Ans. AC 645 Required the hypothenuse and the perpendicular. 2. In the right angled plane triangle ABC, Given { LA 56°.48' Base AB 53.66 Ans. {AC=98 BC 82.01 Required the hypothenuse and the perpendicular. (S) CASE III. Given the angles and perpendicular, to find the base and hypothenuse. Given { The angle A=53°.8′ Required the hypothenuse BY CONSTRUCTION. Draw the line AB of a sufficient length, at any point B erect the perpendicular BC (N. 26.) which make equal to 384 by a scale of equal parts, at c make the angle ACB=36°.52' (R. 27.), the complement of A, from cdraw the hypothenuse, and it will cut the base AB in the point a. Then will ac measure 480, and AB 288. BY CALCULATION. I. The hypothenuse radius, BC will be the sine of a, and ab the cosine thereof. II. The base AB radius, BC will be the tangent of a, and ac the secant thereof. III. The perpendicular BC radius, the base AB will be the tangent of c, or the co-tangent of A; AC will be the secant of C, or co-secant of A. (T) BY GUNTER'S SCALE. 1. Extend the compasses from 53°.8′ to 90° on the line of sines, that extent will reach from 384 to 480, the hypothenuse on the line of numbers. 2. Extend from 53° 8'.to 36°.52', the complement of the angle A, on the line of sines, that extent will reach from 384 to 288 on the line of numbers. PRACTICAL EXAMPLES. 1. In the right angled plane triangle ABC, 439°10′ Given Perp. BC=407.37 Required the hypothenuse and the base. 2. In the right angled plane triangle ABC, LA=56°.48' Given Perp. BC=82.01 Required the hypothenuse and the base. (U) CASE IV. Given the hypothenuse and base, to find the angles and the perpendicular. Given The hypothenuse AC=480 S Required the angles a BY CONSTRUCTION. Draw the base AB equal to 288 from a scale of equal parts, at в erect the perpendicular BC (N. 26); with the distance AC=480, taken from a scale of equal parts, cross BC in the point c. Then BC measured by a scale of equal parts will be 384, and the angles a and c measured by a scale of chords (S. 27.) will be 53°.8' and 36°.52. BY CALCULATION. A C I. The hypothenuse radius, BC will be the sine of the angle A, and AB the cosine thereof. II. The base radius, BC will be the tangent of A, and ac the secant thereof. NOTE. The side BC may be found (by Euclid 47 of 1.) thus, AC-AB2BC=384. (W) From the examples hitherto given, it appears that when we want to find a side, the proportions begin with a given angle; and in the first three cases all the angles are given, therefore any side may be considered as the radius of a circle. But when we want to find an angle, the proportions begin with a given side; and as only two sides are given at once, it follows that these two given sides only can be considered as the radii of circles. (X) BY GUNTER's scale. 1. Extend the compasses from the hypothenuse 480 to the base 288 on the line of numbers, that extent will reach from 90° to 36°.52′, on the line of sines, the complement of the angle A. 2. Extend the compasses from 90° to 53°.8 on the line of sines, that extent will reach from the hypothenuse 480 to BC the perpend. 384. PRACTICAL EXAMPLES. 1. In the right angled plane triangle ABC, Given { Hypoth, AC÷645 Base AB 500 LA 390.10'.: Ans. Lc=50°.50' BC407.37 Required the angles and the perpendicular. 2. În the right angled plane triangle ABC, LA 56°.48" Ans. Lc=33°.12′′ Required the angles and the perpendicular. BC=82.01 (Y) CASE V. Given the hypothenuse and perpendicular, to find the angles and the base. Given { The hypothenuse AC=480 Required the angles a BY CONSTRUCTION. Draw the base AB of an indefinite length, at B erect the perpendicular BC (N. 26.) which make equal to 384 by a scale of equal parts; take AC 480 from the same scale, with this extent in your compasses and centre c cross the base ab in Then AB measured by a scale of equal parts, will be 288, and the angles a and c measured by a scale of chords (S. 27.) will be 53°.8'. and 36°. A. 52'. E B BY CALCULATION. I. The hypothenuse radius, BC will be the sine of the angle ▲, and AB the cosine thereof. II. The perpendicular BC radius, AB will be the tangent of c, or the co-tangent of A, and AC will be the secant of c or co-secant of a. Or AB may be found thus, AC-BC2=AB=288. (Z) BY GUNTER's scale. 1. Extend the compasses from the hypothenuse 480 to the perpendicular 384 on the line of numbers, that extent will reach from 90° to 53°. 8' on the line of sines. 2. Extend the compasses from 90° to 36°. 52', the complement of A, on the line of sines, that extent will reach from 480 to 288 on the line of numbers. PRACTICAL EXAMPLES. 1. In the right angled plane triangle ABC, Given {Hypoth. Ac=645 Perp. BC 407.37 Required the angles and the base. LA=39°.10' Ans. 4c 50°.50' AB = 500 2. În the right angled plane triangle ABC, Given {Perpen. BC=82-01 Required the angles and the base. Ans. LA 56°.48' AB 53.66. (A) CASE VI. Given the base and perpendicular, to find the angles and the hypothenuse. Given {} The base AB=288 Required the angles A and C, The perpend. BC=384 J and hypothenuse ac. BY CONSTRUCTION. Make the base AB equal to 288 by a scale of equal parts, at в erect the perpendicular BC, which make equal to 384 from the same scale; join AC. Then AC measured by a scale of equal parts will be 480, and the angles A and c measured by a scale of chords (S. 27.) will be 53°. 8' and 36°. 52'. Ο C BY CALCULATION. f Ì. The base radius, BC will be the tangent of the angle A, and AC will be the secant thereof. of C, II. The perpendicular radius, AB will be the tangent or the co-tangent of a; and ac will be the secant of c, or the co-secant of a. 1. Extend the compasses from 384 to 288 on the line of numbers, that extent will reach from 45° to 53°.8' on the line of tangents. 2. Extend the compasses from 53°.8 to 90° on the line of sines, that extent will reach from 384 to 480 on the line of numbers. PRACTICAL EXAMPLES. 1. In the right angled plane triangle ABC, Ans. ▲ c=50°.50' Required the angles and the hypothenuse. 2. In the right angled plane triangle ABC, |