An elementary course of mathematics, Volum 2 |
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Resultat 1-5 av 72
Side 3
... consequently the arc ACB is greater than the chord AB . If the arc ACB be not less than the sum of the tangents AD , DB , it must be either equal to that sum or greater than it ; and in either case , EF being less than ED , DF , the arc ...
... consequently the arc ACB is greater than the chord AB . If the arc ACB be not less than the sum of the tangents AD , DB , it must be either equal to that sum or greater than it ; and in either case , EF being less than ED , DF , the arc ...
Side 4
... consequently ( V. 11 ) the perimeter of the polygon ABCDEF is to the perimeter of the poly- gon GHIKLM as the diameter AD to the diameter GK . Also the area of the polygon ABCDEF is to the area of the poly- gon GHIKLM as the square of ...
... consequently ( V. 11 ) the perimeter of the polygon ABCDEF is to the perimeter of the poly- gon GHIKLM as the diameter AD to the diameter GK . Also the area of the polygon ABCDEF is to the area of the poly- gon GHIKLM as the square of ...
Side 5
... consequently ( V. 15 ) as the radii of those circles ; therefore the perimeter of the polygon ABCDEFGH is to the peri- meter of the polygon IKLMNOPQ as AR to RK , that is as RT to RA , the triangles KRA and ART being similar ...
... consequently ( V. 15 ) as the radii of those circles ; therefore the perimeter of the polygon ABCDEFGH is to the peri- meter of the polygon IKLMNOPQ as AR to RK , that is as RT to RA , the triangles KRA and ART being similar ...
Side 6
... consequently the area of ABCD & c . is to the area of IKLM & c . as the square of RA to the square of RK , or as the square of RT to the square ᎡᎪ . PROP . VI . THEOR . of In a given circle an equilateral polygon may be inscribed ...
... consequently the area of ABCD & c . is to the area of IKLM & c . as the square of RA to the square of RK , or as the square of RT to the square ᎡᎪ . PROP . VI . THEOR . of In a given circle an equilateral polygon may be inscribed ...
Side 7
... consequently ( V. 16 , A ) the excess of C above I is less than eight times MR . And because AL is less than AK or HI , OR is greater than OG , and therefore MR is less than GC , that is than DF ; and consequently eight times MR is less ...
... consequently ( V. 16 , A ) the excess of C above I is less than eight times MR . And because AL is less than AK or HI , OR is greater than OG , and therefore MR is less than GC , that is than DF ; and consequently eight times MR is less ...
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ABCD allel altitude angle formed angle of inclination auxiliary plane circle described circumference circumscribed coincide cone consequently construction Descriptive Geometry determined diameter dicular dihedral angle contained distance ellipse equal and similar equal bases equilateral polygon faces ASB figure given angle given plane given point given straight line greater hemisphere horizontal plane horizontal projection horizontal trace inscribed isometric line joining line of level line parallel meets the plane parallel planes parallel to xy parallelepiped parallelogram pendicular perimeter perpen perpendicular to xy plane angles plane MN plane passing plane Prop planes BM planes of projection point of intersection prism Prob PROBLEM projecting plane pyramid rectangle right angles right-angled triangle scale of slope series of cylinders sides solid angle space straight line drawn THEOR third face trihedral vertical plane vertical projection vertical trace Wherefore
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Side 5 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Side 18 - FD. Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF. Because the two parallel planes KL, MN are cut by the plane EBDX, the common sections EX, BD are parallel (Prop.
Side 13 - For the same reason, CD is likewise at right angles to the plane HGK. Therefore AB, CD are each of them at right angles to the plane HGK.
Side 4 - BC above it : and since the straight line AB is in the plane, it can be produced in that plane : let it be produced to D ; and let any plane pass through the straight line AD, and be turned about it until it pass through the point C; and because the points B, C, are in this plane, the straight line* BC is in it: »7Def.1.
Side 9 - Note. (3. 11.) line; let this be BF: therefore the three straight lines AB, BC, BF are all in one plane, viz. that which passes through AB, BC : and because AB stands at right angles to each of the straight lines BD, BE, it is also at right angles (4. 1 1.) to the plane passing through them; and therefore makes right angles (3.
Side 16 - BGH are together equal* to two right angles: and BGH is a right angle; therefore also GBA is a right angle, and GB perpendicular to BA. For the same reason GB is perpendicular to BC. Since therefore the straight line GB stands at right angles to the two straight lines BA, BC, that cut one another in B, GB is perpendicular...
Side 9 - If three straight lines meet all in one point, and a straight line stand at right angles to each of them in that point ; these three straight lines are in one and the same plane. Let the straight line AB stand at right angles to each of the straight lines BC, BD, BE, in B, the point where they meet ; BC, BD, BE are in one and the same plane. If not, let...
Side 1 - A plane is perpendicular to a plane, when the straight lines drawn in one of the planes perpendicular to the common section of the two planes, are perpendicular to the other plane. 5. The inclination of a straight line to a plane...
Side 28 - Cor. 1.) therefore all the angles of the triangles are equal to all the angles of the polygon together with four right angles : (i. ax. 1.) but all the angles at the bases of the triangles are greater than all the angles of the polygon, as has been proved ; wherefore the remaining angles of the triangles, viz. those of the vertex, which contain the solid angle at A, are less than four right angles.
Side 5 - If a straight line stand at right angles to each of two straight lines in the point of their intersection, it will also be at right angles to the plane in which these lines are.