373 the equilateral triangle. Then is AC or BC= 2x, and by right angled triangles the perpendicular CHAC2-AH2 =√/4x3-x2=√3x2=x√/3. 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 = ABX CH = x X x √ 3 = x2 √3, the triangle ABD = ABX DG = xx c = cx, the triangle BCD = BCX DE = x× a = ax, the triangle ACD = ACX D = = x× b = bx. But the three last triangles make up, or are equal to, the whole former, or great triangle ; that is, x2√3 = ax + bx + cx; hence, dividing by x, gives x √3 = a+b+c, and dividing by a+b+c half the side of the triangle sought. , 13 3, gives x3, it is Also, since the whole perpendicular CH 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 ; 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 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 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. 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 opposite 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 PROBLEM XXIV. To determine & 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 à Triangle; having given the Base, the Line bisecting the Vertical Angle, and the Diameter, of the Cir cumscribing Circle. PLANE PLANE TRIGONOMETRY, 1. DEFINITIONS. 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. 57) is supposed to be divided into 360 equal parts, called Degrees; also each degree into 60 Minutes, each minute into 60 Seconds, and so on. Hence a semicircle contains 180 degrees, and a quadrant 90 degrees. 3. The measure of an angle (Def. 58, 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 estimated 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 angles, is equal to 180 degrees. Therefore, in a right-angled triangle, taking one of the acute angles from 90 degrees, leaves the the other acute angle; and the sum of 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. 4. Degrees are marked at the top of the figure with a small, minute 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 compliment of the arc AB; and, reciprocally, AB is the compliment of BD. So that, if AB be an arc of 50°, then its complement BD will be 40°. 6. The Supplement of an arc, is what it wants of a semicircle, or 180°. E I D K 10 A F Thus, if ADE be a semicircle, then BDE is the supplement of the arc AB; and, reciprocally, AB is the supplement of the arc EDE. So that, if AB be an arc of 50°, then its supplement BDE will be 130°. H. The |
