I wrote a program, but the middle algorithm and the correct result are not the same, I found the number of t is incorrectcan you help to see the arithmetic formula where to correct it? Thank you. The correct result is printed on paper. C language

Enter year: 2016 Enter month (1-12): 7 Enter day: 4 Enter hour in 24-hour format: 6 Enter timezone: -5 2.45757e+06 0.165052 73790 Phase of the moon is e degrees. jd 367*year(7*(year+((month+9)/12.0))/4.0)+(275*month/9.8)+day+(hour-timezone)/24.0+1721013.5 printf("%g ", jd); t-(jd-2451545.0)/36525.e; printf("%g ", t); e1-297.8501921+4452671114834*t-(8.0018819*pow(t, 2)1+(pow(t,3)/545868 . )-( pow(t,4)/113965000 ,0); printf("g ", el); ma-134.9633964+477198.8675055*t-0.0087414*pow(t, 2)+pow(t, 3)/69699.0-pow(t, 4)/14712000.0; printf("%g ", ma); a 357.5291092+35999.0582909 t-0.8001536 pow(t, 2)+pow(t, 3)/24490000.0 printf("gIn", sa) 1-188.0-e1-6. 289#5 in (ma*PI/180.0)+2.1+sin( sa*PI/188. )-1.274*sin( (2#81-ma)*PI/186. )-8.658*sin( 2*el+PI/ 180.0)-0.11sin(el PI/180.0); pa2-pa1-( pa1/368 .0)*368 , 0; printf(" Phase of the moon is %d degrees.", pa2); 1. jd is the date converted into a Julian date at UTC. The vertical bars have small bars on the bottom and are floor function brackets (i. e., the largest integer smaller than the quantity). 7year+month+9 jd=367 year _| 12.0-1275 month hour- timezone+1721013.5 +day+ 9.0 24.0 2. t is the proportion of the Julian century t=jd-24515450 36525.0 3. el is the moon's mean elongation in degrees, the angle between the sun and the moon with Earth as the reference point el = 297.850 1921 + 445267.1 1 1 4034 t-0.00 1 88 19 t 545868.0 113065000.0 4. ma is the moon's mean anomaly in degrees, the angular distance from the pericenter of the moon's elliptical orbit which a fictitious moon would have if it moved in a circular orbit, with constant speed, in the same orbital period as the actual moon in its elliptical orbit ma 134.9633964+477198.8675055 0.0087414t+o 69699.0 14712000.0 5. sa is the sun's mean anomaly in degrees sa = 357.5291 092 + 35999.0502909 0.000 1 536t4 24490000.0 6. pa is the phase angle of the moon in degrees. However, the trigonometric library functions in C require their arguments to be in radians (rather than degrees). The formula for converting degrees to radians is radians- degrees180 and is used to convert the various angles into radians as follows al+2. 1 sinlsa sin 2 ma -1.274 sin|Zel-rna )18 al -0, 1 1 sinel. pa = 180.0-el-6.289 sin l ma l 180.0 180.0 180.0 0.658sin 2el 0.2 4. ma is the moon's mean anomaly in degrees, the angular distance from the pericenter of the moon's elliptical orbit which a fictitious moon would have if it moved in a circular orbit, with constant speed, in the same orbital period as the actual moon in its elliptical orbit ma 134.9633964+477198.8675055t-0.0087414 t+ 69699.0 14712000.0 5. sa is the sun's mean anomaly in degrees sa 357.5291092+35999.0502909-0.0001536 t+ 24490000.0 6. pa is the phase angle of the moon in degrees. However, the trigonometric library functions in C require their arguments to be in radians (rather than degrees). The formula for converting degrees to radians is degrees>