Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (anonymous):

tan theta = -4/3 then sin theta =?

OpenStudy (immanuelv):

tan theta = sin theta / cos theta

OpenStudy (anonymous):

yes i found sin theta = -4/3 cos theta

OpenStudy (immanuelv):

yes !

OpenStudy (anonymous):

do u want options

OpenStudy (immanuelv):

what?

OpenStudy (anonymous):

a) -4/5 but not 4/5 b) -4/5 or 4/5 c) 4/5 but not -4/5 d) none of these

OpenStudy (anonymous):

\[sin(\theta)=\pm\frac{4}{5}\]

OpenStudy (anonymous):

option b if you have no further information

OpenStudy (anonymous):

how pls elaborate

OpenStudy (anonymous):

ok if you draw a right triangle and think of one angle as theta, then you can visualize tangent as "opposite over adjacent" label the opposite side 4, the adjacent side 3 and the hypotenuse will be 5 by pythagoras

OpenStudy (anonymous):

CAN U TELL ME WITH THE HELP OF FORMULA

OpenStudy (anonymous):

ok we have one adjacent side 3 and opposite side 4 yes? then pythagoras tells us \[a^2+b^2=c^2\] where the sides are a and b and c is the hypotenuse. \[3^2+4^2=9+16=25\] so the hypotenuse is 5

OpenStudy (anonymous):

if you think of sine as opposite over hypotenuse then you know \[\sin(x)=\frac{4}{5}\] but you do not know what quadrant you are in so you do not know if sine is positive or negative

OpenStudy (anonymous):

I UNDERSTOOD VERY VERY VERY THANKS TO U

OpenStudy (anonymous):

BUT The ratio of the radii of two circles at the centers of which two arcs of the same length subtended a angles of 60 degree and 70 degree is

OpenStudy (anonymous):

welcome

OpenStudy (anonymous):

CAN U ANSWER ME THIS QUESTION ALSO The ratio of the radii of two circles at the centers of which two arcs of the same length subtended a angles of 60 degree and 70 degree is

OpenStudy (anonymous):

actually no because i do not know what the question means

OpenStudy (anonymous):

OK ANYWAY THANKS

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!