evident that AaBb, since ab is parallel to AB and equal thereto, and the angle aBA equal to the angle baв.

EXAMPLE XXXI.

Wanting to know the height of a church BC, I measured 45 yards from the bottom в upon a horizontal plane AB, at A, I took the angle cab=48°. 12', the height of my instrument was 5 feet. Required the height of the church?

Answer. The height bc=50·33 yards, to which add Bb the height of the instrument, then BC=51.99 yards the height of the church.

EXAMPLE XXXII.

Wanting to know the height of a steeple BC, inaccessible on account of several small trees on one side, and uneven ground on the other sides; at D, I took an angle CDB 51o. 30′, and measuring the distance AD= 75 feet, at A,I took the angle

Bas

CAB=26°. 30'. Required the height of the steeple, and the distance of the first station from its base?

SOLUTION.

Subtract the angle CAB from the angle CDB, the remainder 25the angle ACD: for ACD is the difference between ACB and DCB, the complements of the two observed angles, and the difference between the complements of two angles must be equal to the difference between the angles themselves. Then in the oblique triangle ADC all the angles are given to find DC=7918, and in the right-angled triangle DBC, the hypothenuse DC, and angle CDB are given, to find BC=61.97, and DB 49.29.

EXAMPLE XXXIII.

Wanting to know the height of an inaccessible object BC; at A, I took the angle CAB=28°. 34', and measuring in a straight line towards the object from A to D 30 yards, the angle CDB was 50°. 9'. Required the height of the object, and my distance from it at the second station?

Answer. The height BC=30 yards, and distance DB=24.99. *

EXAMPLE XXXIV.

Two observers at the distance of half a mile from each other, on an horizontal plane, took the altitude or angle of elevation of a cloud at the same moment of time, the same point of the

cloud

cloud was observed by both, and in the same direction: the angles were 35°. and 64°. Required the perpendicular height of the cloud, and its distance from each observer?

Answer. The perpendicular height was 935 75, the distance from the observers 1041.1, and 1631-4 yards.

EXAMPLE XXXV.

In the year 1784 two observers on Blackheath, at the exact distance of a mile from each other, observed the angle of elevation of Lunardi's balloon; the angle at the nearest station to it was 36°. 52' and the other angle 30°. 58'. Required the perpendicular height of the balloon, and its distance from each station?

Answer. The balloon's distance from the nearest station was 5.006 miles, from the other station 5.837 miles, and the perpendicular altitude 3.003 miles.

× EXAMPLE XXXVI.

From the top (A) of Flamborough-head light-house, the angle of depression (EAC) of a ship at anchor was 3°. 38'; and at the bottom (B) of the lighthouse, the angle of depression (FBC) was 2°. 43'; required the horizontal distance (DC) of the vessel, and the height (DB) of the promontory above the level of the sea; the light-house (AB) being 85 feet high.

The EAC ACD=3°. 38';

the FBC= 4 BCD=2°. 43'; and AB=85 feet.

Hence CD will be found=5296'4 feet, and DB=251.31 feet. Answer.

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 in a straight line towards the tower, at D the angle CDB was 670.50', and CDE = 16°.50'. Required the height of the tower,

EXAMPLE XXXVII.

E

and

and the height of the hill? The height of the instrument being 5 feet.

SOLUTION.

1. Subtract the angle CAB from CDB, the remainder 23°.50' =ACD, hence all the angles in the triangle CAD are given, and one side AD, to find DC 230-4 yards.

1

2. In the triangle CDE all the angles are given, viz. CDE= 16°.50', DCE 90°67°.50′22°.10', hence the angle CED= 141°. and CE=106 yards; also DE 138 13 yards.

=

3. In the right-angled triangle EDB all the angles are given, viz. EDB CDB-CDE = 51°, and DEB = 39°, also DE is given, hence BE = 107.3 yards, to which add the height of the instrument, and you have 108.96 yards for the height of the hill.

EXAMPLE XXXVIII.

