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Resultat 1-5 av 67
Side 2
... perpendicular to .. RHTP is a rectangle , .. its diagonals 1. xn - ntn - 1 | n2n + 1 PAPER II . - ( n + 1 ) . + 1 | n nan + 1 n2x2 + n2x - - ( n + 1 ) - − ( n + 1 ) xn + n2x2 − - пх - n + n + 1 xn + x ( n + 1 ) + a ( n + 1 ) ( n + 1 ) ...
... perpendicular to .. RHTP is a rectangle , .. its diagonals 1. xn - ntn - 1 | n2n + 1 PAPER II . - ( n + 1 ) . + 1 | n nan + 1 n2x2 + n2x - - ( n + 1 ) - − ( n + 1 ) xn + n2x2 − - пх - n + n + 1 xn + x ( n + 1 ) + a ( n + 1 ) ( n + 1 ) ...
Side 4
... perpendicular to BC . Make the angle GON equal to the angle BẶC . Draw NM parallel to BC , MP parallel to AC . Then MNP is the triangle required . For arc MG = arc GN , .. angle MOG .. angle MPN : = = angle GON . angle BAC & c . 6. Let ...
... perpendicular to BC . Make the angle GON equal to the angle BẶC . Draw NM parallel to BC , MP parallel to AC . Then MNP is the triangle required . For arc MG = arc GN , .. angle MOG .. angle MPN : = = angle GON . angle BAC & c . 6. Let ...
Side 11
... perpendicular to SH . Since CR = CS CH .. SRH is a right angle . Since CS = CR .. CSR = CRS = RSY = .. the triangle SRM Similarly it can be shewn that HM = HZ . Since RM2 SM.MH on the tangent at B. = triangle SRY .. SM = SY . = SY.HZ ...
... perpendicular to SH . Since CR = CS CH .. SRH is a right angle . Since CS = CR .. CSR = CRS = RSY = .. the triangle SRM Similarly it can be shewn that HM = HZ . Since RM2 SM.MH on the tangent at B. = triangle SRY .. SM = SY . = SY.HZ ...
Side 15
... perpendicular to the wall , and CF vertical . Then P , the reaction at B passes through F. Let W be the weight of the rod , and 0 the angle it makes with the horizon , and a its length . Taking moments about A we have AB = BG COS a P.AB ...
... perpendicular to the wall , and CF vertical . Then P , the reaction at B passes through F. Let W be the weight of the rod , and 0 the angle it makes with the horizon , and a its length . Taking moments about A we have AB = BG COS a P.AB ...
Side 19
... perpendicular to dВ and OE respectively . Then since OE bisects the angle OEA , : . 01EE1 = 45 ° = ЕО1Е1 .. O̟1H = EE1 = OE1 · = = " 1 and r .. 71 r- r1 01H = = cot A cot 4 . Socot OH 2 y Te B 13 2 ' ↑ - = cot r3 B C But cot - C ...
... perpendicular to dВ and OE respectively . Then since OE bisects the angle OEA , : . 01EE1 = 45 ° = ЕО1Е1 .. O̟1H = EE1 = OE1 · = = " 1 and r .. 71 r- r1 01H = = cot A cot 4 . Socot OH 2 y Te B 13 2 ' ↑ - = cot r3 B C But cot - C ...
Vanlige uttrykk og setninger
a+b+c ABCD axis bisects the angle Cambridge centre chord circle round circle will go circum-circle Clifton College coef College conic const coordinates cos² cosec Crown 8vo denote diameter directrix draw Edition ellipse envelope equal equation Fcap fixed point geometrical given expression hyperbola inscribed circle intersect Join Let ABC Let the tangent locus Mathematics meet middle point Monday afternoon morning nine-point circle nth term orthocentre PAPER parabola parallel perpendicular plane Professor quadrilateral radical axis radius right angles shew shewn sides Similarly sin² sin³ straight line tan² tangent Todh triangle Trinity College Tripos velocity vertex vertical
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