Pure mathematics, Bind 11874 |
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Resultater 1-5 af 13
Side 100
... AC is equal to AD ( Hyp . ) , The triangle ADC is an isosceles triangle , and the angles ECD , FDC , upon the other side of its base CD , are equal to one another ( I. 5 ) . But the angle ECD is greater than the angle BCD 100 GEOMETRY .
... AC is equal to AD ( Hyp . ) , The triangle ADC is an isosceles triangle , and the angles ECD , FDC , upon the other side of its base CD , are equal to one another ( I. 5 ) . But the angle ECD is greater than the angle BCD 100 GEOMETRY .
Side 101
Edward Atkins. But the angle ECD is greater than the angle BCD ( Ax . 9 ) . Therefore the angle FDC is likewise greater than BCD . Much more then is the angle BDC greater than BCD . Again , because BC is equal to BD ( Hyp . ) , < BDC > Z ...
Edward Atkins. But the angle ECD is greater than the angle BCD ( Ax . 9 ) . Therefore the angle FDC is likewise greater than BCD . Much more then is the angle BDC greater than BCD . Again , because BC is equal to BD ( Hyp . ) , < BDC > Z ...
Side 103
... angle ACD is equal to the angle BCD ( Const . ) ; Therefore the base AD is equal to the base DB ( I. 4 ) . Therefore the straight line AB is divided into two equal parts in the point D. Q. E. F. Proposition 11. — Problem . To draw a ...
... angle ACD is equal to the angle BCD ( Const . ) ; Therefore the base AD is equal to the base DB ( I. 4 ) . Therefore the straight line AB is divided into two equal parts in the point D. Q. E. F. Proposition 11. — Problem . To draw a ...
Side 109
... angle of the triangle ABC , it is greater than the interior and opposite angle ABC ( I. 16 ) . To each of these add ... BCD , PROOF . Because ADB is the exterior angle of the triangle BDC , it is greater than the interior and opposite ...
... angle of the triangle ABC , it is greater than the interior and opposite angle ABC ( I. 16 ) . To each of these add ... BCD , PROOF . Because ADB is the exterior angle of the triangle BDC , it is greater than the interior and opposite ...
Side 110
... < BCD > < BDC . ..DB BC . Proposition 19. - Theorem . The greater angle of every triangle is subtended by the greater side , or has the greater side opposite to it . Let ABC be a triangle , of which the angle ABC is greater than the angle ...
... < BCD > < BDC . ..DB BC . Proposition 19. - Theorem . The greater angle of every triangle is subtended by the greater side , or has the greater side opposite to it . Let ABC be a triangle , of which the angle ABC is greater than the angle ...
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a²b a²b² a²x² a³b ab² ab³ ABCD adjacent angles algebraical algebraical quantity angle ABC angle BAC angle BCD angle EDF angle equal base BC BC is equal bisect brackets cent centim centre circle ABC coefficient common Const cube root decimal figures denominator distance divided divisor equation expression exterior angle factor Find the value fraction given rectilineal given straight line greater Hence join kilom Let ABC logarithm metres millig Multiply opposite angles parallel parallelogram perpendicular PROOF.-Because Q. E. D. Proposition quotient ratio rectangle contained remainder right angles segment side BC square on AC square root subtraction term triangle ABC x²y x²y² x³y xy² xy³
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