. =na? +62 =a 1+C If a? Or, having found the angles, the hypothenuse may be found by Case II. or by Case III. SOLUTIONS OF THE DIFFERENT CASES OF OBLIQUE ANGLED PLANE TRIANGLES. sine B. a a CASE II. Given two sides and an angle opposite to one of them, to find the other angles, and the third side. sine A.O SOLUTION. Sine B= ; sine A = sineca b sine A sine c sine B.C. .C sine c.b ; sine as sinec B•C; sineb= a 6 NOTE. If the given – be obtuse; or if the side opposite to the given be the greater of the two given sides, then the L sought is always acute: in every other case it is ambiguous. Two angles being now given, their sum deducted from 180°, leaves the third 2; hence the remaining side may be found by Case I. Case III. Given two sides and the angle contained between them, to find the other angles; and the third side. anb SOLUTION. Tang ) (A - B) = cot į c (E. 36.) at 6 Then : (A - B)+(90°-C)= Lopposite to the greater side. And 900-10)-1(AB)= Lopposite to the less side. OR THUS, b:a::r : tang 4, and if a be the greater of the two given sides, will be greater than 45o. 3. : tang (°-45°)::cotc: tang } (A-B). ber OR THUS, sine con ANGE COS C COS C. a sine c.6 Tang B= ; tang A= 6 a 7 The remaining side may be found from the angles, by Case I. 19 CASE IV. Given the three sides, to find the angles. e B Solution. Cosļc=rV/1b+a+c).}(6+a+c) - $(b+a+0.06+ a + c)=f(11.37.) Tang {c=rv5(6+a+c)-a.}(b+a+c)-6 =6 one (I. 38.) (6+a+c).} (b+a+c)-c CHAP. II. 1. PRACTICAL RULES FOR THE SOLUTION OF ALL THE DIF FERENT CASES IN RIGHT ANGLED PLANE TRIANGLES, WITH (M) In every right angled plane triangle there must be two given quantities, one of which must be a side, (together with the right angle,) to find the rest. I. TO FIND A SIDE. Call any one of the sides of the triangle radius, and write upon it the word radius; observe whether the other sides be them accordingly. Call the word written upon each side the name of that side. Then say, As the name of the given side, is to the given side; so is the name of the required side, to the required side. II. TO FIND AN ANGLE. Call either of the given sides of the triangle radius, and write upon it the word radius; observe whether the other sides become sines, tangents, or secants, and write these words on them accordingly. Call the word written upon each side the name of that side. Then say, As the side made radius, is to radius; so is the other given side to its name. The angles being found, the remaining side must be found by the first part of this rule. (N) OR; IN ANY RIGHT ANGLED TRIANGLE. I. The sine of any angle is to the side opposite to it, as the sine of any other angle is to its opposite side. And, Any side is to the sine of its opposite angle, as any other side is to the sine of its opposite angle. II. Radius is to the tangent of either of the acute angles, as the side adjacent to that angle is to the side opposite to it. And, The side adjacent to either of the acute angles is to the side opposite to that angle, as radius is to the tangent of that angle. III. Radius is to the co-tangent of either of the acute angles, as the side opposite to that angle is to the side adjacent to it. Also, the side opposite to either of the acute angles is to the side adjacent to that angle, as radius is to the co-tangent of that angle. (O) CASE I. Given the angles* and the hypothenuse, to find the base and perpendicular. The hypothenuse ac=480) Required the base AB G A=53°.8') and perpendicular Bc. BY CONSTRUCTION. Draw the line AB of any length, make the angle CAB=539.8'(R.27) by a scale of chords, draw the hypothenuse Ac=480 from a scale of