It is now required to express all the other lines in terms of OX and trigonometric functions of a and B. 1. By the same process express OQ, QN, OX, NX, NP, and XP in terms of ON and trigonometric functions. 2. Express the same quantities in terms of NP. 3. Express NX separately in terms of OX and XP, and by multiplying the two values prove the geometric theorem that NX is a mean proportional between ОX and XP. 4. In a right triangle the sides which contain the right angle are a and b, (a > b), and ♂ is the difference of the angles at the base (hypothenuse). Express the length of the perpendicular from the vertex upon the base in two ways, and the lengths of the segments into which it divides the base each in one way. The expressions are all to be in terms of a, b and 6. Ans. (in part). One expression for the perpendicular is pa sin (45° — 6). Solution of Right Triangles. Since in a right triangle one angle-the right angle—is given, only two other independent parts are required to solve the triangle. These two parts may be any two of the sides or one side. and one angle. What parts soever are given, the remaining parts may be found by the equations (4) and (5). The following are all the essentially different cases. 37. CASE I. Given the two sides a and b adjacent to the Therefore the quotient of the two sides gives the tangent of the angle opposite the dividend side. From the tangent the angle a is itself found by the trigonometric tables; then sec a or cos a; then the hypothenuse c from the equation cos a Example. Given a 9 metres, 6 12 metres, to find the = b = remaining parts of the triangle. Solution by numbers and measurement. tan α = 12 = 0.75. On the tangent line XN (§22) measure a distance from X equal to 0.75 of the radius OX; join the end of the distance to⚫0, and measure the angle XON which the joining line makes with OX. This angle will be a. The length of the joining line divided by OX will be the secant, which multiplied by b = 12 will give the hypothenuse c. *It is recommended that in commencing this subject the student first solve a few of the problems by his own process, and without the use of any tables but those he may construct for himself by measurement as described in § 18. The first two exercises are made purposely simple, that they may be performed by measurement. 38. CASE II. Given the hypothenuse and one side. Solution. From the equation which may be used to find a when a and c are given. Then the remaining side is found by the equation = b = c cos a. Example. Given a 13, c = 20, to find the remaining parts. Solution by numbers and measurement. iarly to OB, take its ratio to OX, and multiply it by c=5. This Ρ 5. A circle of radius r is drawn with its centre at a distance from a straight line. What length will it cut out from the line? What will be the result if r <p? 39. CASE III. Given an angle and one side, as a and a. NOTE. One angle being given, the remaining one may be found from the equation B = 90° -α. 5. An engineer, desiring to find the distance from a point A on one a†- bank of a river to a point P on the other bank, measured off a base line ACb yards in length, in a direction perpendicular to AP. He then measured the angle ACP, and found 1 1 C 66 P it to be a. What are the expressions for the distances AP and CP? What will be the distance if AC = 80 yards and a = 85° 22′.5 ? 6. An engineer, desiring to determine the height of a vertical wall, measured off a distance P0 = b feet on level ground, and Method of solution. Find the height ПX and add HP = a. 7. Desiring to find the height of an inaccessible rock M above angle of elevation and found it to be y. Express the height BM of the rock. Method of solution. Let the vertical height BM= h and Then we have the two right triangles PBM and QBM, which give QB = x. h = (a + x) tan a; From these two equations we obtain the following values of h Find the height when a 2000 yards, a = 30° 28', and y = 40° 53'. 8. The altitude of a triangle is 7.2648, and the angles at the base are 72° 29.3′ and 40° 30.5′ respectively. Compute the base |