16. Requiring to find the height and distance of an object O, on the top of an inaccessible hill, from the station B situated in a valley. I measured a base BN of 642 yards up the sloping ground BC, directly from the object O, the points O, B, N, being in the same vertical plane; then having set up a staff BS, whose length was equal to the height of the theodolite, I took the following angles of elevation and depression, viz. at the station N, the object O was elevated 3° 59′, that is, the angle OAH; and the top of the staff S was depressed 39', that is, the angle HAS; at the other station B, the elevation of the top of the object O was 5° 52′, that is, the angle PSO. From these observations, the horizontal distance BR, the height RO, and also GN, the height of the station N above B, are required. ... 0.862216 to NG 7.3 yards 17. The mutual distances of three remote objects being given, viz. AB=800, AC=600, BC=400 yards, from a convenient station S, which commanded a view of the three objects, I took the horizontal angles ASC =33° 45′, BSC=22° 30′; it is thence required to find the respective distances of my station from each object. In this, as well as in most other problems in trigonometry, it is necessary to construct the figure; and though we should not take time to delineate the figure off a regular scale, a rough sketch will always be useful before we begin to compute. C H B A K SK From any convenient scale draw the given triangle ABC; from the point A draw a line AH, making with the line AB an angle equal to 22° 30′, and from B draw a line BH, making an angle with AB equal to 33° 45′. Let a circle be described to pass through the points A and B, and through the point of intersection H, of the lines BH and AH, join C and H, and produce the line joining them to the circumference, and where this produced line meets the circumference at S, will be the required station or point. For, connecting AS and BS, the angle HBA is equal to the angle ASH, and HAB equal to HSB, because they stand on the same arcs respectively. Hence S is the point where the angles have the assigned values; and must therefore be the required station from which the angles had been taken. 400+125=525, greater segment of the base. Then to find the angles of the triangle ABC, by means of the known segments of the base AB, made by the perpendicular demited from the point C upon it. |