What is the distance between #(1 ,(3 pi)/4 )# and #(2 , (15 pi )/8 )#?

1 Answer
Dec 2, 2016

#d~~2.95#

Explanation:

Each point on a polar plane is represented by the ordered pair #(r,theta)#.

So lets call the coordinates of #P_1# as #(r_1,theta_1)# and coordinates of #P_2# as #(r_2,theta_2)# . To find the distance between two points on a polar plane use the formula #d=sqrt((r_1) ^2+(r_2)^2-2r_1r_2cos(theta_2-theta_1))#

Thererfore using the points #(1,(3pi)/4)# and #(2,(15pi)/8)#, and the formula

#d=sqrt((r_1) ^2+(r_2)^2-2r_1r_2cos(theta_2-theta_1))#

we have

#d=sqrt((1)^2+(2)^2-2*1*2cos((15pi)/8-(3pi)/4))#

#:. d~~2.95#