Elementary Mathematics: Embracing Arithmetic Geometry, and Algebra1873 |
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Resultat 1-5 av 36
Side 72
... are equal when AB = DE , BC EF , CA - FD . = PROOF . - Apply BE ) DE to AB , and let the point F fall on the side of AB remote from C. Join CF. Then since BC = BF , the angle BCFBFC - ( Prop . 3 ) . Similarly ACF = 72 GEOMETRY .
... are equal when AB = DE , BC EF , CA - FD . = PROOF . - Apply BE ) DE to AB , and let the point F fall on the side of AB remote from C. Join CF. Then since BC = BF , the angle BCFBFC - ( Prop . 3 ) . Similarly ACF = 72 GEOMETRY .
Side 73
... ( Prop . 4 ) . = Proposition 6. - Problem . To bisect a straight line ( AB ) . CONSTRUCTION . - With centres A , B , and equal radiuses , draw circles meeting in C. Join AC , CB , and bisect the angle ACB by the straight line CD ( Prop ...
... ( Prop . 4 ) . = Proposition 6. - Problem . To bisect a straight line ( AB ) . CONSTRUCTION . - With centres A , B , and equal radiuses , draw circles meeting in C. Join AC , CB , and bisect the angle ACB by the straight line CD ( Prop ...
Side 74
... ( Prop . 7 ) . Then it is mani- fest that the sum of the inclina- tions of AB to EB , CB to AB , DB to CB , & c . = the sum of the inclinations of KB to EB and KB to FB . But the inclina- tions of KB to EB and KB to FB are two right ...
... ( Prop . 7 ) . Then it is mani- fest that the sum of the inclina- tions of AB to EB , CB to AB , DB to CB , & c . = the sum of the inclinations of KB to EB and KB to FB . But the inclina- tions of KB to EB and KB to FB are two right ...
Side 75
... ( Prop . 9 ) = the two angles at C. But ABC = ACB ( Prop . 3 ) . Therefore DBC ECB . B 19 Proposition 12. - Theorem . If two angles of a PROPOSITIONS . 75.
... ( Prop . 9 ) = the two angles at C. But ABC = ACB ( Prop . 3 ) . Therefore DBC ECB . B 19 Proposition 12. - Theorem . If two angles of a PROPOSITIONS . 75.
Side 77
... ( Prop . 10 ) . There- = fore the angle BAE the angle ECF . - But ACD is greater than ECF ; therefore ACD is greater than BAE . Similarly we may prove that BCG is greater than ABC . But BCG = ACD ; therefore ACD is greater than ABC ...
... ( Prop . 10 ) . There- = fore the angle BAE the angle ECF . - But ACD is greater than ECF ; therefore ACD is greater than BAE . Similarly we may prove that BCG is greater than ABC . But BCG = ACD ; therefore ACD is greater than ABC ...
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Vanlige uttrykk og setninger
16 Maps a²b ab² ABCD ACD is greater Algebra angle ACB angle equal Arith arithmetical ATLAS base bisect called cent centimes cloth lettered co-efficient concrete quantities decimal fraction denominator digits divide dividend divisor draw a circle equal and parallel equations example expression exterior angle fore four right angles Geometry given straight line gram greater than AC Hence inches inner angles isosceles least common multiple measure metres metrical system mixed number Multiply opposite parallel straight lines parallel to BD parallelogram parallels are equal produced Proposition quotient radius ratio rectilinear recurring decimal remainder result right angles right angles Prop square subtract triangle ABC triangles are equal unit vulgar fraction whole number
Populære avsnitt
Side 85 - If a side of any triangle be produced, the exterior angle is equal to the, two interior and opposite angles ; and the three interior angles of every triangle are equal to two right angles.
Side 110 - If two triangles have two sides of the one equal to two sides of the other...
Side 105 - If one side of a triangle be produced, the exterior angle is greater than either of the interior, and opposite angles.
Side 83 - If a straight line fall on two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite angle on the same side; and also the two interior angles on the same side together equal to two right angles.