## Elements of Geometry: Containing the First Six Books of Euclid: With a Supplement ... to which are Added, Elements of Plane and Sphericale Trigonometry ... From the Last London Ed., Enl |

### From inside the book

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**Cosine**, Cotangent , or Cosecant of that angle . Thus , let CL or DB , which is equal to CL , be the sine of the angle CBH ; HK the tangent , and BK the secant of the same angle : CL or BD is the**cosine**, HK the cotangent , and BK the ... Page 222

... of the sines of the arcs AB and AC ; and KC is the difference of the sines ; also BD is the sum of the arcs AB and AC , and BC the diffe- rence of those arcs D COR . 1. Because EL is the

... of the sines of the arcs AB and AC ; and KC is the difference of the sines ; also BD is the sum of the arcs AB and AC , and BC the diffe- rence of those arcs D COR . 1. Because EL is the

**cosine**of AC 222 PLANE TRIGONOMETRY ' Page 223

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**cosine**of AC , and EH of AB , FK is the sum of these**cosines**, and KB their difference ; for FK = 1FB + EL = EH + EL , and KB = LH = EH - EL . Now , FK : KB :: tan . FDK : tan . BDK ; and tan . DFK = cotan . FDK , because DFK is the ... Page 225

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**cosine**of the angle included by the two sides . Let ABC be any triangle , 2AB.BC is to the difference between AB2 + BC2 and AC2 as radius to cos . B. " From A draw AD perpendicular to BC , and ( 12. and 13. 2. ) the difference be- tween ... Page 227

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**cosine**of half the angle included between the two sides of the triangle . Let ABC be a triangle , of which BC is the base , and AB the greater of the other two sides , 4AB.AC : ( AB + AC + BC ) ( AB + AC − BČ ) : : R2 . ( cos.RAC ) 2 ...### Other editions - View all

### Common terms and phrases

ABC is equal ABCD adjacent angles altitude angle ABC angle ACB angle BAC base BC bisected centre chord circle ABC circumference cosine cylinder demonstrated described diameter divided draw equal and similar equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles given rectilineal given straight line greater Hence hypotenuse inscribed join less Let ABC Let the straight magnitudes meet multiple opposite angle parallel parallelogram parallelopiped perpendicular polygon prism PROB PROP proportional proposition quadrilateral radius ratio rectangle contained rectilineal figure remaining angle right angled triangle SCHOLIUM segment semicircle shewn side BC sine solid angle spherical angle spherical triangle square straight line BC THEOR third touches the circle triangle ABC triangle DEF wherefore