymous
2/3?
Nnesha
  • Nnesha
1/sin sin =-2/3 substitute
anonymous
  • anonymous
1/(-2/3) @Nnesha
anonymous
  • anonymous
3/-2/3
Nnesha
  • Nnesha
\[\huge\rm \frac{ a }{ \frac{ b }{ c } } = a \times \frac{ c }{ b }\] like this !!!
anonymous
  • anonymous
1x-2/3
Nnesha
  • Nnesha
\[\huge\rm \frac{ a }{ \frac{ b }{ c } } = a \times \frac{ c }{ b }\] look at this

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