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+
+clc
+//given
+s=1.125//inch
+e=0.25//inch
+t=2.25//inch
+alpha=35//degrees
+//from 5.2, we know theta+alpha=sininverse(s/t)
+x=asind(s/t)
+y=180-x//sin(x)=sin(180-x)=sin(y)
+//at admission
+p=x-alpha
+//at cutoff
+q=y-alpha
+//from 5.3, theta+alpha=sininnverse(-e/t)
+ang=asind(-e/t)
+angle=abs(ang)
+a=180+angle//lies in the negative region of sine curve
+b=360-angle//lies in hte negative region of sine curve
+//at release
+r=a-alpha
+//at compression
+s=b-alpha
+printf("Angle theta at admission, cut-off,\n release and compression are %.2f, %.2f, %.2f and %.2f degrees respectively",p,q,r,s)
+
+