Part 1b Another Function and Prototype See example 2 for exa

Part 1b: Another Function and Prototype

See example 2 for examples that may help you solve this problem

Complete this program, by adding your mag( ) function from above, and add another function rect2polar (and prototype) that returns the polar form of a rectangular coordinate pair. The data type of the inputs, x, y should be double, as should the data type of the outputs r and theta. Since you already have a function which gets the magnitude of a vector, your rect2polar(...) function can call it to get half the job done.  The commented code (which you\'ll need to un-comment) shows how your function should be called. Name your program pass_by_ref.cpp .

Reminder: The rectangular-to-polar function should take two arguments as input (x, y) and return two values (r, ) where:

r = mag (x,y)

= atan2(y,x)

note: rectangular coordinate (1,1) is equivalent to polar coordinate (1.414, PI/4);  (-2, -2) should give you (2.818, 5PI/4)

note 2:  since your function returns its two values through the reference parameters, the return type of the function should be void .

Name your program pass_by_ref.cpp

===============================THE PROGRAM IS POSTED BELOW==============================

Solution

;>

rect2polar(x,y,r,theta);

cout &xx;&xx; \"the xxxxx xxxx xx ( \"&xx;&xx; x &xx;&xx; \" ,\"&xx;&xx; y &xx;&xx;\" ) xx: ( \" << x << \" , \" <<theta &xx;&xx; \" )\"&xx;&xx;xxxx;xxxxxxxxxx(x,x,x,xxxxx);xxxx &xx;&xx; \"xxx polar xxxx of ( \"&xx;&xx; x << \" ,\"<< x &xx;&xx;\" ) is: ( \" &xx;&xx; x &xx;&xx; \" , \" &xx;&xx;xxxxx &xx;&xx; \" )\"<<endl;

return x;}xxxxxx mag(double x, xxxxxx y){xxxxxx sqrt(x*x + y*y);
}

double xxxxxxxxxx(xxxxxx x, xxxxxx y, xxxxxx&xxx; r, xxxxxx&xxx;xxxxx){xxxxx = xxxx(x/y) * xxx / PI; //PI xx a xxxxxxxx, atan xxxxxxx xxxxxxxxxx xxx return xx radians
r = xxx(x, y);//xxxxx is xxx in xxxxxxx}