## Elements of Geometry and Trigonometry |

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Page 51

If we

If we

**suppose**AB - DE , the angle ACB will be equal to DCE . For , if these angles are not equal ,**suppose**ACB to be the greater , and let ACI be taken equal to DCE . From what has just been shown , we shall have AI - DE : but ... Page 52

**Suppose**, for example , that the angles ACB , DCE , are to each other as 7 is to 4 ; or , which is the same thing ,**suppose**that the angle M , which may serve as a common measure , is contained times in the angle ACB , and 4 times in ... Page 54

Let BAD be an inscribed angle , and let us first

Let BAD be an inscribed angle , and let us first

**suppose**that the centre of the circle lies within the angle BAD . Draw the diameter AE , and the radii CB , CD . E The angle BCE , being exterior to the triangle ABC , is equal to the sum ... Page 63

Take three points , A , B , C , any where in the circumference , or the arc ; draw AB , BC , or

Take three points , A , B , C , any where in the circumference , or the arc ; draw AB , BC , or

**suppose**them to be drawn ; bisect those two lines by the perpendiculars DE , FG : the point O , where these perpendiculars meet , will be ... Page 66

**Suppose**, for instance , we find GB to be contained exactly twice in FD ; BG will be the common measure of the two proposed lines . Put BG = 1 ; we shall have FD = 2 : but EB contains FD once , plus GB ; therefore we have EB = 3 : CD ...### 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