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two angles A and B, whence the sides AC, BC may be found by (39 Trig.)

Example. Let the angle A=86° 52', the angle B=78° 47', and the base AB=10110 feet; what is the distance between A and C?

The proportion in (39 Trig.) gives

As sin (C)* : AB::sin B : AC, or by logarithms arith. com. log sin C (14° 21') 10.605821

log sin B (78° 47') 9.991624
log AB (10110)

4.004751

4.602196 Włence AC=40012.5=7 miles, 4 fur. 137 yds. 1 feet. The perpendicular distance CD may be found by the proportion rad. : sin A:: AC : CD.

log AC

CASE IV.

50. To find the distance between two objects, the distance of each from a third being known, and also the angle at that object which the required distance subtends.

Let A and B be the two objects, whose distance is required, and C the third object. Then in the triangle CAB, we have given the two sides CA, CB, and the included angle ACB ; whence the required distance AB may be found by Case III. Plane Tri

B gonometry.

From the latter part of 41 Plane Trigonometry it appears that AB=VAC? + BC?-2 AC x BC cos C, and by this formula we may find AB without first determining the angles A and B. As we have not before exemplified this formula, we shall apply it to the solution of an example under this case.

Example. In order to determine the distance between two houses at A and B, which could not be done by direct measurement, I ascertained the distance of each from a point C, and found AC=600 yards, BC=700 yards; at C I then took the angle ACB=569 12. From these data I require the distance AB ?

Because sin (A+B)=sin {180-(A+Blasin C

[graphic]

Х

Here AC=(600)=360000 and 2 AC. BC=2 x 600 x 700=840000
BC=(700)=490000 Natural cosine C (56° 12') = .5563

850000
467292

467292 382708 And /(382708)=618.63 yards=AB.

CASE V.

51. To find the distance between two inaccessible objects. Let C and D be two inaccessi

С ble objecte, the distance between which is required. Measure any

DO base as AB; at A find the angles CAD, DAB; and at B find the angles ABC, CBD. In the triangle ABC, in which the angles and side AB are given, find CB; in like manner, in the triangle

A

B ABD, find BD. Then in the triangle CBD, of which the sides CB, BD, and the included angle CBD are given; find the side CD.

Example. Let C and D be two houses on the further side of a river ; required their distance from the following data, viz. ; AB=1000 links, the < CAD=42° 45', [ DAB=54° 12', _ ABC=57° 33, and Z CBD=50° 19'.

1. In ABC, to find BC
arith. com. log sin C (25° 30') = 10.366016

log ein A (96° 57') 9.996797
log AB (1000)

3.
log BC=2305.75 8.362813
2. In ABD, to find BD.
arith. com. log sin D (17° 56') = 10.511576

log sin A (54° 12') 9.909055
log AB (1000)

3. log BD-2634.09 3.420631 Now in the triangle CBD we know the two sides BC and BD, and the included angle CBD ; whence as in the last case we easily find CD=2120.95 links, the distance required.

=

CASE VI.

52. Given the distances between three objects A, B, C, ana the angular distances between these objects at a station S in the same

B

horizontal plane ; to find the distance between that station and each of the objects.

1st. Let the station S be without the triangle ABC, formed by lines connecting the three objects.

Construct the triangle ABC; at A make the angle DAB=the given angle CSB; at B, make the angle ABD=ASD. Then through the

D points A, B, D, describe a circle;

С join DC, and produce it till it meet the circle in S, the station of the observer, and join SA, SB.

Then in the triangle ABC, of which the three sides are known, find the angle BAC. In the triangle ABD, in which the angles and side AB are known, find the side AD. In the triangle CAD, of which two sides AC, AD, and the included angle CAD are known, find the angle ACD. In the triangle ASC, of which the angles and the side AC are known, find the sides SA, SC. And in the triangle ASB, of which the angles and the side AB are known, find SB.

Example. There are three objects A, B, C, (above figure) the distances of which from one another are as follows: AB=424, AC=213, BC=262 yards; I desire to know my distance from each of these objects, from a station S, without the triangle ABC, where I ob served the angle ASC to be=13° 30', and BSC=29° 50'?

