Now, since the area or space of a rectangle, is expressed by the product of the base and height (cor. 2, th 81, Geom.), and that a triangle is equal to half a rectangle of equal base and height (cor. 1, th. 26), it follows that, the whole triangle ABC is = AB X CH = r × x √3 = x2 √3, the triangle ABD ABX DG = x X c = cr, -- the triangle BCD BC X DE x x a = ax, or are equal to, the that is, r3/3 = ax + bx + cx; hence, dividing by x, gives T√3 = a + b + c, and dividing by 3, gives a+b+c 1= √3 half the side of the triangle sought. Also, since the whole perpendicular cн is = x√3, it is therefore a+b+c. That is, the whole perpendicular CH, is just equal to the sum of all the three smaller perpendiculars DE+DF + DG taken together, wherever the point D is situated. PROBLEM VI. In a right-angled triangle, having given the base (3`, and the difference between the hypothenuse and perpendicular (1); to find both these two sides. PROBLEM VII. In a right-angled triangle, having given the hypothenuse (5), and the difference between the base and perpendicular (1); to determine both these two sides. PROBLEM VIII. HAVING given the area, or measure of the space, of a rectangle, inscribed in a given triangle; to determine the sides of the rectangle. PROBLEM IX. IN a triangle, having given the ratio of the two sides, together with both the segments of the base, made by a perpendicular from the vertical angle; to determine the sides of the triangle. PROBLEM X. In a triangle, having given the base, the sum of the other two sides, and the length of a line drawn from the vertical angle to the middle of the base; to find the sides of the triangle. PROBLEM XI. In a triangle, having given the two sides about the vertical angle, with the line bisecting that angle, and terminating in the base; to find the base. PROBLEM XII. To determine a right-angled triangle; having given the lengths of two lines drawn from the acute angles, to the middle of the opposite sides. PROBLEM XIII. To determine a right-angled triangle; having given the perimeter, and the radius of its inscribed circle. PROBLEM XIV. To determine a triangle; having given the base, the perpendicular, and the ratio of the two sides. PROBLEM XV. To determine a right-angled triangle; having given the hypothenuse, and the side of the inscribed square. PROBLEM XVI. To determine the radii of three equal circles, described in a given circle, to touch each other and also the circumference of the given circle. PROBLEM XVII. In a right-angled triangle, having given the perimeter, or sum of all the sides, and the perpendicular let fall from the right angle on the hypothenuse; to determine the triangle, that is, its sides. PROBLEM XVIII. To determine a right-angled triangle; having given the hypothenuse, and the difference of two lines drawn from the two acute angles to the centre of the inscribed circle. TO GEOMETRY. PROBLEM XIX. To determine a triangle; having given the base, the perpendicular, and the difference of the two other sides. PROBLEM XX. To determine a triangle; having given the base, the perpendicular, and the rectangle or product of the two sides. PROBLEM XXI. To determine a triangle; having given the lengths of three lines drawn from the three angles, to the middle of the oppo. site sides. PROBLEM XXII. In a triangle, having given all the three sides; to find the radius of the inscribed circle. PROBLEM XXIII. To determine a right-angled triangle; having given the side of the inscribed square, and the radius of the inscribed circle. PROBLEM XXIV. To determine a triangle, and the radius of the inscribed circle; having given the lengths of three lines drawn from the three angles, to the centre of that circle. PROBLEM XXV. To determine a right-angled triangle; having given the hypothenuse, and the radius of the inscribed circle. PROBLEM XXVI. To determine a triangle; having given the base, the line bisecting the vertical angle, and the diameter of the circum. scribing circle. 97 PLANE TRIGONOMETRY. DEFINITIONS. 1. PLANE TRIGONOMETRY treats of the relations and calculations of the sides and angles of plane triangles. 2. The circumference of every circle (as before observed in Geom. Def. 56) is supposed to be divided into 360 equal parts, called Degrees; also each degree into 60 Minutes, and each minute into 60 Seconds, and so on. Hence a se. micircle contains 180 degrees, and a quadrant 90 degrees. 3. The Measure of an angle (Def. 57, Geom.) is an arc of any circle contained between the two lines which form that angle, the angular point being the centre; and it is esti mated by the number of degrees contained in that arc. Hence, a right angle, being measured by a quadrant, or quarter of the circle, is an angle of 90 degrees; and the sum of the three angles of every triangle, or two right an gles, is equal to 180 degrees. Therefore, in a right-angled triangle, taking one of the acute angles from 90 degrees, leaves the other acute angle; and the sum of the two angles, in any triangle, taken from 180 degrees, leaves the third angle; or one angle being taken from 180 degrees, leaves the sum of the other two angles. D LH 4. Degrees are marked at the top of the figure with a small, minutes with', seconds with ", and so on. Thus, 57° 30′ 12", denote 57 degrees 30 minutes and 12 seconds. 5. The Complement of an arc, is what it wants of a quadrant or 90°. Thus, if AD be a quadrant, then BD is the complement of the arc AB; and, reciprocally, AB is the complement of BD. So that, if AB be an arc of 50°, then its complement BD will be 49°. 6. The Supplement of an arc, is what it wants of a semicircle, or 180°. Thus, if ADE be a semicircle, then E K F BDE is the supplement of the arc AB; and, reciprocally, AB is the supplement of the arc Bde. So that, if AB be an arc of 50°, then its supplement BDE will be 130°. 7. The Sine, or Right Sine, of an arc, is the line drawn from one extremity of the arc, perpendicular to the diameter which passes through the other extremity. Thus, BF is the sine of the arc AB, or of the supplemental arc BDE. Hence the sine (BF) is half the chord (BG) of the double arc (BAG). 8. The Versed Sine of an arc, is the part of the diameter intercepted between the arc and its sine. So, AF is the versed sine of the arc AB, and EF the versed sine of the arc EDB. 9. The Tangent of an arc, is a line touching the circle in one extremity of that arc, continued from thence to meet a line drawn from the centre through the other extremity; which last line is called the Secant of the same arc. Thus, AH is the tangent, and cir the secant, of the arc AB. Also, EI is the tangent, and cr the secant, of the supplemental are BDE. And this latter tangent and secant are equal to the former, but are accounted negative, as being drawn in an opposite or contrary direction to the former. 10. The Cosine, Cotangent, and Cosecant, of an arc, are the sine, tangent, and secant of the complement of that arc, the Co being only a contraction of the word complement. Thus, the arcs AB, BD, being the complements of each other, the sine, tangent, or secant of the one of these, is the cosine, cotangent, or cosecant of the other. So, BF, the sine of AB, is the cosine of BD; and BK, the sine of BD, is the cosine of AB in like manner, AH, the tangent of AB, is the cotangent of BD; and DL, the tangent of DB, is the cotangent of AB; also, CH, the secant of AB, is the cosec.nt of BD; and CL, the secant of BD, is the cosecant of ab. Corol. Hence several important properties easily follow from these definitions; as, 1st, That an arc and its supplement have the same sine, tangent, and secant; but the two latter, the tangent and secant, are accounted negative when the arc is greater than a quadrant or 90 degrees. 2d, When the arc is 0, or nothing, the sine and tangent are nothing, but the secant is then the radius CA, the least it can be. As the arc increases from 0, the sines, tangents, and secants, all proceed increasing, till the arc becomes a whole quadrant AD, and then the sine is the greatest it can be, being the radius CD of the circle; and both the tangent and secant are infinite. 3d, Of any arc AB, the versed sine AF, and cosine BK, or CF, together make up the radius ca of the circle.-The |