Case 3. When the three sides of a triangle are given, and the angles required. Rule. Let fall a perpendicular from the greatest angle to the longest side or base, which divides it into two segments, and the whole triangle into two right angled triangles. Then, as the base or sum of the two segments is to the sum of the other two sides, so is the difference of those sides to the difference of the segments of the base. Half the difference of the segments thus found, added to the half base, gives the greater segment, or subtracted from the half base gives the less segment, and the whole triangle will be divided into two right angled triangles, with two sides and one angle given to each; the remaining sides and angles of which may be found by the rule to the first case. Problem 69. Fig. 9, Plate 28. In the triangle ABC; given A B 6880, BC 6756, CA 4960, to find the angles A B and C. By Geometrical Construction. Set off the base line A B from a scale of equal parts; take the length of each side from the scale, and strike arcs intersecting in C; then from C let fall a perpendicular. By Gunter's Scale. To find the difference of segments. As A B BC+AC:: BC - CD: diff. of segments Extend the compasses on the line of lines from 6880 A B to 11716, the sum of the sides BC and AC; that extent will reach from 1796, the difference of the two sides, to 3058, the difference of the segments, or fourth number. To find the angle B. As BC radius :: BD: the co-s. B Extend the compasses from 6756 BC to 4960 BD, the greater segment; that extent from the radius on the sines will reach to 47° 14' the BCD, the complement of which is Extend the compasses from 4960 AC to 1911, the less segment; that extent from radius on the sines will reach to 22° 39′ ACD, the complement of which is 67° 21' the A: the supplement of the sum of the angles A and B equal to 69° 53', the angle A C B. By Logarithms. As 6880 11716 :: 1796 :: 3058 diff. of segments Half diff. of segments 1529 Half the base 3440+the diff. 1529 4969 greater segment To find the ▲ A. = = 1911 lesser segment : the co-sine B 42° 46' The supplement of the angles A and B equal to 69° 53' = ≤C Problem 70. Example for Practice. In the triangle ABC; given the side A B 930, the side A C 600, and the side CD 650; required the angles. First, by geometrical construction. Second, by Gunter's scale. Third, by logarithmic calculation. Problem 71. Miscellaneous Questions for Practice. Fig. 10, Plate 28. It is proposed to throw a bridge across the river; what is the exact distance from C to D? Given the line A B 1017, the 2 A 41° 29', the B 35° 01', to find the line CD. Half the base 508.5 58.438566.938 the greater segment دو Fig. 11, Plate 28. The Problem 72. distance is required between two points, A B, on the summit of two hills; given the angle of declivity at A 41° 29', and length from A to C 600; the angle of acclivity from C to B 35° 01', and length 692. Fig. 12, Plate 28. This and the preceding figure will show the principle of obtaining not only a correct distance, but also a tolerable section of the ground from one eminence to another, and the valley between them, and if carefully performed, the heights from the base or datum line A E are ascertained with tolerable effect. Care should always be taken that the object fixed to take the angles should be of the same height as the telescope of the theodolite. Commencing at A, the angle of acclivity is 24° 10′, and length to B 1350; from B the angle of declivity is 28° 0', the length 1140; from C the angle of acclivity is 25° 20′, and length 1560; and so on for any distance. Plot this to a large scale, and set off the angles by a good protractor. Problem 74. Fig. 13, Plate 28. When a station or other object falls within one of the large triangles formed by the three objects. Let A B C represent three towers, whose distance from each. other is known; to find the distance from the tower D, measure the angles ADC, BDC, ADB; plot the angles by the protractor, the point of intersection will be the point D. Problem 75. Fig. 14, Plate 28. To find the height of a building as A C. At B measure the angle FBC; set off any distance B D as a base; measure the angle B DC; CBD is the supplement of F B C, and B C D is the supplement of C B D + B D C. Then as sine BCD: BD:: CDB: BC; BC being found, we have in the right angled triangle F B C, the side B C, and the angle FB C, to find F C, which is found by this proportion: As radius is to sine of the angle FB C, so is BC: FC; FC added to A F, the height of the instrument, gives the height of the tower. Problem 76. Fig. 15, Plate 28. To take the map of a country. First, choose two places so remote from each other that their distance may serve as a common base for the triangle to be observed, in order to form the map. Let ABCDEFGHIK be several remarkable objects, whose situations are to be laid down in the map. Make a rough sketch of these objects, according to their positions in regard to each other; on this sketch the different measures taken in the course of the observations are to be set down. Measure the base A B, whose length should be proportionate to the distance of the extreme objects from A to B; from A, the extremity of the base, measure the angles EA B, F A B, GAB, CA B, D A B, formed at A with the base A B. From B, the other extremity of the base, observe the angles EBA, FBA, GBA, CBA, DBA. If any object cannot be seen from the points A and B another point must be found, or the base changed, so that it may be seen, it being necessary for the same object to be seen at both stations, because its positions can only be ascertained by the intersections of the lines from the ends of the base with which they form the triangles. It is evident from what has been already said, that having the base A B given, and the angles observed, it will be easy to find the sides, and from them lay down, with a scale of equal |