Page images
PDF
EPUB

2. = Given b 51° 30', and B = 58° 35', to find a, c,

and C.

Because b< B, there are two solutions.

[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][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

As a check, to test the accuracy of the above work

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

As the test is satisfied, the work is probably correct.
Other cases may be treated in like manner.

3. Given a = 86° 51', and B = 18° 03' 32", to find b, c, and C.

Ans. b 18° 01' 50", c = 86° 41′ 14′′, C = 88° 58' 25".

=

4. Given b = 155° 27′ 54′′, and c = 29° 46′ 08′′, to find a, B, and C.

Ans. a 142° 09′ 13", =

142° 09′ 13′′, B = 137° 24′ 21′′,
137° 24′ 21′′, C 54° 01′ 16′′.

=

5. Given c 73° 41′ 35′′, and B 99° 17′ 33′′, to find = = a, b, and C.

Ans. a = 92° 42′ 17′′, b = 99° 40′ 30′′, C = 73° 54′ 47′′.

[ocr errors]

6. Given b = 115° 20', and B 91° 01' 47", to find a, c, and C.

a

=

64° 41' 11", с = 177° 49′ 27′′, C = 177° 35′ 36′′. a' 115° 18' 49", c' = 2° 10' 33", C' = 2° 24' 24".

7. Given B = 47° 13′ 43", and C 126° 40' 24", to find a, b, and c.

Ans. a = 133° 32′ 26′′, b = 32° 08′ 56", c = 144° 27' 03".

QUADRANTAL SPHERICAL TRIANGLES.

77. A QUADRANTAL SPHERICAL TRIANGLE is one in which one side is equal to 90°. To solve such a triangle, we pass to its supplemental polar triangle, by subtracting each side and each angle from 180° (B. IX., P. VI.). The resulting polar triangle will be right-angled, and may be solved by the rules already given. The supplemental polar triangle of any quadrantal triangle being solved, the parts of the given triangle may be found by subtracting each part of the supplemental triangle from 180°.

[blocks in formation]

Passing to the supplemental polar triangle, we have

A = 90°,

b 104° 18',
=

and C =

= 161° 23'.

Solving this triangle by previous rules, we find

a = 76° 25' 11", c = 161° 55′ 20′′, B 94° 31' 21";

hence, the required parts of the given quadrantal triangle are,

A' 103° 34′ 49", C' 18° 04' 40", b' 85° 28' 39",

Other quadrantal triangles may be solved in like

manner.

FORMULAS

USED IN SOLVING OBLIQUE-ANGLED SPHERICAL TRIANGLES.

78. To show that, in a spherical triangle, the sines of the sides are proportional to the sines of their opposite angles.

Let ABC represent an oblique-angled spherical triangle. From any vertex, as C, draw the arc of a great circle, CB', perpendicular to the opposite side. The two triangles ACB' and BCB' will be rightangled at B'.

From the triangle ACB', we have, formula (2) Art. 74,

sin CB' sin A sin b.

From the triangle BCB', we have

sin CB' = sin B sin a.

Equating these values of sin CB', we have

sin A sin b

sin B sin ɑ;

from which results the proportion,

[ocr errors]

C

B'

B

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

are proportional to the sines of their opposite angles.

Had the perpendicular fallen on the prolongation of AB,

the same relation would have been found.

79. To find an expression for the cosine of any side of a spherical triangle.

a

B

Let ABC represent any spherical triangle, and O the centre of the sphere on which it is situated. Draw the radii OA, OB, and OC; from C draw CP perpendicular to the plane AOB; from P, the foot of this perpendicular, draw PD and PE respectively perpendicular to OA and OB; join CD and CE, these lines will be respectively perpendicular to OA and OB

[ocr errors]

C

(B. VI., P. VI.), and the angles CDP and CEP will be equal to the angles A and B respectively. Draw DL and PQ, the one perpendicular, and the other parallel to OB. We then have

[blocks in formation]

In the right-angled triangle OLD,

OL OD cos DOL = cos b cos c.

The right-angled triangle PQD has its sides respectively perpendicular to those of OLD; it is, therefore, similar to it, and the angle QDP is equal to c, and we have

QP = PD sin QDP = PD sin c.

The right-angled triangle CPD gives

PD = CD cos CDP = sin b cos A;

substituting this value in (2), we have

QP sin b sin c cos A;

[ocr errors]
[ocr errors][merged small]
« PreviousContinue »