## Elements of Geometry and Trigonometry |

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Page 45

Sch . ) ; and hence that the arc AH is greater than AD ; since the

Sch . ) ; and hence that the arc AH is greater than AD ; since the

**whole**is greater than its part . Scholium . The arcs here treated of are each less than the semicircumference . If they were greater , the reverse property would have ... Page 48

If the two parallels DE , IL , are tangents , the one at H , the other at K , draw the parallel secant AB ; and , from what has just been shown , we shall have MH = HP , MK = KP ; and hence the

If the two parallels DE , IL , are tangents , the one at H , the other at K , draw the parallel secant AB ; and , from what has just been shown , we shall have MH = HP , MK = KP ; and hence the

**whole**arc HMK = HPK . Page 51

... AB is equal to DE ; hence AI must be equal to AB , or a part to the

... AB is equal to DE ; hence AI must be equal to AB , or a part to the

**whole**, which is absurd ( Ax . 8. ) ... if two angles at the centre are to each other in the proportion of two**whole**numbers , the intercepted arcs will be to each ... Page 52

Therefore the

Therefore the

**whole**arc AB will be to the**whole**arc DE , as 7 is to 4. But the same reasoning would evidently apply , if in place of 7 and 4 any numbers whatever were employed ; hence , if the ratio of the angles ACB , DCE , can be ... Page 53

The arcs AB , AI , will be to each other as two

The arcs AB , AI , will be to each other as two

**whole**numbers , and by the preceding theorem , we shall have , the angle ACB : angle ACI :: arc AB arc AI . Comparing these two proportions with each other , we see that the antecedents ...### 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