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°.

Example.

Let A'B'C' be a quadrantal triangle, in which

B

and

B'C' = 90°,

B' = 75° 42',

c' 18° 37'.

Passing to the supplemental polar triangle, we have

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", in like

Other quadrantal triangles may be solved in

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 a;

a

B

с

B

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 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

P

B

(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

OE cos ɑ, DC = sin b,

We have from the figure,

OD = cos b.

OE OL + QP.

(1.)

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;

and now substituting these values of OE, OL, and QP, in (1), we have

COS C = cos a cos b + sin a sin b cos C. ·

That is, the cosine of any side of a spherical triangle is equal to the rectangle of the cosines of the two other sides, plus the rectangle of the sines of these sides into the cosine of their included angle.

80. To find an expression for the cosine of any angle of a spherical triangle.

If we represent the angles of the supplemental polar triangle of ABC, by A', B', and C', and the sides by a', b', and c', we have (B. IX., P. VI.),

Substituting these values in equation (3), of the preceding

or, changing the signs and omitting the primes (since the preceding result is true for any triangle),

That is, the cosine of any angle of a spherical triangle is equal to the rectangle of the sines of the two other angles into the cosine of their included side, minus the rectangle of the cosines of these angles.

The formulas deduced in Arts. 79 and 80, for cos a, cos A, etc., are not convenient for use, as logarithms can not be applied to them; other formulas are, therefore, derived from them, to which logarithms may be applied.

81. To find an expression for the cosine of one half of any angle of a spherical triangle.

From equation (3), Art. 79, we deduce,

If we add this equation, member by member, to the number 1, and recollect that 1+ cos A, in the first member, is equal to 2 cos2 A (Art. 66), and reduce, we have