equal parts, from c let fall the perpendicular CB (P. 26) then abc is the triangle required. AB measured by the same scale of equal parts, which the hypothenuse was measured by, will be 288, and BC will be 384. A4 'B * The given parts of a triangle are generally marked with a dasă 1, and the required parts with an 0, as in the figures annexed to the constructions. BY CALCULATION. I. The hypothenuse radiús, BC will be the sine of the angle A, and as the cosine. To find Bc. To find AB. radius, sine of 90°, 10-00000 || radius, sine of 90°, 10.00000 : hypoth. Ac=480 2.68124 : hypoth. Ac=480 2.68124 ; : sine LA=530.8' 9.90311 : : cosine LA 530.81 9.77812 : perpendicular Bc=384 2.58435 : base AB - 288 2.45996 II. The base radius, bc will be the tangent of the angle A, and ac will be the secant thereof. To find Bc. To find AB. secant LA=539.8' 10.22188 secant LA=53o.8' 10.22188 : hypoth. Aca 480 2.68124 : hypoth. AC=480 2.68124 : : tang. LA=530.8° 10.12499 : : radius, sine of 90°, 10.00000 : perpendicular BC=384 2.58435 : base AB= 288 2.45936 III. The perpendicular radius, the base AB will be the tangent of the angle c, or co-tangent of A, and the hypothenuse AC will be the secant of c, or the co-secant of A. "To find BC. To find as. co-secant LA= = 59°.8' 10.09689 co-secant LA=53o.8' 10.09689 : hypoth. AC=480 2•68124 || : bypoth. AC=480 2-68124 : : radius, sine of 90° 10-00000 :: co-tangent <A = 530.8! 9.87501 : perpendicular BC=384 2.58435 : base AB=288 2.45936 (P) BY GUNTER'S SCALE. In working the several cases by Gunter's scale, we shall always suppose the hypothenuse radius, (where it can be done,) being the most simple of the three. 1. Extend the compasses from 90° to 53'.8' on the line of sines, that extent will reach from 480 to 384 on the line of numbers. 2. Extend the compasses from 90° to 360.52' the complement of the angle A, that extent will reach from the hypothenuse AC=480, to the base AB=288, on the line of numbers. PRACTICAL EXAMPLES. 1. In the right angled plane triangle ABC, Given { Hypoth. Ac=645 AB=500 Ans. { BC=407037 Sab=53•66 Ans. { BC=82:01 (Q) Case II. Given the angles and the base, to find the hypothenuse and the perpendicular. The angle A=539.8' Required the hypoth. AC, Given {The base AB=288 BY CONSTRUCTION. Draw the base ab, which make=288, from a scale of equal parts, at B erect the perpendicular BC (N. 26), make the angle A=539.8'. (R. 27), and draw the hypothenuse ac to cut the perpendicular Bc in the point c. Then will ac mea.sure 480, and BC 384. В BY CALCULATION. 1. The hypothenuse radius, BC will be the sine of the angle A, and as the cosine. To find ac. To find BC. cosine LA=539.8' 9.77812 | cosine LA=530.8' 9677812 : base AB= 288 2.45939|| : base ab=288 mm 2-45939 : : radius, sine of 90° 10.00000 : : sine La=539.8' 9.90311 : hypoth. AC=480 2.68127 : perpend. BC =384 2.58438 II. The base AB radius, Bc will be the tangent of A, and ac the secant thereof. To find ac. To find BC. radius 10.00000 | radius 10.00000 : base AB=288 2.45939 : base AB = 288 2.45939 : : secant LA=53°.8 10.22188 10:12499 : hypoth. AC-480 2:68127 2.58438 III. The perpendicular Bc radius, AB will be the tangent of C, or co-tangent of , and ac will be the secant of c, or cosecant of a. To find ac. co-tangent LA=539.8' To find BC. 530.8 9.87501 2.45939 10.00000 9.58438 (R) BY GUNTER'S SCALE. 1st. Extend the compasses, from 36o.52' the complement of A to 90, on the line of sines; that extent will reach from 288 to 480=AC on the line of numbers. 2d. Extend from 360.52 to 53o. 8' on the line of sines; that extent will reach from 288 to BC=384 on the line of numbers. |