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formula (20),

(b+c+a) (b+c-a)

cos. A =

1.

2 bc

In applying the logarithmic formulas in Rule 2 to the determination of any particular angle, it will generally be best, when this angle is opposite to the longest side of the triangle, to use the first formula, and when it is opposite to the shortest side to use the second; the third may be used when the required angle is opposite to the mean side. If two sides of the triangle are equal, then, of course, neither of these formulas will be used, as the unknown parts will be more readily found as in Example 3, p. 15.

EXAMPLES.

1. The three sides of the triangle ABC are

AB=1637, AC=2065, BC=3387·974;

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Using the third formula in the second rule, we have

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4. The three sides of a plane triangle are

AB=137, AC=153, BC= 101.616;

required the three angles,

A=40° 33′ 12′′, B=78° 13′ 1′′, C= 61° 13′ 47′′.

5. The three sides of a plane triangle are

AB=1886, BC= 960, AC=2400·364;

required the angle B,

B=128°4'.

6. Required the angles when the sides are 4, 5, and 6. The angles are 41°24′ 35′′, 55° 46′ 16′′, and 82° 49′ 9′′.

CHAPTER III.

APPLICATION OF PLANE TRIGONOMETRY TO THE MENSURATION OF HEIGHTS AND DISTANCES.

PROBLEM I.

A person on one side of a river observes an obelisk on the opposite side, and, being desirous to ascertain its height, he took with a quadrant the angle B=55° 54′, which the obelisk subtended at the place where he stood, then going back the distance BA=100 feet, he again measured the subtended angle, and found it to be A=33° 20'; what was the height of the obelisk?

C

D

In the triangle of ABC are given the angle A=33° 20′, the angle ACB=55° 54′ — 33° 20′ — 22° 34′, and the side AB; and, therefore, BC may be found by Case 1. of oblique angled triangles. Again, in the triangle BCD, we shall have given the side BC, and the angle B to find CD, which belongs to Case 1. of right angled triangles.

The actual computation, however, will be shortened by combining these two rules in a single formula, thus from the first

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The problem may be solved still more readily, as follows.

If we take CD for radius, DB will be the tangent of the angle DCB, and DA the tangent of DCA, therefore, AB is the difference of those tangents; but by referring to the table of natural tangents, we find that to radius 1

nat. tan. 56° 40′ = 1·5204261

nat. tan. 34° 6′ = 6770509

difference = .8433752

... ·8433752 : 1 :: 100: 118.57, as before.

* To the height of the object thus determined the height of the observer's eye, or of the instrument, must be added.

PROBLEM II.

B

A person at A wishes to know his distance from an inaccessible object at C, but he has no instrument for taking angles. He, therefore, sets up a staff at A, from which he measures the distance AA'=60 feet, so that when he stands at A' the staff and the object appear in the same straight line A'C; he, in like manner, measures another distance BB'86 feet, from a second station B, 38 feet from the former A, and he finds the diagonal distances AB', BA', to be respectively 97 feet and 81 feet. From these data it is required to determine the distance of A from the object C

B'

All the three sides of the triangle A'AB are given, therefore to find the angle A'AB we have, by using the first formula at (25),

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..

A'AB 109° 23′52′′ .. CAB 70° 36' 8".

Again, by applying the same formula to the triangle B' BA, we have

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