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

and C.

=

Because b< B, there are two solutions.

As a check, to test the accuracy of the above work, formula (2) may be used. Thus, from that formula,

log sin c = log sin a log sin C 10.

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

a = 86° 51', and B 18° 03' 32", to find b,

3. Given a = c, and C.

4.

=

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

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

Ans. a 142° 09′ 13′′,

B =

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′′.

=

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

Ans. α = 133° 32' 26", b = 32° 08′ 56", c =

and C 126° 40' 24", to =

144° 27′ 03′′.

89

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°.

Passing to the supplemental polar triangle, we have

b

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,

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 a;

from which results the proportion,

B

B'

a

B

That is, in any spherical triangle, the sines of the sides 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.

Let ABC represent any spherical triangle, and O the centre of the sphere on which it

is situated.
OB, and OC; from C draw CP per-
pendicular to the plane AOB; from
P, the foot of this perpendicular,
draw PD and PE respectively per-
pendicular to OA and OB; join CD
and CE, these lines will be respect-
ively perpendicular to OA and OB

Draw the radii OA,

(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

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

= PD sin c.

QP PD sin QDP =

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;

(2.)