A Treatise on Plane and Spherical Trigonometry

distance is a quarter of a mile: the elevation of the top of the castle, seen from the remote station, is 16° 28'; the elevations of the top and bottom, seen from the near station, are 52° 24′ and 48° 38′ respectively: (1) what is its height, and (2) what its elevation above the sea?

Ans. (1) 60.82 feet; (2) 445.23 feet.

124. To find the Distance of an Object on a Horizontal Plane, from Observations made at Two Points in the Same Vertical Line, above the Plane.

Let the points of observation A and B be in the same vertical line, and at a given distance from each other; let C be the point observed, whose horizontal distance CD and vertical distance AD are required.

Measure the angles of depression, bBC, aAC, equal to a and ẞ respectively, and denote AB by a.

1. From the top of a house, and from a window 30 feet below the top, the angles of depression of an object on the ground are 15° 40′ and 10°: find (1) the horizontal distance of the object, and (2) the height of the house.

Ans. (1) 288.1 feet; (2) 80.8 feet.

2. From the top and bottom of a castle, which is 68 feet high, the depressions of a ship at sea are observed to be 16° 28' and 14°: find its distance. Ans. 570.2 yards.

125. To find the Distance between Two Inaccessible Objects on a Horizontal Plane.

Let C and D be the two inac

cessible objects.

Measure a base line AB, from whose extremities C and D are visible. At A observe the angles CAD, DAB; and at B observe the angles CBA and CBD.

know two angles and the In the triangle ABD we .. AD may be found.

Then, in the triangle ABC, we side AB. ... AC may be found. know two angles and the side AB. Lastly, in the triangle ACD, AC and AD have been determined, and the included angle CAD has been measured and thus CD can be found.

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EXAMPLES.

= 76° 30'

1. Let AB = 1000 yards, the angles BAC, BAD: and 44° 10', respectively; and the angles ABD, ABC= 81° 12′ and 46° 5', respectively: find the distance between C and D. Ans. 669.8 yards.

2. A and B are two trees on one side of a river; at two stations P and Q on the other side observations are taken, and it is found that the angles APB, BPQ, AQP are each equal to 30°, and that the angle AQB is equal to 60°. If PQa, show that

126. The Dip of the Horizon.

Since the surface of the earth is spherical, it is obvious that an object on it will be visible only for a certain distance depending on its height; and, conversely, that at a certain height above the ground the visible horizon will be limited.

Let O be the centre of the earth, P a point above the surface, PD a tangent to the surface at D. Then D is a point on the terrestrial horizon; and CPD, which is the angle of depression of the most distant point on the horizon seen from P, is called the dip of the horizon at P. The angle DOP is equal to it.

Denote the angle CPD by 0, the height AP by h, and the radius OD by r. Then

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Since, in all cases which can occur in practice, h is very small compared with 2r, we have approximately

Let n the number* of miles in PD, h = the feet in PA,

*It will be noticed that n is a number merely, and that the result will be in feet,

since the miles have been reduced to feet.

That is, the height at which objects can be seen varies as the square of the distance.

Thus it appears that an object less than 8 inches above the surface of still water will be invisible to an eye on the surface at the distance of a mile.

Example. From a balloon, at an elevation of 4 miles, the dip of the sea-horizon is observed to be 2° 33' 40": find (1) the diameter of the earth, and (2) the distance of the horizon from the balloon.

Ans. (1) 8001.24 miles; (2) 178.944 miles.

127. Problem of Pothenot or of Snellius. To determine a point in the plane of a given triangle, at which the sides of the triangle subtend given angles.

Let ABC be the given triangle, and P the required point. Join P with A, B, C.

Let the given angles APC, BPC be denoted by α, B, and the unknown A angles PAC, PBC by x, y respectively; then a and ẞ are known; and when x and y are found, the position of P can be determined, for the distances PA and PB can be found by solving the triangles PAC, PBC.

We have x+y=2π-α-B-C

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