Calculation.

In the right angled triangle DEC, we have the angle CDE=47° 30', and the base DE-AB-100 feet to find CE=109.13 feet; to CE add EB=DA=5 feet, the height of the quadrant, and it will give BC=114.13 feet, the required height of the steeple.

EXAMPLE 8, Fig. 62.

Wishing to know the height of a tree situated in a bog, at a station D, which appeared to be on a level with the bottom of the tree, I took the angle of elevation BDC=51° 30'; I then measured DA--75 feet in a direct line from the tree, and at A, took the angle of elevation BAC=26° 30'. Required the height of the

tree.

Calculation.

1. Because the exterior angle of a triangle is equal to the sum of the two interior and opposite ones, the angle BDC=DAC+ACD; therefore ACD=BDC-DAC=25°: now in the triangle ADC we have DAC=26° 30′, ACD =25°, and AD=75, to find DC=79.18.

2. In the right angled triangle DBC are given DC= 79.18, and the-angle BDC=51° 30′ to find BC=61.97 feet, the required height of the tree.

EXAMPLE 9, Fig. 63.

Wanting to know the height of a tower EC, which stood upon a hill, at A, I took the angle of elevation CAB=44°; I then measured AD 134 yards, on level ground in a straight line towards the tower; at D the angle CDB was 67° 50′ and EDB 51°. Required the height of the tower and also of the hill.

Calculation.

1. In the triangle ADC we have the angle DAC=44°, the angle ACD-BDC-DAC=23° 50', and the side AD, to find DC-230.4.

2. In the triangle DEC all the angles are given, viz. CDE=BDC-BDE=16°50′, DCE=90°— BDC=22°10′, DEC-180° the sum of the angles CDE and DCE,= 141°, and CD=230.4, to find CE=106 yards, the height of the tower.

3. In the right-angled triangle DBC, we have the angle BDC-67° 50', and the side DC-230.4, to find BC= 213.4; then BE-BC-CE=213.4-106-107.4 yards, the height of the hill.

EXAMPLE 10, Fig. 64.

An obelisk AD standing on the top of a declivity, I measured from its bottom a distance AB-40 feet, and then took the angle ABD 41°; going on in the same direction 60 feet farther to C, I took the angle ACD=23° 45': what was the height of the obelisk?

Calculation.

1. In the triangle BCD, we have given the angle BCD, =23° 45', the angle BDC=ABD-BCD-17° 15', and side BC=60, to find BD-81.49.

2. In the triangle ABD are given the side AB-40, BD=81.49, and the angle ABD 41°, to find AD=57.64 feet, the height of the obelisk.

EXAMPLE 11, Fig. 65.

Wanting to know the height of an object on the other side of a river, but which appeared to be on a level with the place where I stood, close by the side of the river; and not having room to go backward on the same plane, on account of the immediate rise of the bank, I placed a mark where I stood, and measured in a direct line from the object, up the hill, whose ascent was so regular that I might account it a right line, to the distance of 132 yards, where I perceived that I was above the level of the top of the object; I there took the angle of depression of the mark by the river's side equal 42°, of the bottom of the object equal 27°, and of its top equal 19°: required the height of the object.

Calculation.

1. In the triangle ACD, are given the angle CAD= EDA=27°, ACD=180°-CDE (FCD)=138° and the side CD=132, to find AD=194.55 yards.

2. In the triangle ABD, we have given ADB-ADEBDE=8°, ABD—BED+BDE=109° and AD=194.55, to find AB=28.64 yards, the required height of the object.

EXAMPLE 12, Fig. 66.

A May-pole whose height was 100 feet standing on a horizontal plane, was broken by a blast of wind, and the extremity of the top part struck the ground at the distance of 34 feet from the bottom of the pole: required the length of each part.

Construction.

Draw AB=34, and perpendicular to it, make BC= 100; join AC and bisect it in D, and draw DE perpendicular to AC, meeting BC in E; then AE-CE=the part broken off.*

Calculation.

1. In the right-angled triangle ABC, we have AB=34 and BC=100, to find the angle C=18° 47'.

2. In the right-angled triangle ABE, we have AEB= ACE+CAE=2ACE=37° 34', and AB=34, to find AF =55.77 feet, one of the parts; and 100-55.77-44.23 feet the other part.

PRACTICAL QUESTIONS.

1. At 85 feet distance from the bottom of a tower, the angle of its elevation was found to be 52° 30': required the altitude of the tower. Ans. 110.8 feet.

2. To find the distance of an inaccessible object, I measured a line of 73 yards, and at each end of it took

* DEMONSTRATION. In the triangles AED, DEC, the angle ADE=CDE, the side AD=CD, and DE is common to the two triangles, therefore (4.1) AE=CE.

Note. This question may be neatly solved in the following manner without finding either of the angles. Thus, draw DF perpendicular to BC, then (31.3 and cor. 8.6) FC: DC :: DC: CE; conseDC2 AC2 AB2+ BC2

quently CE=FC

; but DC

and FC=BC;

there

the angle of position of the object and the other end, and found the one to be 90°, and the other 61° 45': required the distance of the object from each station. Ans. 135.9 yards from one, and 154.2 from the other.

3. Wishing to know the distance between two trees C and D, standing in a bog, I measured a base line AB= 339 feet; at A the angle BAD was 100° and BAC 36° 30'; at B the angle ABC was 121° and ABD 49°: required the distance between the trees. Ans. 697 feet.

4. Observing three steeples, A, B and C, in a town at a distance, whose distances asunder are known to be as follows, viz. AB=213, AC=404, and BC=262 yards, I took their angles of position from the place D where I stood, which was nearest the steeple B, and found the angle ADB=13° 30'; and the angle BDC=29° 50'. Required my distance from each of the three steeples. Ans. AD=571 yards, BD=389 yards, and CD=514 yards.

5. A May-pole, whose top was broken off by a blast of wind, struck the ground at 15 feet distance from the foot of the pole: what was the height of the whole Maypole, supposing the length of the broken piece to be 39 Ans. 75 feet.

feet?

6. At a certain place the angle of elevation of an inaccessible tower was 26° 30'; but measuring 75 feet in a direct line towards it, the angle was then found to be 51° 30': required the height of the tower and its distance from the last station. Ans. Height 62 feet, distance

49.

7. From the top of a tower by the sea side, of 143 feet high, I observed that the angle of depression of a ship's