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, and let fall the perpendicular CH on the base. Let DG=a, DE=b, and DF c: put one of the equal sides AB = or therefore, A =2x; hence AH=x, and CH=√AC2_AH2=√4x2—xa √3x2=x√3. Now since the area of a triangle is equal to half its base into the altitude, (Bk. IV. Prop. VI.) AB × CH=x×x √3=x2 √3=triangle ACB AB×DG=x × α =αx =triangle ADB BCX DE=xxb =bx = triangle BCD ACX DF=xx c =cx =triangle ACD But the three last triangles make up, and are consequently equal to, the first; hence, x2√3=ax+bx+cx=x(a+b+c); F E PROBLEM VI. H G B REMARK. Since the perpendicular CH is equal to x √√3, ıt 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 all from any point within the triangle on the sides respectively. PROBLEM VII. In a right angled triangle, having given the base and the difference between the hypothenuse and perpendicular, to find the sides. In a right angled triangle, having given the hypothenuse and the difference between the base and perpendicular, to deter mine the triangle. PROBLEM VILL. 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, togeth er with both the segments of the base made by a perpendic ular 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. 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 circumference 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 nated by the characters: o,',": thus the expression 16° 6′ 15′′ represents an arc, or an angle, of 16 degrees, 6 minutes, and 15 seconds. III. The complement of an angle, or of an arc, is what remains after taking that angle or that are from 90°. Thus the complement of 25° 40' is equal to 90°-25° 40′-64° 20′; and the complement of 12° 4' 32" is equal to 90°-12o 4_32′′=77° 55′ 28′′. In general, A being any angle or any arc, 90°-A is the complement of that angle or arc. If any arc or angle be added to its complement, the sum will be 90. Whence it is evident that if the angle or arc is greater than 90°, its complement will be negative. Thus, the complement of 160° 34′ 10′′ is —70° 34' 10". In this case, the complement, taken positively, would be a quantity, which being subtracted from the given angle or arc, the remainder would be equal to 90°. The two acute angles of a right-angled triangle, are together equal to a right angle; they are, therefore, complements of each other. IV. The supplement of an angle, or of an arc, is what remains after taking that angle or are from 180°. Thus A being any angle or arc, 180°-A is its supplement. In any triangle, either angle is the supplement of the sum of the two others, since the three together make 180°. If any arc or angle be added to its supplement, the sum will be 180°. Hence if an arc or angle be greater than 180°, its supplement will be negative. Thus, the supplement of 200° is -20°. The supplement of any angle of a triangle, or indeed of the sum of either two angles, is always positive. |