5. Wanting to ascertain the altitude AB of a moun

Here BCA= DBC + D (32.1); hence DBC=

BCA-D-46°-31°=15°.

of the natural tangents. In the annexed figure, let AB =80 feet, and the angle B=70°; let ab=1; draw be parallel to BC. The triangles ABC and Abc are similar, and making Ab radius, Ac will be the tangent of the angle B or b, which is 70°. Now the natural tangent of 70° corresponding to the radius 1, is 2.7473774 = Ac. Hence, 1:2.7473774:80: AC = 2.7473774 × 80 =

2.15578

6. Wanting to find the height of an obelisk, standing on the top of a regularly sloping hill, I first measured from its bottom a distance of 36 feet, and then found the angle formed by the inclined plane and a line from the centre of the instrument to the top of the obelisk 41°; but after measuring on downwards in the same sloping direction 54 feet farther, I found the angle formed in like manner to be only 23° 45′. What was the height of the obelisk, and what angle did the sloping ground make with the horizon?

The figure being constructed as here described, there

are given in the triangle ABC, all the angles. For ACB=CBD-CAB=41°-23° 45′ 17° 15′. Then

In the triangle BDC are known BC 73.339, BD 36, and CBD 41°; then CB+BD:CB-BD::tan. (BCD+BDC): tan. (BDC-BCD); that is, 109.339:37.339 :: tan. (139°): tan. (BDC

BCD.)

tan. (BDC-BCD)=42°24′ 9.50499

1 2

Hence 69° 30'+42°24′=112° 54'1=CDB; and 69°30′-42°24′= 26° 5'1=BCD.

Then, sin. BCD (26°5′1⁄2) : BD (36°) : : sin. CBD (41°) : CD=51.86, the height of the obelisk.

The angle of inclination DAE=HDA=CDB-90° =22°54′1.

Note. When the distance BD cannot be measured, the angle DAE of the sloping ground must be taken, as well as the angles CAB and CBD. In that case the angle DAE+90°=CDB; so that after having found CB from the known parts of the triangle ACB, CD may be found from the known parts of the triangle CBD.

7. Being on one side of a river, and wanting to know the distance of a fort on the other side, I measured

500 yards along the side of the river, in a straight line AB, and found the two angles between this line and the object to be, viz. CAB 74° 14', and CBA 49° 23'; required the distance between each station and the object. ACB=180-(A+B); that is,

C

A

B

E

180-(74° 14′+49° 23′)=56° 23′; then

Note. The distance AC or CB may be easily found without any calculation; thus, measure Ba in the pro

duction of AB and equal to it; make the angle a = CAB, then BH being measured in a right line with CB, will be equal to CB (26.1).

Those distances can be ascertained without any instrument to measure the angles A and B.

Take AG and BT of any length whatever, in the same directions with CA and CB; and measure the sides GA, GB, AB, from which obtain the angle GAB, and hence its supplement CAB; likewise measure the sides AB, BT, AG, from which the angle ABT may be discovered, and therefore its supplement ABC; then having the angles A, B, and AB, the sides AC and CB may be obtained.

8. Wanting to find the height of an object AB on sloping ground, I measured a base AC=250 feet; and at C found the angles of elevation of the top of the object and slope of the ground, to be 48° 16′, and 11° 20′ respectively; required the height of the object.

Construction.- Make CD equal to the height of the instrument; draw DF parallel to the horizon; make the angle BDF equal to 48° 16′, and EDF

equal to 11° 20′. On CA, drawn parallel to DE, make CA equal to 250, and draw AB perpendicular to DF; then AB is the height required.

Here BDE=BDF-EDF=48° 16′—11°20′=36° 56', and DBE=90°-BDF=90°-48°16′=41°44′. Then in the triangle DBE, there are given DE=CA =250 feet, and the angles BDE and DBE. Hence