## Elements of Geometry and Trigonometry |

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Page 7

Book I.

Book I.

**Prop**. Cor . of Cor . 1 . & 2 . ~ 4**Prop**. 5 Legendre . Book I. 5 11 5 Cor . of 11 6 12 8 10 13 14 15 15 Sch . of 29 30 Cor.1 . of 32 1 3 4 16 Cor . of 25 17 25 18 19 20 21 24 9**Prop**. 25 9 26 6 27 Cor . 1. of 19 28 Cor . Page 15

Now , since ACD is a straight line , the angle FCD will be a right angle (

Now , since ACD is a straight line , the angle FCD will be a right angle (

**Prop**. I. Cor . 1. ) ; and since ACE is a straight line , the angle FCE will likewise be a right angle . Hence , the angle FCD is equal to the angel FCE ( Ax ... Page 19

1 " triangle GAC is equal to DEF , since , by construction , they have an equal angle in each , contained by equal sides , (

1 " triangle GAC is equal to DEF , since , by construction , they have an equal angle in each , contained by equal sides , (

**Prop**. V. ) ; therefore CG is equal to EF . Now , there may be three cases in the proposition , according as ... Page 21

... B and the side BC common : therefore , the two triangles , BDC , BAC , have two sides and the included angle in the one , equal to two sides and the included angle in the other , each to each : hence they are equal (

... B and the side BC common : therefore , the two triangles , BDC , BAC , have two sides and the included angle in the one , equal to two sides and the included angle in the other , each to each : hence they are equal (

**Prop**. V. ) . Page 22

CBF are right angles , the side CB is common , and the side AB equal to BF , by construction ; therefore , the triangles are equal , and the angle ACB = BCF (

CBF are right angles , the side CB is common , and the side AB equal to BF , by construction ; therefore , the triangles are equal , and the angle ACB = BCF (

**Prop**. V. Cor . ) . But the angle ACB is a right angle , by hypothesis ...### What people are saying - Write a review

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### Common terms and phrases

ABCD adjacent altitude base become Book called centre chord circle circumference circumscribed common cone consequently contained Cosine Cotang cylinder described determine diameter difference distance divided draw drawn equal equations equivalent expressed extremities faces feet figure follows formed four frustum give given gles greater half hence homologous hypothenuse included inscribed intersection less let fall logarithm manner means measured meet middle multiplied number of sides opposite parallel parallelogram pass perpendicular plane polygon prism PROBLEM Prop proportional PROPOSITION pyramid quadrant quantities radii radius ratio reason rectangle regular remaining right angles Scholium segment sides similar Sine solid solid angle sphere spherical triangle square straight line suppose taken Tang tangent THEOREM third triangle triangle ABC unit vertex whole