Question

Real Quick Math question: http://prnt.sc/d87gdd

Answer 1

Remark: Cosine is negative (so the opposite side is negative), but Sine
is positive (so the adjacent side is positive. This means we are in the Q4.
~~~~~~~~~~~~~~~~~~~~~~~~~~~
\(\large {\rm opposite~side}=\displaystyle \sqrt{4^2-(\sqrt{3})^2}= \sqrt{16-9}=\sqrt{7}\)

Answer 2

Oh excuse me, we are in the 3rd quadrant because and \(\sin\theta<0\).

Answer 3

(I missread that) ...
So your opposite side is and \(-\sqrt{7}\),
and the hypotenuse you already know...

Answer 4

16? '-'

Answer 5

you mean \(\sin\theta=16\) ?

Answer 6

that is not possible.

Answer 7

aw dang, i stink at math :(

Answer 8

You have the adjacent side = \(-\sqrt{3}\)
the opposite side = \(-\sqrt{7}\)
the hypotenuse = 4

Answer 9

you should be able to find \(\sin \theta\).

Answer 10

does that mean the answer is c?

Answer 11

i know B is wrong

Answer 12

\(\displaystyle \sin\theta=\frac{-\sqrt{7}}{4}\)

Answer 13

your options are all wrong.

Answer 14

oh wow I see. is there any one i should go with, because i cant put anything else

Answer 15

oh I did it incorrectly.

Answer 16

oh wheww haha

Answer 17

\(\color{black}{\displaystyle \sqrt{4^2-\sqrt{3}^2}=\sqrt{16-3}=\sqrt{13} }\)

Answer 18

So the opposite side is \(-\sqrt{13}\).

Answer 19

Then, the sine is \(\displaystyle \frac{\sqrt{-13}}{16}\).

Answer 20

ok, thanks so much - and thanks for explaining the work

Answer 21

(it was marked incorrect)