EXERCISE XVI. Solve the following triangles, taking the three sides as the 14. Given a 16. Given a = 2, 6, c=10; 6, c=6; find the angles. find the angles. 5, c = 12; find the angles. b = √6, c=√3−1; find the angles. b = √6, с √3+1; find the angles. 17. The distances between three cities A, B, and C are as follows: AB=165 miles, AC=72 miles, and BC= 185 miles. B is due east from A. In what direction is C from A? What two answers are admissible? 18. Under what visual angle is an object 7 feet long seen. by an observer whose eye is 5 feet from one end of the object and 8 feet from the other end? 19. When Formula [28] is used for finding the value of an angle, why does the ambiguity that occurs in Case II. not exist? 20. If the sides of a triangle are 3, 4, and 6, find the sine of the largest angle. 21. Of three towns A, B, and C, A is 200 miles from B and 184 miles from C, B is 150 miles due north from C; how far is A north of C? . 844. AREA OF A TRIANGLE. If F denote the area of the triangle ABC (Fig. 30 or 31, page 50), then, by Geometry, And, in like manner, Fab sin C and Fbc sin A. [33] That is: The area of a triangle is equal to half the product of two sides and the sine of the included angle. By Formula [33] the area of a triangle may be found directly when two sides and the included angle are given; in the other cases the formula may be used when these parts have been computed. When the three sides of a triangle are given, as in Case IV., a formula for its area may be found as follows: By § 33, sin B = 2 sin 1 BX cos B. By substituting for sin B and cos B their values in terms of the sides given in § 43, sin B 2 ас ̧√s ( s − a) ( s − b ) (s — c). By substituting this value of sin B in [33], F= √s (s a) (s—b) (s—c). [34] If R denote (as in § 36) the radius of the circumscribed circle, we have, from § 36, By substituting this value of sin B in [33], If r denote the radius of the inscribed circle, and we divide the triangle into three triangles by lines from the centre of this circle to the vertices, the altitude of each of the three triangles is equal to r. Therefore, Fr(a+b+c) = rs. [36] By substituting in this formula the value of F given in [34], whence, in [31] § 43, is equal to the radius of the inscribed 1. Given a 4474.5, b = 2164.5, C=116° 30′ 20′′. 12. Obtain a formula for the area of a parallelogram in terms of two adjacent sides and the included angle. -- 13. Obtain a formula for the area of an isosceles trapezoid in terms of the two parallel sides and an acute angle. 14. Two sides and included angle of a triangle are 2416, 1712, and 30°; and two sides and included angle of another triangle are 1948, 2848, and 150°; find the sum of their areas. 15. The base of an isosceles triangle is 20, and its area is 100÷√3; find its angles. EXERCISE XVIII. 1. From a ship sailing down the English Channel the Eddystone was observed to bear N. 33° 45′ W.; and after the ship had sailed 18 miles S. 67° 30′ W. it bore N. 11° 15' E. Find its distance from each position of the ship. 2. Two objects, A and B, were observed from a ship to be at the same instant in a line bearing N. 15° E. The ship then sailed north-west 5 miles, when it was found that A bore due east and B bore north-east. Find the distance from A to B. 3. A castle and a monument stand on the same horizontal plane. The angles of depression of the top and the bottom of the monument viewed from the top of the castle are 40° and 80°; the height of the castle is 140 feet. Find the height of the monument. 4. If the sun's altitude is 60°, what angle must a stick make with the horizon in order that its shadow in a horizontal plane may be the longest possible? 5. If the sun's altitude is 30°, find the length of the longest shadow cast on a horizontal plane by a stick 10 feet in length. 6. In a circle with the radius 3 find the area of the part comprised between parallel chords whose lengths are 4 and 5. (Two solutions.) 7. A and B, two inaccessible objects in the same horizontal plane, are observed from a balloon at C and from a point B directly under the balloon, and in the same horizontal plane with A and B. If CD = 2000 yards, ▲ ACD = 10° 15′ 10′′, ▲ BCD=6° 7′ 20′′, ▲ ADB = 49° 34′ 50′′, find AB. 8. A and B are two objects whose distance, on account of intervening obstacles, cannot be directly measured. At the summit C of a hill, whose height above the common horizontal plane of the objects is known to be 517.3 yards, ▲ ACB is found to be 15° 13′ 15′′. The angles of elevation of C viewed from A and B are 21° 9′18′′ and 23° 15' 34" respectively. Find the distance from A to B. MISCELLANEOUS PROBLEMS. [Selected by permission from "Problems in Plane Trigonometry," prepared by Prof. C. J. White, of Harvard College, and published by Charles W. Sever, Cambridge.] 1. The angular distance of any object from a horizontal plane, as observed at any point of that plane, is the angle which a line drawn from the object to the point of observation makes with the plane. If the object observed be situated above the horizontal plane (that is, if it is farther from the earth's centre than the plane is), its angular distance from the plane is called its angle of elevation. If the object be below the plane, its angular distance from the plane is called its angle of depression. These angles are evidently vertical angles. If two objects are in the same horizontal plane with the point of observation, the angular distance of one object from the other is called its bearing from that object. If two objects are not in the same horizontal plane with cither each other or the point of observation, we may suppose vertical lines to be passed through the two objects, and to meet the horizontal plane of the point of observation in two points. The angular distance of these two points is the bearing of either of the objects from the other. It may also be called the horizontal distance of one object from the other. NOTE. "Problems in Plane Trigonometry" can be obtained in pamphlet form of Charles W. Sever, Cambridge, Mass. |