Given the hypothenuse and either side about the right angle, to find the remaining parts.

41. The angle at the base may be found by one of Formulas (10) and (11), and the remaining side may then be found by one of Formulas (7) and (8).

EXAMPLES.

1. Given a = 2391.76, and b = 385.7, to find B, C, and C.

OPERATION.

Applying logarithms to Formula (11), we have,

log cos C =

log b + 10 – log a = log b + (a. c.) log a ;

Ans. B = 8° 43′ 54′′, C′ = 81° 16′ 6′′, and b = 19.3 yds.

3. Given a = 100, 100,

and b = 60, to find B, C, and e.

Ans. B = 36° 52′ 11′′, С = 53° 7′ 49′′, and c

= 80.

SOLUTION OF

OBLIQUE-ANGLED

TRIANGLES.

42. In the solution of oblique-angled triangles, four cases may arise. We shall discuss these cases in order.

CASE I.

Given one side and two angles, to determine the remaining

sides, and A and B the .

Since a and b are any two are any two sides, and A angles lying opposite to them, we have the following princi ple:

The sides of a plane triangle are proportional to the sines of the opposite angles.

It is to be observed that Formula (13) is true for any value of the radius. Hence, to solve a triangle, when a side and two angles are given:

First find the third angle, by subtracting the sum of the given angles from 180°; then find each of the required sides by means of the principle just demonstrated.

EXAMPLES.

1. Given B = 58° 07′, c = 22° 37′, and a = 408, to find A, b, and C.

To find b, write the proportion,

sin A: sin B :: a : b;

that is, 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.

Applying logarithms, and reducing, we have,

log blog a + log sin B + (a. c.) log sin A 10;

log c

log a

log a log sin C+ (a. c.) log sin A 10;

log sin C

Ans. A = 99° 16′, b
99° 16′, b = 351.024,
351.024, and c = 158.976.

2. Given A

= 38° 25′, B = 57° 42′, to find C, α, and b.

and c = 400,

3. Given A = 15° 19′ 51′′, c = 250.4 yds, to find B, α,

Ans. C 83° 53', α 249.974,

b = 340.04.

C = 72° 44′ 05′′, and

and b.

=

262.066 yds.

C

37° 21′ 25′′, and

and C.

Ans. B 91° 56' 04", a = 69.332 yds., b

4. Given B = 51° 15′ 35′′, a = 305.296 ft., to find A, b,

Ans. A = 91° 23′, b =238.1978 ft., c = 185.3 ft.

CASE II.

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

44. The solution, in this case, is commenced by finding a second angle by means of Formula (13), after which we may proceed as in CASE I.; or, the solution may be completed by a continued application of Formula (13).

ɑ : b :: sin A: sin B ;

that is, the side opposite the given angle, is to the side op posite the required angle, as the sine of the given angle is to the sine of the required angle.