EXAMPLES FOR EXERCISE. 1. Required the height of a wall, whose angle of elevation is observed, at the distance of 463 feet, to be 16° 21' ? Answer, 135'8 feet. 2. The angle of elevation of a hill is, near its bottom, 31° 18', and 214 yards farther off, 26° 18', required the perpendicular height of the hill, and the distance of the perpendicular from the first station? Answer, the height of the hill is 565.2, and the distance of the perpendicular from the first station is 929.6. 3. The wall of a tower which is 149.5 feet in height, makes, with a line drawn from the top of it to a distant object on the horizontal plane, an angle of 57° 21', what is the distance of the object from the bottom of the tower ? Answer, 233.3 feet. 4. From the top of a tower, whose height was 138 feet, I took the angles of depression of two objects which stood in a direct line from the bottom of the tower, and upon the same horizontal plane with it. The depression of the nearer object was found to be 48° 10, and that of the farther 19° 52', what was the distance of each from the bottom of the tower ? Answer, distance of the nearest 123.5, and of the farthest 403.8 feet. 5. Being on the side of a river, and wishing to know the distance of a house on the other side, I measured 312 yards in a right line by the side of the river, and then found that the two angles, one at each end of this line, subtended by the other end and the house, were 31° 15' and 86° 27', what was the distance between each end of the line and the house ? Answer, 35107, and 182:8 yards. 6. Having measured a base of 260 yards in a straight line, close by one side of a river, I found that the two angles, one at each end of the line, subtended by the other end and a tree close to the opposite bank were 40° and 80°, what was the breadth of the river ? Answer, 190'l yards. 7. From a ship a headland was seen, bearing N EL N, the vessel then stood away N WI W 20 miles, and the same headland was observed to bear from her E IN, required the distance of the headland from the ship at each station ? Answer, distance at the first station 19.09, and at the second 26.96 miles. 8. A cape was observed to bear from us NW, and another headland to bear NNEE; standing away ENE E 23 miles, we found the first bore from us W NW, and the second Nb W / W, required the bearing and distance of the cape from the headland ? Answer, W 46.46 miles, K 9. From an eminence of 268 feet in perpendicular height, the angle of depression of the top of a steeple which stood on the same horizontal plane was found to be 40° 3', and of the bottom 56° 18', what was the height of the steeple ? Answer, 117.8 feet. 10. Wanting to know the distance between two objects which were separated by a morass, I measured the distance from each to a point where I could see them both; the distances were 1840 and 1428 yards, and the angle which, at that point, the objects subtended, was 36° 18' 24", required their distance ? Answer, 1090.85 yards. 11. From the top of a mountain, 3 miles in height, the visible horizon appeared depressed 2° 13' 27", required the diameter of the earth, and the distance of the boundary of the visible horizon ? Answer, diameter of the earth 7958 miles, distance of horizon 154.54 miles. 12. A statue 12 feet high stands on the top of a column, whose height is 48 feet, at what distance from the base of the column, on the same horizontal plane, will the statue appear under the greatest possible vertical angle, and what will that angle be ? Answer, distance 53.6656 feet, angle 6° 22' 46". 13. Wanting to know the distance of two objects, A and B, from each other, and from another object D, all in the same horizontal plane, in B A produced on the side of A, a point C was taken, and CD being measured was found to be 549.36 yards, and the angle C 57o. At D, the angle C D A was observed to be 14°, and the angle B D A 41° 30', required the distances of A, B, and D from each other? Answer, A B 349:52, A D 487.27, and B D 349:52 yards. APPLICATION OF THE FORMULÆ INVESTIGATED IN THE ELEMENTARY PRINCIPLES OF SPHERICAL TRIGONOMETRY, TO THE NUMERICAL COMPUTATION OF THE MEASURES OF THE SIDES AND ANGLES OF RIGHT ANGLED SPHERICAL TRIANGLES. SPHERICAL triangles present six parts for consideration, three sides, and three angles ; but in right angled triangles the measure of the right angle can never be the subject of inquiry; therefore the sides, the hypothenuse, and the two oblique angles, are all the parts of the triangle to which it is necessary to advert; and any two of these being given, the others may be computed. Now if three of these five parts be taken, one of the three will either lie between the other two, or be separated from them by the two remaining parts of the triangle. This part is called the middle part; and, when it is adjacent to the other two parts, they are called, with respect to it, the adjoining extremes; otherwise, they are called With this understanding respecting the denominations of the different parts, as middle or extremes, let the sides about the right angle, the complement of the hypothenuse, and the complements of the öblique angles, be considered as the five circular parts ; then the two following equations, called Napier's Rulės FOR THE CIRCULAR Parts, obtain universally, and they are sufficient for the solution of any case that can arise from considering different parts of the triangle as given. 