Wanting to know the height of a castle standing upon an eminence, I took an angle of elevation CAD=40°, and measur ing 40 yards in a straight line towards the castle, the angle CDB was 63°. 20', and CDE 14°. 30': what was the height of the castle EC, and the distance DB?

Answer. Ec=24 69 yards, and DB=29 13 yards.

ther, the angle CAB was 23°. 45': Required the height of the object?

SOLUTION.

1. In the triangle ADC are given the angles, viz. CAB = 23°. 45', ADC=180°-41°139°, hence ACD=17°. 15', or CDB CAB=ACD=17°. 15' and AD being=60 feet, DC will be found to be 81-488.

2. In the triangle CDB are given CD=81.488, DB=40 and the angle CDB=41°, to find BC=57.64.

EXAMPLE XL.

An object standing on a declivity whose height I wished to determine,-I measured 150 yards from the base of it, and then

took

took an angle CDB=47°. 50', and in the same direction I measured 80 yards farther, and then took the angle CAB = 38°.30': Required the height of the object?

H

Answer. DC=307'1, and BC the height 234-4 yards.

XEXAMPLE XLI.

Wanting to know the height of a castle CE standing upon a hill,

at D the angle

CDB was 58°, CDE

=25°, and ADC

=72°. 10'. The

ground not permitting me to retreat in astraight

line towards F,

I

therefore mea- F

sured from D to

A 52 yards; at A the angle CAD was 64°. 30': Required the height of the castle EC, and the height of the hill BE, above the level of the first station D?

SOLUTION.

1. In the triangle ADC are given the angles ADC=72°. 10', CAD=64. 30′ and consequently ACD=43°. 20', and the side AD=52 yards, to find DC=68.392 yards.

2. The remaining part of the solution is the same as in Example xxxvII; CE will be found=34-463 and BE=23.536 yards.

having set up a staff AC, whose height was equal to that of the theodolite, I measured 1809.5 feet up the sloping ground AB, as a base, in a direct line with the tower, keeping the points K, E, C, B in the same vertical plane. At B I took the following angles, viz. FDC=BAI=1°.54'; and EDF=1°. 33'. Required the horizontal distance AH, the height HE of the hill, the height EK of the tower, and IB the elevation of the station B above that at A.

SOLUTION.

SOLUTION.

1. In the triangle Enc, the LEDC=3°27′, ECD=175°.23', DEC=1°. 10', and the side DC=1809.5 feet, hence CE=5348.08 feet.

2. In the triangle GCE, the GCE=2°. 43', and CE=5348·08 feet, hence CG=AH=5342·07 feet, and GE=253.48 feet.

3. In the triangle KCE, the KCE=0° 55′, CEK=92° 43′, CKE =86° 22′, and CE=5348'08 feet, hence EK=85'731 feet.

4. In the triangle BAI, AB=1809.5 feet, and the angle BAI= 1°. 54', hence B1=59.994 feet.

Lastly, if to the height GE, the height AC, of the instrument be added, it will give HE, the height of the hill.

25019-7024 miles : 360° :: 25 miles : 21', 34′′ ▲ DAB. In the right-angled triangle ABD, AB and the DAB are given, to find AD=3982.078446 miles, from which take AC AB, the remainder DC=078446 miles=414-19 feet, the height of the

tower.

EXAMPLE XLIV.

Supposing it were possible to see a light-house or other object D, in the horizon, at the distance of 200 miles, it is required to find its height CD, the diameter EC of the earth being 7964 miles, and its circumference 25019-7024 miles.

Answer. The DAB = - 2o. 52′. 39′′, aD = 3987·0272 miles, and DC=5.0272 miles=26543 616 feet, being 5910616 feet higher than Chimboraço, the highest of the Andes, and perhaps the highest mountain in the world.

EXAMPLE XLV.

The Peak of Teneriffe is said to be 24 miles above the level of the sea; at what distance can it be seen, supposing the radius of the earth to be 3982 miles?

Answer.

EDX DC = DB (Euclid III. and 36) 141.12 miles.

EXAMPLE