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From the right-angled triangles BAD and CAD,

have,

AD2 = AB2
AB2 – BD2, and AD2 =A02

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(DC + BD) (DC – BD).

we

Converting this equation into a proportion (B. II., P. II.), we have,

DCBD AC+ AB :: AC-AB DC-BD;

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that is, if in any plane triangle, a line be drawn from the vertex of the vertical angle perpendicular to the base, dividing it into two segments; then,

The sum of the two segments, or the whole base, is to the sum of the two other sides, as the difference of these ides is to the difference of the segments.

The half difference added to the half sum, gives the greater, and the half difference subtracted from the half sum gives the less segment We shall then have two rightangled triangles, in each of which we know the hypothenuse and the base; hence, the angles of these triangles may be found, and consequently, those of the given triangle.

EXAMPLES.

1. Given a = 40, b = 34, and e 25, to find A, B, and C.

OPERATION.

Applying logarithms to Formula (15), we have,

log (ss') = log (b+c) + log (bc) + (a. c.) log (s+s')-10;

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log cos C = log s + (a. c.) log b .. C = 38° 25′ 20′′,

and

log cos B = logs' (a. c.) log c .'. B= 57° 41′ 25′′

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Ans. A 82° 49′ 09", B = 55° 46' 16", C= 41° 24′ 35".

3. Given a = 71.2 yds., b = 64.8 yds., and

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c = 37.4

Ans. A 83° 44' 32", B = 64° 46′ 56", C= 31° 28' 30".

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2. At what horizontal distance from a column, 200 feet high, will it subtend an angle of 31° 17' 12" ?

3. Required the height of a hill D above a horizontal plane AB, the distance between A and B

being equal to 975 yards,

Ans. 329.114 ft.

D

and the angles of elevation at A and B being respectively 15° 36' and 27° 29'.

Ans. DC 587.61 yds.

4. The distances AQ and BC are found by measurement to be, respectively, 588 feet and 672 feet, and their included angle 55° 40'.

ed the distance AB.

Requir

Ans. 592.967 ft.

B

5. Being on a horizontal plane, and wanting to ascertain the height of a tower, standing on the top of an inaccessible hill, there were measured, the angle of elevation of the top of the hill 40°, and of the top of the tower 51°; then measuring if a direct line 180 feet farther from the hill, the

;

angle of elevation of the top of the tower was 33° 45' required the height of the tower. Ans. 83.998 ft.

6. Wanting to know the horizontal distance between two inaccessible objects E and W, the

following measurements were made:

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Ans. 939.634 yds.

Required the distance EW.

F

7. Wanting to know the horizontal distance between two inaccessible objects A and B, and not finding any station from which both of them could be seen, two points C and D, were chosen at a distance from each other equal to 200 yards; from the former could be seen, and from the latter, B; points C and D, a staff was set up. From C, a distance CF was measured, not in the direction DC, equal to 200 yards, and from D, a distance DE, equal to 200 yards, and the following angles taken :

of these points, A and at each of the

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Required the distances AP, BP, and CP.

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This problem is used in locating the position of buoys in maritime surveying, as follows. Three points A, B, and C, on shore are known in position. The surveyor stationed at a buoy P, measures the angles APC and BPC. The distances AP, BP, and CP, are then found as follows: Suppose the circumference of a circle to be described through the points A, B, and P. Draw CP, cutting the circumference in D, and draw the lines DB and DA. The angles CPB and DAB, being inscribed in the same segment, are equal (B. III., P. XVIII., C. 1); for a like reason, the angles CPA and DBA are equal: hence, in the triangle ADB, we know two angles and one side; we may, therefore, find the side DB. In the triangle ACB, we know the three sides, and we may compute the angle B. Subtracting from this the angle DBA, we have the angle DBC. Now, in the triangle DBC, we have two sides and their included angle, and we can find the angle DCB. Finally, in the triangle CPB, we have two angles and one side, from which data we can find CP and BP. In like manner, we can find AP.

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