Spherical Trigonometry: For the Use of Colleges and SchoolsMacmillan and Company, 1863 - 132 σελίδες |
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Σελίδα 6
... sin POα . O a II . SPHERICAL TRIANGLES . 15. Spherical Trigonometry investigates the relations which subsist between the angles of the plane faces which form a solid angle and the angles at which the plane faces are inclined to each ...
... sin POα . O a II . SPHERICAL TRIANGLES . 15. Spherical Trigonometry investigates the relations which subsist between the angles of the plane faces which form a solid angle and the angles at which the plane faces are inclined to each ...
Σελίδα 16
... sin b sin c cos A. cos a - COS b COS C COS A sin b sin c 38. We have supposed , in the construction of the ( 16 )
... sin b sin c cos A. cos a - COS b COS C COS A sin b sin c 38. We have supposed , in the construction of the ( 16 )
Σελίδα 17
... sin b sin c ' cos B'AC ; but a ' = π - α , c ' = π C , BAC = -- = π - А ; thus cos a = cos b cos c + sin b sin c cos A. ( 2 ) Suppose both the sides which contain the angle A to be greater than quadrants . Produce AB and AC to meet at A ...
... sin b sin c ' cos B'AC ; but a ' = π - α , c ' = π C , BAC = -- = π - А ; thus cos a = cos b cos c + sin b sin c cos A. ( 2 ) Suppose both the sides which contain the angle A to be greater than quadrants . Produce AB and AC to meet at A ...
Σελίδα 18
... sin CD sin BD cos CDB , π and cos CDB = 0 , cos CD = cos ( ~ 6 ) = sin b , cos BD = cos A ; " COS α = sin b cos A ; thus and this is what the formula in Art . 37 becomes when c = π ( 4 ) Suppose that both the sides which contain the ...
... sin CD sin BD cos CDB , π and cos CDB = 0 , cos CD = cos ( ~ 6 ) = sin b , cos BD = cos A ; " COS α = sin b cos A ; thus and this is what the formula in Art . 37 becomes when c = π ( 4 ) Suppose that both the sides which contain the ...
Σελίδα 19
... sin b sin c cos a cos b cos therefore sin2 A = 1 – sin b sin c c ) 2 - - ( 1 − cos2 b ) ( 1 — cos3c ) — ( cos a - cos b cos c ) 2 2 sin ' b sin2 c 1 - cos2 a cos2b - cos2c + 2 cos a cos b cos c 2 sin ' b sinc therefore sin A = √ ( 1 ...
... sin b sin c cos a cos b cos therefore sin2 A = 1 – sin b sin c c ) 2 - - ( 1 − cos2 b ) ( 1 — cos3c ) — ( cos a - cos b cos c ) 2 2 sin ' b sin2 c 1 - cos2 a cos2b - cos2c + 2 cos a cos b cos c 2 sin ' b sinc therefore sin A = √ ( 1 ...
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a+b+c angular points approximately bisecting calculated centre circle described circular measure cos b cos cos² cos³ cosb cosines Crown 8vo deduce denote equal equation Fcap Fellow of St formed formulæ given greater Hence inscribed John's College late Fellow less Let ABC limp cloth lune M.A. Fellow middle point Napier's analogies Napier's Rules obtain perpendicular plane triangle Plane Trigonometry polar triangle pole polygon preceding article primitive triangle quadrant radius regular polyhedron respectively right angles right-angled triangles School Second Edition Sermons preached shew shewn Similarly sin b sin sin² sin³ sine small circle solid angle solution sphere spherical excess spherical triangle Spherical Trigonometry St John's College suppose surface tangent tetrahedron Third Edition Treatise triangle ABC Trinity College University of Cambridge values vols
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