Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
OpenStudy (anonymous):
I want mine in silver tho
OpenStudy (anonymous):
idk how to give silver medals
OpenStudy (anonymous):
There are only gold medals, that's why I'm always sad
OpenStudy (anonymous):
oh
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
do you know how to use sin and cos?
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
This is much easier with 30-60-90 triangle rules and 45-45-90 triangle rules
OpenStudy (anonymous):
I get weird answers with those rules :(
OpenStudy (anonymous):
ok so were going to assume a=12 because algebra is easier than numbers lol. so to find z, we need the angle between a and z. do you know what that angle is?
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
30 degrees
OpenStudy (anonymous):
\[\large \sf y\sqrt{3}=12\]\[\large \sf y=\frac{12}{\sqrt{3}}\] \[\large \sf z=2y\] Ez applications
OpenStudy (anonymous):
so z would be 2 times\[12\sqrt{3}\]
OpenStudy (anonymous):
Well \(\large \sf 2\frac{12}{\sqrt{3}}\)
OpenStudy (anonymous):
the easiest method would be z*cos30=a where z = a / cos30 = 12/ cos30 or z= a/sin60
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
I'm so confused
OpenStudy (anonymous):
this based on the relationships with right triangles and the unit circle
OpenStudy (anonymous):
have you learned the unit circle yet?
OpenStudy (anonymous):
or trig cirle. forget what its called
OpenStudy (anonymous):
yes I have, but I barely understood it
Still Need Help?
Join the QuestionCove community and study together with friends!