C language to calculate the radius of the meridian circle and the unitary circle at a certain latitude

1. Enter the latitude and calculate the radius of the meridian circle and the unitary circle.

Enter 1 and choose the international 75 ellipsoid. Enter the latitude of 12°34'56″, and the calculated radius of the meridian circle is 6338193.296 meters, and the radius of the unitary circle is 6379063.053 meters. The

input azimuth angle is 21°43′6″, and the radius of curvature at this point is 6343412.766 meters.

2. Part of the source code

#include<stdio.h>
#include<math.h>
#define guoji75_a 6378140.00
#define guoji75_b 6356755.2881575287
#define PI 3.1415926535897932
#define keshi_a 6378245.0
#define keshi_b 6356863.0187730473

double zwqqlbj (int w[],double a,double b)
{
	double B,M,e1;
	e1 = (sqrt((a * a) - (b * b))) / a;
	B = (w[0] + (w[1] / 60) + (w[2] / 3600)) * PI /180;
	M=...... 
	return M;
}

double myqqlbj(int w[],double a,double b)
{
	double B,N,e1;
	e1 = (sqrt((a * a) - (b * b))) / a;
	B = (w[0] + (w[1] / 60) + (w[2] / 3600)) * PI /180;
	N = ...
	return N;
}
double ryfjhqlbj(int w[],double M,double N)
{
	double A,R;
	A = (w[0] + (w[1] / 60) + (w[2] / 3600)) * PI/ 180;
	R = ...
}
double zwxhc(int w[],int tuoqiu)
{
	double X,B;
	B = (w[0] + (w[1] / 60) + (w[2] / 3600)) * PI /180;
	if(tuoqiu==1)
		X=111133.005*B*180/PI-16038.528*sin(2*B)+16.833*sin(4*B)-0.022*sin(6*B);
	if(tuoqiu==2)
		X=111134.861*B*180/PI-16036.48*sin(2*B)+16.828*sin(4*B)-0.022*sin(6*B);
	return X;
}

main()
{
	double a, e1,e2,M,N,R,_a,_b,A,B,X,P,x1,x2;
	int weidu[3],fangweijiao[3],L1[3],L2[3],B1[3],B2[3],weidu1[3],weidu2[3];
	int i,fg;
	printf("请输入计算所用的椭球（国际75椭球请输1，克氏椭球请输2，回车结束。）：\n\n");
	scanf_s("%d",&fg);
	if(fg==1)
	{
		_a=guoji75_a;
		_b=guoji75_b;
	}
	else if(fg==2)
	{
		_a=keshi_a;
		_b=keshi_b;
	}

	printf("请输入某点的纬度（度分秒之间用空格隔开，按回车结束）：\n\n");
	for(i=0;i<3;i++)
		scanf_s("%d",&weidu[i]);

	M = zwqqlbj(weidu,_a,_b);
	printf("纬度为%d°%d′%d″处子午圈曲率半径M=%f米\n\n",weidu[0],weidu[1],weidu[2],M);

	N = myqqlbj(weidu,_a,_b);
	printf("纬度为%d°%d′%d″处卯酉圈曲率半径N=%f米\n\n",weidu[0],weidu[1],weidu[2],N);
	
	X=zwxhc(weidu,fg);
	printf("从赤道到纬度为%d°%d′%d″的子午线弧长为%f米\n\n",weidu[0],weidu[1],weidu[2],X);

	printf("请输入一个方位角（度分秒之间用空格隔开，按回车结束）：\n\n");
	for(i=0;i<3;i++)
		scanf_s("%d",&fangweijiao[i]);
	
	R = ryfjhqlbj(fangweijiao,M,N);
	printf("纬度%d°%d′%d″处方位角为%d°%d′%d″方向的曲率半径为：%f米\n\n",weidu[0],weidu[1],weidu[2],fangweijiao[0],fangweijiao[1],fangweijiao[2],R);
}