CASE VII. The Bafe BA, and the Hypothenufe BC being given, to find the Perpendicular. 1. Operation, Hypothenufe Radius. This Cafe may be otherwise refolved, thus. Find the fum and difference of the Bafe, and Hypothenufe; then multiply the Sum and Difference together, and out of the Product extract the fquare Root, and that fhall be the Perpendicular. By Logarithms you you may add the Logarithms of the faid Sum and Difference together, half the Sum thereof fhall be the Log. of the Perpendicular, thus. This is the Log, of 81.113 the Perpendicular, which is the fame as before. By Scale and Compasses. For the first Operation, extend the Compasses from 146. to 121.394 in the Line of Numbers, that extent will reach from 90° to 56° 15' in the Sines. Then for the second Operation, extend from 90° to 33° 45′ in the Sines, that extent will reach from 146 to 81.113 in the Line of Numbers. By the Sliding-Rule. For the firft, fet 146 on A to 121.394 on B, then against 90° on SS, you will find 56° 15′ on S. Then for the fecond, fet 90° on SS to 33° 45′ on S, and against 146 on A, is 81.113 on B. By Geometrical Protraction. Draw the Line BA, and from your Scale take 121. 394, and fet from B to A, and upon A raise a Perpen * dicular: dicular: then from your Scale take 146, and fet one foot of the Compaffes in B, crofs the Perpendicular in C, and measure AC upon your Scale, and you will find it to be 81.113. IN N the Solution of Oblique angled plain Triangles, there are but Five Cafes, the refolving of which depends upon the three following Axioms. The first of which ferves for the refolving of the first and fecond Cafes. The fecond for refolving the third and fourth Cafes. And the third is appropriated only to the fifth and laft Cafe, I fhall firft explain and demonftrate the faid Axioms, and then proceed to the Refolution of the Cafes thereby. AXIOM I. In every plain Triangle (as well right as oblique angled) the Sides are in proportion one to the other, as are the Sines of the Angles oppofite to thofe Sides, & contra. Cons Conftruction. Let the Triangle ABC be infcribed in a Circle, and from E the Centre D let C there be alfo drawn the Radius DC. Demonftration. Now because the Angle at the Center EDC, is equal to the Angle in the Periphery ABC; and CDF at the Center, equal to CAB (by Eucl. 3. Lib. 20 Pr.) therefore fhall the halves of the Sides be as Sines: and what proportion the Side CA hath to the Side CB, the fame fhall the Sine CH have to the Sine CK: for what proportion the whole hath to the whole, the fame proportion hath the half to the half, which was, &c. And from this Axiom follow thefe Confectaries. 1. If the Angles of a Triangle be given, the reafon of the Sides is alfo given. And confequently, If one Side be given befides the Angles, both the other Sides are alfo given. 2. If two Sides of the Triangle be given, with an Angle oppofite to one of them, the Angle oppofite to the other of them is alfo given. CASE CASE I. The Angles and one Side being given, to find the other two Sides. In the Triangle ABC (in the Figure above) there are given the Angles CAB 62° 30', and CBA 37° 30′, and the Side AC 350; and it is required to find the Side BC. As the s. of the CBA 37° 30′ Ar. Com. 0.2155529 2.5440680 9.9479289 2.7075498 For the Side AB. As the s. of the CBA 37° 30' Ar. Com. 0.2155529 2.5440680 9.9933515 2.7529724 By Scale and Compaffes. For the first Proportion, extend the Compaffes from 37° 30′ to 62° 30' in the Sines; that extent will reach from 350 to 509.976 in the Line of Numbers. For the fecond Proportion, extend from 37° 30' to 80° oo' in the Sines; that extent will reach from 350 to 566.203 in the Numbers. By |