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Distance of inaccessible object.

observer. Measure off the distance AC of any convenient length, and observe the angles A and C.

Calculation. AB and BC are found by $ 73 of Pl. Trig.

71. Corollary. The perpendicular distance BP of the point B from the line AC, and the distances AP and PC are found in the triangles ABP and BPC, by $ 32 of Pl. Trig.

72. Corollary. Instead of directly observing the angles A and C, the bearings of the lines AB, AC, and BC, may be observed, when the plane ABC is horizontal, and the angles A and Care easily determined,

73. EXAMPLES.

1. An observer sees a cape, which bears N. by E.; after sailing 30 miles N. W., he sees the same cape bearing east; find the distance of the cape from the two points of observation.

Ans. The first distance = 21.63 miles.

The second dist. = 25.43 miles,

2. Two observers, stationed on opposite sides of a cloud, observe the angles of elevation to be 44° 56', and 36° 4', their distance apart being 700 feet; find the distance of the cloud from each observer, and its perpendicular altitude. Ans. Distances from observers=417.2 feet, and=500.6 ft.

Height=294.7 feet.

3. The angle of elevation of the top of a tower at one station is observed to be 68° 19', and at another station, 546 Height of inaccessible object.

feet farther from the tower, the angle of elevation is 32° 34'; find the height and distance of the tower, the two points of observation being supposed to be in the same horizontal plane with the foot of the tower. Ans. The height

= 467.45 ft. The dist. from the nearest point of observ. = 185.86 ft.

74. Problem. To find the distance of an object from the foot of a tower of known height, the observer being at the top of the tower.

Solution. Observation. Let the tower be AB (fig. 23), and the object C. Measure the angle of depression HAC. Calculation. Since

ACB = HAC, we know in the triangle ACB the leg AB and the opposite angle C, as in 33 of Pl, Trig.

EXAMPLE.

Given the height of the tower = 150 feet, and the angle of depression = 17° 25'; to find the distance from the foot of the tower.

Ans. 478.16 feet.

75. Problem. To find the height of an inaccessible object above a horizontal plane, and its distance from the observer. [B. p. 96.]

Solution. Observation. Let A (fig. 25) be the object. At two different stations, B and C, whose distance apart and

Distance of two objects.

bearing from each other are known, observe the bearings of the object, and also the angle of elevation at one of the stations, as B.

Calculation. In the triangle BCD, the side BC and its adjacent angles are known, so that B D is found by $ 73 of Pl. Trig. In the right triangle ABD, the height AD is, then, computed by $ 34 of Pl. Trig.

EXAMPLE.

At one station the bearing of a cloud is N. N. W., and its angle of elevation 50° 35'. At a second station, whose bearing from the first station is N. by E., and distance 5000 feet, the bearing of the cloud is W. by N.; find the height of the cloud.

Ans. 7316.3 feet.

76. Problem. To find the distance of two objects, whose relative position is known. [B. p. 90.]

Solution. Observation. Let B and C (fig. 1) be the two known objects, and A the position of the observer. Observe the bearings of B and C from A.

Calculation. In the triangle ABC, the side BC and the two angles are known. The sides of AB and AC are found by $ 73 of Pl. Trig.

EXAMPLE.

The bearings of the two objects are, of the first N. E. by E., and of the second E. by S. ; the known distance of the Distance apart of two objects.

first object from the second is 23.25 miles, and the bearing N. W.; find their distance from the observer. Ans. The distance of the first object is = 18.27 miles.

That of the second object = 32.25 miles.

77. Problem. To find the distance apart of two objects separated by an impassable barrier. [B. p. 91.]

Solution. Observation. Let A and B (fig. 1) be the objects; the distance of which from each other is sought. Measure the distances and bearings from any point C, to both A and B.

Calculation. In the triangle ABC, the two sides AC and BC, and the included angle C, are known. The side AB and the angles A and B may be found by $ 82 of Pl. Trig.

EXAMPLE.

Two ships sail from the same port, the one N. 10° E. a distance of 200 miles, the second N. 70° E. a distance of 150 miles; find their bearing and distance. Ans. The distance

= 180.3 miles. The bearing of the first ship from the second = N. 36° 6' W.

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78. Problem. To find the distance apart of two inaccessible objects situated in the same plane with the observer, and their bearing from each other. [B. p. 92.]

Solution. Observation. Let A and B (fig. 26) be the two inaccessible objects. At two stations, C and E, observe the bearings of A and B, and the bearing and distance of C from E.

Distance apart of two objects.

Calculation. In the triangle AEC, we have the side CE, and the angles ACE and AEC, so that AC is found by $ 73 of Pl. Trig.

In the same way BC is calculated from the triangle BCE.

Lastly, in triangle ABC, we know the two sides AC and BC, and the included angle, for ACB

BCE. Hence AB and the angles BAC and CBA are found by 82.

= ACE

EXAMPLE

An observer from a ship saw two headlands; the first bore E. N. E., and the second N. W. by N. After he had sailed N. by W. 16.25 miles, the first headland bore E. and the second N. W. by W.; find the bearing and distance of the first headland from the second,

Ans. Distance = 55.89 miles.

Bearing = S. 80°42' E.

79. Problem. To find the distance of an object of known height, which is just seen in the horizon.

Solution. I. If light moved in a straight line, and if A (fig. 27) were the eye of the observer, and B the object, the straight line APB would be that of the visual ray. The point P, at which the ray touches the curved surface CPD of the earth, is the point of the visible horizon at which the object is seen. The distances PA and PB may be calculated separately, when the heights AC and BD are known. For this purpose, let O be the earth's centre, let B D be produced to E, and let

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