A Companion to Wrigley's Collection of Examples: Being Illustrations of Mathematical Processes and Methods of Solution |
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Page 158
... string = CD + C'D ' + arc CGC ' + arc DHD ' . G A C E D ' B H Now CD = ( A'D2 – CA'3 ) * = ( a3 — c2 ) ‡ , CA ' 10 = 40 cos = AC cos AC A'D AC = arc CF AC cos - I AE .. arc CGC ' = 24C ( π — cos ̄ ) . Similarly , arc DHD ′ = 2BD ( · π ...
... string = CD + C'D ' + arc CGC ' + arc DHD ' . G A C E D ' B H Now CD = ( A'D2 – CA'3 ) * = ( a3 — c2 ) ‡ , CA ' 10 = 40 cos = AC cos AC A'D AC = arc CF AC cos - I AE .. arc CGC ' = 24C ( π — cos ̄ ) . Similarly , arc DHD ′ = 2BD ( · π ...
Page 251
... string BC = tension in the string AC - P ; and these two tensions are balanced by the weight Px3 , which we may regard as their resultant . Therefore if be the angle ACB , we have A 3P2 = P2 + P2 + 2P3 cos 0 , W whence cos 0 = cos 60 ...
... string BC = tension in the string AC - P ; and these two tensions are balanced by the weight Px3 , which we may regard as their resultant . Therefore if be the angle ACB , we have A 3P2 = P2 + P2 + 2P3 cos 0 , W whence cos 0 = cos 60 ...
Page 252
... string CD , A CE = 0 , BDF = $ ; E F B D then P : T = sin ( 90 ° + 0 ) : sin ( 180 ° — 0 ) = cos 0 : sin ✪ = I : tan 0 . Similarly T Q = sin : cos = tan & : 1 ; : = .. P : Qtano : tan 0 = y : x . Now AC - AE CE DF2 — DB2 — BF2 ...
... string CD , A CE = 0 , BDF = $ ; E F B D then P : T = sin ( 90 ° + 0 ) : sin ( 180 ° — 0 ) = cos 0 : sin ✪ = I : tan 0 . Similarly T Q = sin : cos = tan & : 1 ; : = .. P : Qtano : tan 0 = y : x . Now AC - AE CE DF2 — DB2 — BF2 ...
Page 253
... string is equal in length to the arc of a quadrant , the angle PCQ will be a right angle . Therefore taking the mo- ments of P and Q about C , we have B C Pxr cos PCA Qxr cos QCD Qxr sin PCA , = whence tan PCA = 19. The reactions R , R ...
... string is equal in length to the arc of a quadrant , the angle PCQ will be a right angle . Therefore taking the mo- ments of P and Q about C , we have B C Pxr cos PCA Qxr cos QCD Qxr sin PCA , = whence tan PCA = 19. The reactions R , R ...
Page 256
... string BC , the reaction R of the wall . Let CA = x . Then , resolving perpendicular to the string , and taking moments about C , we obtain R cos 0 W sin o , = Rx = WxAB sin & . 2 sin 0 cos - • 3 sin But cos 0 - - 2 X 315 = - 215 3 5 5 ...
... string BC , the reaction R of the wall . Let CA = x . Then , resolving perpendicular to the string , and taking moments about C , we obtain R cos 0 W sin o , = Rx = WxAB sin & . 2 sin 0 cos - • 3 sin But cos 0 - - 2 X 315 = - 215 3 5 5 ...
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A Companion to Wrigley's Collection of Examples: Being Illustrations of ... John Thompson Platts,Alfred Wrigley No preview available - 2016 |
Common terms and phrases
a+b+c a²+b² a²b³ AB² ABCD AC² angle ABC axis BC² beam centre of gravity chord circle co-ordinates coefficient cos² cos³ cosec² curve diameter draw ellipse equal equilateral Eucl Hence hyperbola Join latus rectum Let ABC middle point MULTINOMIAL THEOREMS parabola parallel perpendicular plane point of contact point of intersection r₁ radius required locus right angle roots S₁ sec² segment sides Similarly sin² sin³ square straight line tan² tangent term transformed equation triangle ABC values vertical weight whence X₁