3. To find the width of a river, two points, A and C, are taken on one bank 100 feet apart. If B is the point on the other bank directly opposite C, and if CAB equals 72°.16, how wide is the river? 4. From the top of a lighthouse 35 feet high, the angle of depression of a ship is 11°.38. How far is the ship from the lighthouse? 5. What is the angle of elevation of the sun if a pole 49 feet high casts a shadow of 11 feet? 6. An iron wedge for splitting rails is to have a base two inches wide and a vertex angle of 15°. How long will each side be? 7. A rustic summer house, or shelter, is to be built with an octagonal floor, 6 feet on a side. Determine the amount of flooring necessary, making an allowance of 25% on account of the "tongue and groove” and for waste. 8. Solve Exercise 7 if the floor is to be pentagonal. Solve Exercise 7 if the floor is to be a regular seven-sided polygon. Can this Exercise be solved by plane geometry? 9. If the radius of a regular polygon of n sides is r, find (a) the perimeter in terms of n and r; (b) the area. 10. If a side of a regular polygon is a, express the area as a function of a, assuming n to be constant. 11. What is the angle of inclination of an ordinary gable roof, if its pitch (the ratio of its height to its entire width) is ?? ? 12. A point P moves with uniform speed around a circle of radius one foot, making a complete revolution every 36 seconds. The projection M of P on a diameter AOB, moves along the diameter. Find OM for the values 50°, 60°, Find the average velocity increases from 50° to 60° and from P 70°, if 0 = LAOP. B A M FIG. 104. of M as 60° to 70°. Find the average velocity of M as increases from 59° to 60° and from 60° to 61°. From 59°.9 to 60° and from 60° to 60°.1. Approximately what is the velocity of M at the instant when 0 = 60°? 13. Two life saving stations are 10 miles apart on a beach running 22°.5, east of north. A lightship anchored off the beach lies 33°.75 north of east from one station, and 11°.25 east of south from the other. How long would it take a boat to run from one station to the other if its speed is 10 miles per hour, and if it passes outside the lightship? How long would it take the boat to go from the ship to the nearest point on the beach? 14. (a) Two sides of a triangle, not a right triangle, are 12 and 20, and the included angle is 28°.48. Find the altitude on the latter side and the area. (b) Find the area of any triangle ABC in terms of the sides b and c and the included angle A. 15. Find the area of a parallelogram in terms of two adjacent sides and the included angle. 16. Two ships are on a line with a lighthouse, which is 40 feet high. At the top of the lighthouse, the angles of depression of the ships are respectively 4°.26 and 6°.31. How far apart are the ships? 17. A regular octagonal tower is 4 feet on each side. The roof is 10 feet high. How long should the rafters be? 18. A bridge is to be built across a ravine between two points A and B on the same level. Two stations C and D are chosen in the ravine which lie in the vertical plane through A and B. At A, the angle of depression of C is 42°.37, and AC 50 feet. At C the angle of depression of D is 4°.52, and CD 40 feet. At D the angle of elevation of B is 37°.89, and Find the length of the bridge. DB = 60 feet. 19. Check the accuracy of the measurements in Exercise 18, by finding the difference in altitude of A and D, and also of B and D. 20. To find the width of a river, a tree is selected on one shore. At a point 50 feet from the tree, the angle of elevation of the top is 48°.72. At the point on the other shore directly opposite the tree, the angle of elevation of the top is 17°.39. How wide is the river? 64°.37870.24 Ă 1000 ft. B FIG. 105. D C 21. From a station A the horizontal angle between a mountain top C and a second station B, 1000 feet from A and at the same altitude, is 64°.37. At B the horizontal angle between the mountain top and A is 90°, and the angle of elevation of the mountain top is 37°.24. How high is the mountain? 66. Parallelogram Law - Velocities, Accelerations, Forces. The law to be considered in this section is illustrated in B EXAMPLE 1. The current in a river flows at the rate of 2 miles an hour. A man rows across at the rate of 4 miles an hour, keeping his boat at right angles to the shore. Show that the boat moves in a straight line. Find how fast it moves, and in what direction. Take the starting point for the origin of a system of coördinates, and let the x-axis lie along the bank of a river. At the time t, the boat will be at a point P, whose coördinates give the distance the boat has been carried by the current, x = 2t, 2t, and the distance the man has rowed from shore, y = 4t. Eliminating t, we AM obtain y = 2x, which is true for all values of t. Hence the boat moves in a straight line (Corollary 1, page 57). Let C denote the position of the boat one hour after starting, and let OA and OB be the coördinates of C. The line OA represents the velocity of the water with reference to the earth, the direction of the line being that in which the water flows, and the length of the line being the number of miles per hour the water moves. Similarly, OB represents, in direction and magnitude, the velocity of the boat with reference to the water. (If the water were at rest, OB would represent