PROBLEM III. In a rectangle, having given the diagonal and perimeter, to find the sides. Let ABCD be the proposed rectangle. D x2 + y2=d? А. y=}a+ Vida?=8 or 6, and x=aF Vžda?=6 or 8. PROBLEM IV. Having given the base and perpendicular of a triangle, to find the side of an inscribed square. Let ABC be the triangle and HEFG the inscribed square. Put AB=b, CD=a, and HE or GH=x: then CI=2-X. TA А H DE B В b:@:: 3:03 Hence, abbx=ax ab or = the side of the inscribed square ; a+b which, therefore, depends only on the base and altitude of the triangle. or X PROBLEM V. In an equilateral triàngk, having given the lengths of the three perpendiculars drawn from a point within, on the three sides: to determine the sides of the triangle. Let ABC be the equilateral triangle; DG, DE and DF the given perpendiculars let fall from D on the sides. Draw DA, DB, DC to the vertices of the angles, F and let fall the perpendicular CH on DE the base. Let DG=a, DE=b, and DF=c: put one of the equal sides AB A H G B В =2x; hence AH=x, and CH=VĀC2_AH'=4x?_22 =V3x?=xV3. Now since the area of a triangle is equal to half its base into the altitude, (Bk. IV. Prop. VI.) AB CH=Xxx V3=z? V3=triangle ACB =triangle ADB =triangle BCD ZAC x DF=XXC =triangle ACD But the three last triangles make up, and are consequently equal to, the first ; hence, 22 V3=ax+bx+cx=x(a+b+c); x V3=a+b+c therefore, a+b+c V3 or REMARK. Since the perpendicular CH is equal to x V3, it is consequently equal to a+b+c: that is, the perpendicular let fall from either angle of an equilateral triangle on the opposite side, is equal to the sum of the three perpendiculars let fall from any point within the triangle on the sides respectively. PROBLEM VI. In a right angled triangle, having given the base and the difference between the hypothenuse and perpendicular, to find the sides. PROBLEM VII. In a right angled triangle, having given the hypothenuse and the difference between the base and perpendicular, to determine the triangle. PROBLEM VIII. Having given the area 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 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, together 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 the 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 within and tangent to, a given circle, and also tangent to each other. PROBLEM XVII. In a right angled triangle, having given the perimeter and the perpendicular let fall from the right angle on the hypothenuse, to determine the triangle. 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. 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 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 opposite sides. PROBLEM XXII. In a triangle, having given 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 right angled triangle, having given the hypothenuse and radius of the inscribed circle PROBLEM XXV. To determine a triangle, having given the base, the line bisecting the vertical angle, and the diameter of the circumscribing circle. 1 PLANE TRIGONOMETRY. In every triangle there are six parts: three sides and three angles. These parts are so related to each other, that if a certain number of them be known or given, the remaining ones can be determined. Plane Trigonometry explains the methods of finding, by calculation, the unknown parts of a rectilineal triangle, when a sufficient number of the six parts are given. When three of the six parts are known, and one of them is a side, the remaining parts can always be found. If the three angles were given, it is obvious that the problem would be indeterminate, since all similar triangles would satisfy the conditions. It has already been shown, in the problems annexed to Book III., how rectilineal triangles are constructed by means of three given parts. But these constructions, which are called graphic methods, though perfectly correct in theory, would give only a moderate approximation in practice, on account of the imperfection of the instruments required in constructing them. Trigonometrical methods, on the contrary, being independent of all mechanical operations, give solutions with the utmost accuracy. These methods are founded upon the properties of lines called trigonometrical lines, which furnish a very simple mode of expressing the relations between the sides and angles of triangles. We shall first explain the properties of those lines, and the principal formulas derived from them; formulas which are of great use in all the branches of mathematics, and which even furnish means of improvement to algebraical analysis. We shall next apply those results to the solution of rectilineal triangles. DIVISION OF THE CIRCUMFERENCE. I. For the purposes of trigonometrical calculation, the cir cumference of the circle is divided into 360 equal parts, called degrees; each degree into 60 equal parts, called minutes; and each minute into 60 equal parts, called seconds. The semicircumference, or the measure of two right angles, contains 180 degrees; the quarter of the circumference, usually denominated the quadrant, and which measures the right angle, contains 90 degrees. II. Degrees, minutes, and seconds, are respectively desig |