. Elements of plane and spherical trigonometry . - 0J are > 180°, thereis but one solution. If both of these conditions are fulfilled,there is no solution. The formulse, derived by means ofNapiers Eules, are as follows: cos A = tan tp cot. cos b = cot 6 cot A , cos^l tan tp = , cot b cos b cot 0 = cos a = cos tpx cos p, cos cot ACOS tp cos p («) (JO cos p cos a cos b or COS tpx COS tpx COS tp cos a cos tp cos 6 cos bcot b tan p, cos 6X = (r) cot a tan p : tanp cos 6 cos 0t or cot b cot <2cos 0 cot a cos #, = cot b Check. c = <p ±<p1; C=0± B = cot a tan prcos 0 = sin B cos ^r
. Elements of plane and spherical trigonometry . - 0J are > 180°, thereis but one solution. If both of these conditions are fulfilled,there is no solution. The formulse, derived by means ofNapiers Eules, are as follows: cos A = tan tp cot. cos b = cot 6 cot A , cos^l tan tp = , cot b cos b cot 0 = cos a = cos tpx cos p, cos cot ACOS tp cos p («) (JO cos p cos a cos b or COS tpx COS tpx COS tp cos a cos tp cos 6 cos bcot b tan p, cos 6X = (r) cot a tan p : tanp cos 6 cos 0t or cot b cot <2cos 0 cot a cos #, = cot b Check. c = <p ±<p1; C=0± B = cot a tan prcos 0 = sin B cos ^r « SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 159 Example: a = 67° 35 12, b = 58° 36 6, A log cos A ={—) log tan = (—) log cos a = log cos =(—) colog cos b = log cos 0X =(—) = 162° 12 24<px = 134° 10 28 c1= 296° 2252c,= 28° 156 56 6, A = 101° 17 48. log cos b = log cot A = (-) log cot 6 = (-) log cot a = log cos 0 = (-) colog cot b = log COS 0j = (-) 0 = 159° 119 0i = 129° 7 15 Ci = 288° 8 34 C2 = 29° 54 4 But the values of cx and Cx are impossible. Hence there is butone so
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