the actual motion of the boat, in one hour, with reference to the earth.) With reference to the earth, the boat moves along the line OC, and in one hour moves from 0 to C. Hence OC represents, in direction and magnitude, the actual, or resultant, velocity of the boat. The resultant velocity is readily computed. Its magnitude is OC = √ÕA2 + AC2 = √22 + 42 Its direction may be given by ZAOC. 4.47 miles per hour. Since tan ZAOC = AC/OA = 4/2 = 2, we have ZAOC = 63°.43. This illustration exemplifies the law known as the Parallelogram of velocities. If a body is subjected to two different velocities, represented in direction and magnitude by two lines OA and OB, the actual or resultant velocity is represented by the diagonal OC of the parallelogram determined by OA and OB. If a body is moving through the air in any way, it has an acceleration of 32 feet per second per second directed toward the center of the earth (see Section 22, page 63). This means that second by 32 feet its vertical velocity is increased or decreased each per second, according as it is falling or rising. Its horizontal velocity is uniform, and is not affected by the action of gravity. No account is taken here of the resistance of the air, which would make the discussion very much more complicated. The acceleration due to gravity may be represented by a vertical line running downward whose length is 32. In like manner, any acceleration may be represented, in direction and magnitude, by a line. A force may also be represented, in direction and magnitude, by a line OA, O representing the point of application of the force. Accelerations and forces may also be combined according to the parallelogram law. Finding the resultant of two velocities is known as the composition of velocities, the two given velocities being called components of the resultant. The converse problem of determining two component velocities which have a given resultant is called the resolution of velocities. The same terms are also used with reference to accelerations and forces. If the components are at right angles, the parallelogram is a rectangle, and the composition or resolution may be effected by solving a right triangle. If the parallelogram is not a rectangle, it is necessary to use the methods to be developed in Sections 71-73. Problems of this sort will be found in the exercises following Section 73. EXAMPLE 2. A ball rolls down a plane whose inclination is 30°. Resolve the acceleration due to gravity into two components parallel and perpendicular to the plane. Let OC = 32 represent the acceleration due to gravity. Through O and C draw lines parallel and perpendicular to the plane DE, forming the rectangle OABC. Then OA and OB represent the required components, for the resultant of OA and E B F FIG. 107. As there is no motion in a direction perpendicular to the plane, the component OB is neutralized by the plane. The component parallel to the plane, OA 16 feet per second per second, gives approximately the effective acceleration with which the ball rolls down the plane. If the ball starts to roll from rest, how fast will it be moving at the end of one second? At the end of two seconds? At the end of four seconds? If a ball is rolled up the plane, how will its velocity be affected during any second? If it is started up with a velocity of 50 feet per second, how long will it roll up the plane? 67. Conditions of Equilibrium of a Particle. In measuring a force we shall use the pound as the unit. If a number of forces act on a particle, which we take as the origin of a system of coördinates, each of the forces may be resolved into two components, one acting along each axis. If the particle is in equilibrium, there is no motion in the direction of the x-axis, and hence the algebraic sum of the components along that axis must be zero. Similarly, the sum of the components along the y-axis must be zero. And conversely, if each of these sums is zero the particle must be in equilibrium. Hence we have the Theorem. A particle is in equilibrium under the action of any number of forces if and only if the sum of the components in each of two perpendicular directions is zero. In applying this theorem, first determine all the forces acting on the particle, and then choose the perpendicular directions. If two of the forces are at right angles to each other, choose these directions. EXAMPLE 1. A particle weighing 4 ounces is supported on a smooth plane whose inclination is 30° by a cord parallel to the plane. Find the tension of the cord and the pressure on the plane. 30° R 4 sin 30 FIG. 108. 130 W 30 4 cos 30° Τ are The forces acting on the particle (1) Its weight, W = 4, acting vertically. (2) The tension of the cord, T, acting parallel to the plane. (3) The resistance of the plane, R, acting perpendicular to the plane. Since R and T act at right angles, we proceed to resolve all the forces into components parallel and perpendicular to the plane. Choosing these directions facilitates the work, because W is then the only force whose components must be determined. By the parallelogram law and the solution of a right triangle, the component of W parallel to the plane is found to be 4 sin 30°, and that perpendicular to the plane is 4 cos 30°, both acting downward. Then by the theorem, taking the directions of R and T as positive, we have |