Trigonometry ...1866 |
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Side 3
... Draw OD'D perpendicular to A'B ' and AB . Then from the similar triangles AOB , A'OB ' we have AB A'B ' AO that is therefore == A'O ' AB A'B ' n . AB n . A'B ' r Now , n . AB and n . A'B ' are the perimeters of the two polygons , and ...
... Draw OD'D perpendicular to A'B ' and AB . Then from the similar triangles AOB , A'OB ' we have AB A'B ' AO that is therefore == A'O ' AB A'B ' n . AB n . A'B ' r Now , n . AB and n . A'B ' are the perimeters of the two polygons , and ...
Side 8
... draw PM perpendicular to X'X ; then if we put the angle POX = A , we have the following definitions : PM OM sin A = COSA = OP ' OP ' PM OM tan A = cot A = OM ' PM ' OP OP sec A = OM ' cosecД = PM ' vers A = 1- cosД , covers A1 - sinA ...
... draw PM perpendicular to X'X ; then if we put the angle POX = A , we have the following definitions : PM OM sin A = COSA = OP ' OP ' PM OM tan A = cot A = OM ' PM ' OP OP sec A = OM ' cosecД = PM ' vers A = 1- cosД , covers A1 - sinA ...
Side 21
... Draw the indefinite straight line BD perpen- dicular to AB , and inflect AD BC ( that is , from = the centre A at a distance equal to BC describe a circle cutting BD in D , A and join AD ) . B AB AB Then cos A = = AD BC ' and sinD = sin ...
... Draw the indefinite straight line BD perpen- dicular to AB , and inflect AD BC ( that is , from = the centre A at a distance equal to BC describe a circle cutting BD in D , A and join AD ) . B AB AB Then cos A = = AD BC ' and sinD = sin ...
Side 21
... Draw BD perpendicular to AB , and = BC , and join AD . AB AB BDBC BD BC ' AB AB = BD BC Then cotA = and tanD = cotA = AB Hence , if the cotangent be given = BC A is the required AB = BC ' D is the required angle ; and if the tangent be ...
... Draw BD perpendicular to AB , and = BC , and join AD . AB AB BDBC BD BC ' AB AB = BD BC Then cotA = and tanD = cotA = AB Hence , if the cotangent be given = BC A is the required AB = BC ' D is the required angle ; and if the tangent be ...
Side 24
... draw PM perpendicular to OX and PQ perpendicular to OC ; draw QN perpendicular to OX and QR perpendicular to PM . D P R B -X MN Then the angle QPR = 90 ° - PQR = RQO = A . Now sin ( A + B ) = OP PM RM + PR QN + PR OP OP QN OQ , PR PQ OM ...
... draw PM perpendicular to OX and PQ perpendicular to OC ; draw QN perpendicular to OX and QR perpendicular to PM . D P R B -X MN Then the angle QPR = 90 ° - PQR = RQO = A . Now sin ( A + B ) = OP PM RM + PR QN + PR OP OP QN OQ , PR PQ OM ...
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acute called Cambridge centre CHAPTER circle circular measure coincides College containing cosA cosec cotA cotangent decreases determine difference dividing draw Elementary English equal Euclid Examination Examples expressed Fellow formulae four Fourth Geometrical given given angle greater Greek half Hence increases indefinitely inscribed L sin Late Fellow Latin less loga Mathematics negative Notes numerically opposite perpendicular Plautus positive positive angle Problems Professor prove quadrant radius respectively revised right angles right-angled triangles Schools secant Second Edition sides signs Similarly sinA sinB sine and cosine solution solve the triangle Students subtended subtract tanA tangent Text Third Edition Translation Treatise triangle triangle ABC Trigonometrical Ratios Trinity College University values
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