BC 112 poles, and the angle C 57° 58'; required the third side and the remaining angles. Answ. AB 112.65 poles, the angle A 57° 27', and the angle B 64° 34′. 4. In the triangle ABC, there is given AB 345 yards, AC 232, BC 174.07 yards; required the angles А, В, С. Answ. A 27° 4′, В 37° 20′, С 115° 36′. 5. In the triangle ABC, there is given AB 120, AC 112.6, BC 112; required the angles A, В, С. Answ. A 57° 27′, В 57° 59′, С 64° 34′. 6. In the triangle ABC, there is given AC 246.5, BC 370.5, and the angle A 101° 21'; required the side AB, and the angles C, B. Answ. AB 232.3, В 40° 43′, C 37° 56′. 7. In the triangle ABC, there is given AB 462, AC 384, CD 169; required the angle C. 70 SECTION V. On the Measurement of Heights and Distances of Objects. The instruments generally employed to measure angles are quadrants, sextants, theodolites, &c. A good pocket sextant, and an accurate micrometer attached to a telescope, are highly useful to engineers and military men. Such persons ought also to be furnished with a portable box of graduated tape, and 50 feet and 100 feet chains-all of which will be found useful for measuring small distances, as bases. 1. Wishing to ascertain the distance between two buildings A and B, which could not be directly mea sured because of a lake which occupied much of the intermediate space; I measured the distance of each of them from a third object C, viz. AC=588 yards, BC=672 yards; and then at the point C took the angle ACB between the two buildings=55° 40'; required their distance asunder. AC+CB=588+672=1260 180°-55°40′-124.20=sum of A and B. 124°20′÷2=62°10′-half sum of A and B. to tan. 7° 12′ diff. of A, B, 9.10129 Now, to find AB, we say, by Prop. 4, sec. 3, object AB, on a horizontal plane I measured from A to C 160 feet, and then standing at C, took the angle of elevation BDE 56° 30'; required the height of the object AB from these conditions, allowing the height of the theodolite to be 5 feet; that is, DC=5 feet. Then EB+AE=241.7+5=246.7, the height re quired. 3. Wanting to know the distance between two inaccessible objects, which lay in a direct line from the bottom of a tower on whose top I stand, I took the angles of depression of the two objects, viz. of the most remote, 25° 30', of the nearest, 57°. What is the distance between them, the height of the tower being 120 feet? The figure being constructed as here given, we have AB-120 feet, and having drawn the horizontal line AH parallel to BD, we have the angle HAD= ADB-25°30′. Hence BAD=BAH-HAD=90°-25°30′=64° 30'. And HAC=ACB=57°. Hence BAC=BAH-HAC=90°-57°=33°. Then BD-BC-252-77.93=174.07-CD. 4. From the top of a ship's mast, which was 80 feet above the water, the angle of depression of another ship's hull, at a distance on the water, is 20°; what is their distance? Here AB-80 feet. And ABC ABH-HBC=90° 20°-70°. Then making AB radius, we have *As 1: 2.7473774(=tan.70)::80: 2.7473774× 80-219.790192 feet-AC. * It is often useful to solve some of the cases in Trigonometry by means |