What is the Cartesian form of $(99,(17pi )/12)$ ?

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Answer 1 · 2017-03-19T18:10:24

$(-95.63, -25.62)$

polar coordinates $(r, theta) = (99, (17pi)/12)$ is in quadrant IV

rectangular coordinates ( $rsin theta, r cos theta)$

Make sure your calculator MODE is in Radians:

$99 sin (17pi/12) ~~-95.6266568$

$99 cos (17pi/12) ~~ -25.62308547$

Rectangular coordinates: $(-95.63, -25.62)$

Answer 2 · 2017-03-19T18:33:46

$-(99 (sqrt3 -1))/(2sqrt2),-(99 (sqrt3 +1))/(2sqrt2)$

In polar coordinates it is $r=99 and theta=(17pi)/12$ . In cartesean form it would be $x= rcos theta$ and $y = r sin theta$

Accordingly cartesean coordinates would be $x=99 cos ((17pi)/12)=99 cos(pi +(5pi)/12)= -99 cos((5pi)/12)= -99 cos 75^o=-(99 (sqrt3 -1))/(2sqrt2)$

similarly, $y= 99 sin ((17pi)/12)= -99 sin((5pi)/12)= -99 sin 75^o= -(99(sqrt3 +1))/(2sqrt2)$

What is the Cartesian form of (99,(17pi )/12)(99,(17pi )/12)?