1. Radius x sine of the middle part = rectangle of the tangents of the adjoining extremes. 2. Radius X sine of the middle part = rectangle of the cosines of the opposite extremes. In the solution of problems, there will in every čaše be two parts given and one required; hence, in applying these equations, of the three quantities concerned take that as the middle part which is either adjacent to the other two, or is separated from them both by the remaining parts of the triangle, and observe whether the other parts become adjoining or opposite extremes, and form the equation accordingly. Transform the equation into a proportion, so that the required term may be the last; and, on solving this proportion, the required term will be obtained. For example, in the triangle A B C let the hypothenuse A C and the angle A be given to find the other parts. To find BC. Here of the three parts A C, _ A, and B C, A C cannot be the middle part, for the angle A would then be an adjoining extreme, and B C being separated from A C by the angle C, would be an opposite extreme. Neither can the angle A be the middle part; for A C would be an adjoining extreme, and B C, being separated from the angle A by the side A B, would be an oppo А B site extreme; B C is therefore the middle part; and A C and the angle A are opposite extremes. But, in forming the equations, the complements of 2 A and A Care used. Hence from Equation 2. we have rad . sin B C = sin A. sin A C and this equation, resolved into a proportion, gives rad : sin AC:: sin A : sin B C. To find A B. Here 2 A is the middle part, and A C and A B are adjoining extremes ; but as in forming the equations the complements of L A and A C are used, we have from the first equation rad . cos A = tan A B. cot AC and consequently cot A C : rad :: cos A : tan À B. To find the angle C. Here A C is the middle part, and the angles A and C being adjacent to A C, are adjoining extremes with respect to it. But in forming the equation, the complements of all the three parts here concerned are used. Hence Equation 1 gives rad . cos A C= cot A, cot C and the equation being resolved into a proportion, we have cot A : rad : : cos A C: cot C To determine whether the different parts of right angled spherical triangles are obtuse or acute, we have the following rules. 1. The sides about the right angle are of the same affection with their opposite angles, and the oblique angles are of the same affection with their opposite sides. 2. When the sides about the right angle are of the same affection, the hypothenuse is acute; but when the sides about the right angle are of different affections, the hypothenuse is obtuse. 3. And hence also when the oblique angles are of the same affection, the hypothenuse is acute ; but when they are of different affections, the hypothenuse is obtuse. As an example of the application of these rules, let us suppose that, in the last figure, A C and the angle A are both given obtuse. Then B C, being of the same affection as the angle A, will also be obtuse ; and as A C is obtuse, the sides A B and B C will be of different affections ; consequently as B C is obtuse, A B must be acute; and hence its opposite angle C is acute also. EXAMPLE I. In the right angled spherical triangle ABC (see the last figure) given A B 46° 18' 23', and the angle A 34° 27' 39", to find the other parts? Here, as the angle A is acute, B C is acute; and as A C is acute, A B and B C must be of the same affection, therefore A B is also acute, and consequently the angle C is acute. To compute A C. In this case the angle A is the middle part, and A C and A B are adjoining extremes; hence the equation is rad . cos A = tan . A B. cot A C; and the resulting proportion to find A C is tan A B : rad : : cos A: cot A C. Or tan AB 46° 18' 23" 10.019811 10.000000 34 27 39 9.916198 : rad To compute B C. Here A B is the middle part, and angle A and B C are adjoining extremes, hence the equation is rad . sin A B = tan B C.cot A, and the proportion to find B C is cot A : rad : : sin A B : tan B C. Or cot A 34° 27' 39" 10:163503 : rad 10.000000 :: sin A B 46 18 23 9:859165 : tan B C 26 23 97 9.695662 To compute C. Here angle C is the middle part, and Z A and A B are opposite extremes ; hence the equation is rad . cos C = cos A B. sin A, and the proportion to find 2 C is rad : sin A : : cos A B : cos C. Or rad 10.000000 9.752695 9.839354 9:592049 EXAMPLE II. In the right angled spherical triangle A B C (see the last figure) given A B 29° 41' 32", and B C 116° 30' 43", to find the other parts ? Here as A B and B C are of different affections, AC is obtuse ; and the angle C, being of the same affection with A B, is acute ; and A C, being of the same affection with B C, is obtuse. To compute A C. In this case A C is the middle part, and A B and B C are opposite extremes, the equation therefore is rad. cos A C = cos A B . cos BC, and the proportion to find A C is rad : cos A B :: cos B C ; cos A C. 10.000000 :cos A B 9.938870 :: cos B C 116 30 43 9•649708 : cos AC 112 48 58 9:588578 Or rad ..... 29° 41' 32''... To compute A. Here A B is the middle part, and B C and angle A are adjoining extremes ; hence the equation is rad. sin A B = tan B C .cot A, and the proportion to find 2 A is tan B C : rad : : sin A B : cot A. Or tan B C 116°3043 10.000000 :: sin A B 29 41 32 9.694904 : cot A 103 52 47 9.392866 ....10:302038 : rad |