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Either side adjacent to the right angle is equal to the tangent of the opposite angle multiplied by the other side. ... pb tan P.

Introducing radius and applying logarithms, as in the preceding cases, we find p 183.95.

Either side adjacent to the right angle is equal to the co-sine of the adjacent acute angle multiplied by the hypotenuse.

b

cos P

... b

3. Given

Introducing radius and applying logarithms, as above, we shall find h = 239.05.

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h cos P; .. h

Sp= 3963.35 miles

2. Given

the earth's radius.

\ P=57′ 2.3′′= the moon's horizontal parallax. Required h, the distance of the moon from the earth. Ans. h238889 miles.

p

- 3963.35 miles the earth's radius. P8.9" the sun's horizontal parallax. Required h, the distance of the sun from the earth. Ans. h 91852000 miles.

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1. Given P

{

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8.9

60

...log sin 8.9"-log sin l'+ log 8.9 + a.c. log 60-10.

Scholium. Sin 8.9" sin 1' X

68. Case IV.

Given the two sides adjacent to the right angle, required the remaining parts.

29.37. b 37.29.

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B

H

The tangent of either acute angle is equal to the opposite side divided by the adjacent side.

.. tan P:

Introducing radius and applying logarithms, we shall find that P= 38° 13′ 28′′.

B=90° — P = 90° — 38° 13′ 28′′ = 51° 46′ 32′′.

Either side adjacent to the right angle is equal to the sine of the opposite angle multiplied by the hypotenuse.

p h sin P.

... h

2. Given

3. Given

Introducing radius and applying logarithms, we find h = 47.466.

4. Given

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p

b

1728.

1575.

p sin P

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A

47° 39′ 07′′.

} Required B=42° 20′ 53′′.

h

=

2338.1.

OBLIQUE TRIANGLES.

69. Case I.

Given one side and two angles, required the remaining

parts.

Let ABC be an oblique triangle, and let the sides opposite the angles A, B, and C be denoted respectively by a, b and c.

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p

a

B

Let the angles A and B and the side a be given, and the angle C and the sides b and c be required. We find C from the formula,

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Draw the perpendicular p from the vertex C to the side c, thus forming two right triangles. There are two cases:

1st. When the perpendicular falls on the side c. From the principles of the right triangle we have,

p=b sin A and pa sin B.
... b sin A a sin B.

A

.. (1) sin A: sin B:: a: b.

But CBD is the supplement of CBA, or B of the triangle. Since the sine of an angle is equal to the sine of its supplement,

A

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2d. When the perpendicular falls on e produced.

p=b sin A and p a sin CBD.

sin CBD sin B; .. p =a sin B.
... b sin A a sin B.

(1) sin A sin B a b.

:

sin A sin C: a c.

C

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B

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In like manner we may find,

(2)

Hence, The sine of the angle opposite the given side is to the sine of the angle opposite the required side as the given side is to the required side.

Introducing radius by substituting for the function to the radius 1, the corresponding function to the

radius R divided by R, and reducing, the proportions (1) and (2) will be of the same form as before substitution, and hence are true for any radius. From proportions (1) and (2), we find,

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(5) log blog a

(6) log clog a

c=

Applying logarithms to (3) and (4), we have,

(4) c

70. Examples.

A=35° 45'.

1. Given B=45° 28′. a = 7985.

log sin B+ a. c. log sin A-10. log sin C+ a. c. log sin A-10.

C.

Req.b.

a sin C

sin A

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C. A

81° 13': 98° 47'.

C=180°

(A + B) = 180°

Since the sine of the angle opposite the given side is to the sine of the angle opposite the required side as the given side is to the required side, we have the proportion,

sin A sin B::a: b, .. b

sin A sin C::a: c, .'. c =

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a sin B

sin A

... log blog a log sin B+ a. c. log sin A-10.

log a (7985)

= 3.90227

log sin B (45° 28′) = 9.85299

a. c. log sin A (35° 45') = 0.23340

log b

3.98866

In like manner we have the proportion,

a sin C
sin A

B

... b 9742.25.

=

.. log clog a log sin C+ a. c. log sin A-10..

log a (7985)

3.90227

log sin C (98° 47')

9.99488

a. c. log sin A (35° 45′)

0.23340

log c

4.13055

2. Given

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In finding log sin 98° 47', take the supplement of 98° 47, which is 81° 13', and find log sin 81° 13'.

50° 30′ 40′′. B=70° 45′ 30′′. a = 478.35 yd.

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A

13506.88.

58° 43′ 50′′.

585.2 yd.
529.8 yd.

71. Case II.

Given two sides and an angle opposite one of them, required the remaining parts.

54° 05′ 51′′.

11.72 miles.

10.91 miles.

1. WHEN THE GIVEN ANGLE IS ACUTE.

Let the sides a and b and the angle A be given, and the remaining parts be required.

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Let the perpendicular p be drawn from C to the opposite side. Then we shall have,

p=b sin A.

1st. If a> p and a <b, there will be two solutions. For, if with C as a center and a as radius a circumference be described, it will intersect the side opposite. C in two points, B and B', and either triangle, ABC or ABC will fulfill the conditions of the problem, since

B'

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0

p

a

B

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