Page images
PDF
EPUB
[blocks in formation]

10. Given a = 3.019, b = 6.731, c=4.228.

==

CASE IV.

124. Given two sides and the angle opposite to one of them.

It was stated in Art. 106 that a triangle is in general completely determined when three of its elements are known, provided one of them is a side. The only exceptions occur in Case IV.

To illustrate, let us consider the following:

1. Given a = 52.1, b = 61.2, A=31° 26'. Required B, C, and c.

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small]

But in determining the angle corresponding, attention must be paid to the fact that an angle and its supplement have the same sine (Art. 47).

Therefore another value of B will be 180° - 37° 47.5', or 142° 12.5'; and calling these values B1 and B2, we have

B1 = 37° 47.5', and B, 142° 12.5'.

=

Note. The reason for this ambiguity is at once apparent when we attempt to construct the triangle from the data.

[merged small][ocr errors][merged small][merged small][merged small][merged small][merged small]

We first lay off the angle DAF equal to 31° 26', and on AF take AC 61.2. With C as a centre, and a radius equal to 52.1, describe an arc cutting AD at B1 and B. Then either of the triangles AB1C or ABC satisfies the given conditions.

The two values of B which were obtained are the values of the angles ABC and ABC respectively; and it is evident geometrically that these angles are supplementary.

To complete the solution, denote the angles ACB, and ACB by C1 and C2, and the sides AB, and AB, by c1 and c. Then, C1=180° — (A+B1) = 180° - 69° 13.5'=110° 46.5', C2=180°- (4+B) = 180°-173° 38.5' 6° 21.5'.

and

[blocks in formation]

=

[merged small][merged small][merged small][merged small][ocr errors][merged small][merged small]

c1 = a sin C1 csc A,

log sin C1 = 9.9708

log csc A = 0.2827

log c1 = 1.9703

.*. C1 = 93.4.

=

log a =

=

a sin C2 csc A.

= 1.7168

log sin C 9.0443
log csc A= 0.2827

[blocks in formation]

125. Whenever an angle of an oblique triangle is determined from its sine, both the acute and obtuse values must be retained as solutions, unless one of them can be shown by other considerations to be inadmissible; and hence there may sometimes be two solutions, sometimes only one, and sometimes none, in an example under Case IV.

I. Let the data be a, b, and A, and suppose b<a. Since, by Geometry, B must be < 4, only the acute value of B can be taken; in this case there is but one solution.

II. Let the data be a, b, and A, and suppose b> a. Since B must be > A, the triangle is impossible unless A is acute.

Again, since sin B

sin A

[blocks in formation]

Hence both the acute and obtuse values of B are > A, and there are two solutions, except in the following cases:

If the data are such as to make log sin B = 0, then sin B1 (Art. 87) and B = 90°, and the triangle is a right triangle; if log sin B is positive, then sin B is > 1, and the triangle is impossible.

126. The results of Art. 125 may be stated as follows:

If, of the given sides, that adjacent to the given angle is the less, there is but one solution, corresponding to the acute value of the opposite angle.

If the side adjacent to the given angle is the greater, there are two solutions unless the log sine of the opposite angle is 9 or positive; in which cases there are one solution (a right triangle), and no solution, respectively.

127. We will illustrate the above points by examples:

2. Given a = 7.42, b=3.39, A= 105° 13'; find B. Since bisa, there is but one solution, corresponding to the acute value of B.

[blocks in formation]

3. Given b=3, c=2, C=100°; find B.

Since b is >c, and C is obtuse, the triangle is impossible. 4. Given a = 22.764, c=50, A= 27° 4.8'; find C.

[blocks in formation]

log c = : 1.6990

colog a = = 8.6428 log sin 49.6582

log sin C=0.0000

... sin C= 1, and C = 90°.

Here there is but one solution; a right triangle.

5. Given a = .83, b=.715, B= 61° 47'; find A.

[merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small]

Since log sin A is positive, the triangle is impossible.

[blocks in formation]
[blocks in formation]

128. 1. Given a = 18.063, A = 96° 30', B = 35°; find K.

By Art. 119, 2K=

Whence,

a2 sin B sin C

sin A

=a2 sin B sin C csc A.

log (2K)= 2log a + log sin B + log sin C+ log csc A. From the data,

C = 180° - (A+B) = 48° 30'.

log a = 1.2568; multiply by 2 = 2.5136

[blocks in formation]
« PreviousContinue »