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What is the polar form of #( 36,48 )#?

Answer 1

Emily Ammerman

Polar form of #(36,48)# is #(60,53.13^o)#

Polar coordinates #(r,theta)# are related to Cartesian coordinates #(x,y)# by the following:

#x=rcostheta#, #y=rsintheta# and #r^2=x^2+y^2#

Hence, let polar coordinates of #(36,48)# be #(r,theta)#

Hence #r=sqrt(36^2+48^2)=sqrt(1296+2304)=sqrt3600=60#

#costheta=36/60=0.60# and #sintheta=48/60=0.80#

and from tables #theta=53.13^o#

Hence, polar form of #(36,48)# is #(60,53.13^o)#

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Answer 2

Paisley Johnson

To find the polar form of the point ( (36, 48) ), we use the formulas:

[ r = \sqrt{x^2 + y^2} ] [ \theta = \arctan\left(\frac{y}{x}\right) ]

where ( r ) represents the distance from the origin to the point, and ( \theta ) represents the angle between the positive x-axis and the line connecting the origin to the point.

Given ( (x, y) = (36, 48) ), we calculate:

[ r = \sqrt{36^2 + 48^2} = \sqrt{1296 + 2304} = \sqrt{3600} = 60 ]

[ \theta = \arctan\left(\frac{48}{36}\right) = \arctan\left(\frac{4}{3}\right) ]

Using a calculator to find the arctan value, we get ( \theta \approx 53.13^\circ ) (rounded to two decimal places).

Therefore, the polar form of ( (36, 48) ) is ( (60, 53.13^\circ) ).

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Answer from HIX Tutor

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