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48° 12', and C AB = 89° 18'; at B the angle A B C was 46° 14', and A B D 87° 4'; it is required from these data to compute the distance between C and D?

From the angle CAB take the angle D A B, the remainder, 41° 6', is the angle CAD. To the angle D BA add the

D angle D A B, and 44° 44', the supplement of the sum, is the angle A D B. In the same way the angle A CB, which is the supplement of the sum of C AB and C B A, is found to be 44° 28'.

Hence in the triangles A B C and A B D we have,

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Then, in the triangle C AD, we have given the sides C A and A D, and the included angle CAD, to find CD; to compute which we proceed thus: The supplement of 2 CAD is the sum of the angles ACD and ADC;

A CD + A DC hence

= 69° 27', and, by proportion, we have, 2 As A D + AC

940.8

2.937497 : AD - AC....

149

2.173186 ACD + ADC

69° 27' ... 10*426108
2

12.599294
ACD - ADC
: tan

22 54

9.625797
2
L ACD.

sum 92 21
LADC..

diff. 46 33

:: tan

As sin A D C 46° 33'
: AC 395.9 yards
: : sin CAD 41° 6'...

9.860922 2:597585 9.817813 12415398

EXAMPLE III. To determine the altitude of a lighthouse, I observed the elevation of its top above the level sand on the seashore to be 15° 32' 18", and measuring directly from it along the sand 638 yards, I then found its elevation to be go 56' 26"; required the height of the lighthouse?

Let C D represent the height of the lighthouse above the level of the sand, and let B be the first station, and A

с the second ; then the angle C B D is 15° 32' 18", and the angle CAB is 9° 56' 26'' ; therefore the angle A CB, which is the difference of the angles CBD and C AB, is 5° 35' 52''.

A B
Hence as sin ACB 5° 35' 52"

8.989201
: A B 638 ...

2.804821 :: sin 4 A 9° 56' 26"

9:237107

12:041928 :BC 1129'06 yards

3.052727

As radius

10.000000 : B C 1129:06

3:052727 : : sin C B D 15° 32' 19" 9:427945

12:480672 :D C 302.46 yards

2:480672

EXAMPLE IV. Wanting to know the height of a steeple, I measured 210 yards from the bottom of it, and then found the elevation of its top above the level of my instrument to be 33° 28' 40'; required its height, the instrument standing five feet above the ground?

Let CE represent the steeple, DE the ground, and AD the instrument. Draw A B parallel to D E, then A B and DE will be equal, and B E will be equal to AD, the height of the instrument. Now in the right angled triangle A B C,

A

B В there are given A B, 210 yards, and the angle B A C 33° 28' 40%. Hence As radius

10:000000 : A B 210 yards

2.322219 : : tan B A C 33° 28' 40". 9.820427

12:142646 : BC 138:39

2:142646

5 The required height B E 143.89

D

EXAMPLE V.

Coming from sea, at the point D, I observed two headlands, A and B, and inland at C a steeple, which appeared between the headlands; I found from a map that the headlands were 5:35 from each other, that the distance from A to the steeple was 2.8 miles, and from B to the steeple 3:47 miles; and I found with a sextant that the angle ADC was 12° 15', and the angle B D C 15° 30'; required my distance from each of the headlands, and from the steeple.

BY CONSTRUCTION. On A C describe the segment of a circle to contain the angle ADC (Prob. 12. Geo.;) and on B C describe the segment of a circle to contain the angle BDC, and these circles will intersect in D, the place of the ship.

BY CALCULATION. Let ADB be the segment of a circle described an A B to contain the sum of the two given angles. Join D C and let it meet the circumference of the circle in E, and join A E, B E. Then the angles E AB, E DB, being in the same segment, are equal to each other; and so, for the same reason, are the angles A B E and A D E. Hence in the triangle A B E, all the angles and the side A B are given to find the side A E; and as all the sides of the triangle A B C are given, the angle B A C may be computed ; and the difference of the angles B A C and B A E is the angle C AE, which, therefore, becomes known.

Now, in the triangle C AE, the two sides A C and A E, and the included angle CAE, being known, the angle C may be determined ; and hence, as the angle A D C is given, the angle CAD, and consequently B AD, the difference of CAD and CAB. Hence all the angles of the triangle A D C and the side AC are given, whence A D and C D become known; and A B being known, as well as the angles B AD, BD, A of the triangle A B D, the side B D is also determined.

The computation at length is as follows:

L E AB 15° 30%
LEBA 12 15

27 45

To find A E.
As sin A E B 1520 15'.... 9•668027
: A B 5:35

•728354 : : sin A B E 12° 15' .... 9:326700

180 o

10:055054

To find the angle B A C.

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BC 3:47
AB 5.35 log 28354
AC 2.80 log •447158
2)11.62 1:175512

5.81 log 764176
BC 2:34 log •369216

20
21.133392

2 19.957880
17° 41' 58"

COS 9.978940

2
LBAC 35 23 56
LE AB 15 30
2 E AC 19 53 56

180
2) 160 6 4
80 3 2 AEC + ACE

2

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LAEC

101 33 14 sum
LACE or A CD 58 32 50 diff.
L

CDA 12 15
70 47 50 supplement 109° 12' 10' CAD

35 23 56 2 САВ
73 48. 14. < BAD

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The elevation of a spire at one station was 23° 50' 17'', and the horizontal angle at this station between the spire and another station was 93" 4' 20''; the horizontal angle at the latter station between the spire and the first station was 54° 28'36", and the distance between the two stations 416 feet, required the height of the spire ?

Let CD be the spire, A the first station, and B the second ; then the vertical angle CAD is 23° 50' 17"; and as the horizontal angles C A B and C B A are 93° 4' 20', and 54° 28' 36' respectively; the angle A C B, the supplement of their sum, is 32° 27' 4". B

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