Answer, BS=524.326, AS=605.712, CS=429.682 yards. 2. When the station S is within the triangle ABC.

Construct the triangle ABC; make the angles BAD, ABD, equal to the supplements of the given angles BSC,

D ASC, respectively. Through the points A, B, D, describe a circle,

С join DC, which will cut the circle in S, the station of the observer.

B This done, the calculation will be exactly as before.

Example. The distances between three objects A, B, C, (above figure) are as follows: AB=267 yards ; AC=346, BC=209; and the angles at a point S within the triangle ABC, subtended by those distances, are 128° 40', 140°, 91° 20', respectively; what is the distance of the point S from each of these objects ?

Answer, AS=189.33, CS=178.85, BS=104.049 yards. 3. When the three objects are in one straight line.

Construct the circle as in the first example of this case. Then in the triangle ADC, of which the angles and the side AC are known; find AD. In the triangle ABD, of which

B the sides AB, AD, and the included angle BAD are known, find the angle ABD=SBC, the supplement of which ABS=DBC. In the triangle ABS, of which the angles and side AB are known, And SA, SB. And in the triangle ASC, of which the angles and AC are known, find SC

Example If three objects, A, B, C, as seen from S, are situate in one straight line, (as in the above figure) and AB=2550 yards, AC=7000 yards, the angles ASB=250 15, and BSC=35° 30'. Required the distances SĂ, SB, SC?

Answer, AS=5936.16, CS=7609.58, BS=5669.92. N. B. When the station S is in one of the sides of the triangle, or in one of the sides produced, the solution presents no difficulty.. The above examples are not worked at length, as after the sull explanation given it was not thought necessary to do so,

MISCELLANEOUS EXAMPLES ON HEIGHTS

AND DISTANCES.

1. Suppose the breadth of a well at the top to be 6 feet, and the angle formed by its side, and a visual diagonal line from the edge at the top, to the opposite side at the bottom to be 18° 30'; what is the depth of the well ?

Answer, 17.932 feet. 2. From the top of a tower 180 feet high, I took the angle of depression* of two trees, A and B, which stood in a direct line, on the same horizontal plane with the bottom of the tower, and found that of the nearer to be 55°, and that of the farther 3]0; what was the distance between the trees ? Answer, 173.532 feet.

3. Let A and B be two stations 500 yards apart, and C a church separated from AB by a river; also let the angle A be 75°. 19', and the angle B 59° 43'; how far is the church from each station?

Answer, 610.972 yards, and 684.412 yards. 4. Required the distance between two objects, A and B, separated from each other by a marsh, from the following data : A point C is taken, from which both A and B are accessible, and AC, is found=500 yards, BC=450, and the angle ACB=66° 30'?

Answer, 522.555 yards.

• The angle of depression of an object, is the angle formed by two lines one drawn from the eye parallel to the horizon, and the other drawn from the eye to the object.

5. Wanting to know the distance between a house D, and . church C, both on the further side of a river, and separated from each other by a wood; I pitched upon two stations A and B, 300 yards apart, and found the angles to be as follows: ABC=53° 30', CBD=45° 15', CAD=37°, and DAB=58° 20'; what is the disa tance between the house and church ? Answer, 479.304 yards.

6. To determine nearly the distance between two ships at sea, I carefully observed the interval of time between the flash and report of a gun from each, and measured also the angle which the two ships subtended. The intervals were 4 seconds and 6 seconds, and the angle 48, 12. Now supposing that sound moves uniforma ly at the rate of 1142 feet per second, and that the flash strikes the eye at the instant of its production ; required the distance of the ships?

Answer, 5147.9 feet. 7. At the top of a castle which stood on a hill near the sea shore, the angle of depression of a ship at anchor was 4° 52', and at the bottom of the castle its depression was 4° 2'; required the height of the top of the building above the level of the sea, sup. posing the castle itself 54 feet high; required also the horizontal distance of the vessel ?

Answer, height 314.2 feet; distance 3690.3 feet.

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