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 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 polar triangle of any quadrantal triangle being solved, the parts of the given triangle may be found by subtracting each part of the polar triangle from 180°.

Solving this triangle by previous rules, we find,

B

α = 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 a similar manner, other quadrantal triangles may be solved.

FORMULAS USED IN

SOLVING OBLIQUE-ANGLED SPHERICAL TRI

ANGLES.

78. Let ABC represent an oblique-angled spherical tri angle. From either vertex, 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,

C

B

c

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. Let ABC represent any spherical triangle, and O

CD and CE, these lines will be respectively perpendicular to OA and OB (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,

substituting this value in (2), we have,

and now substituting these values of OE, OL, and QP,

in (1), we have,

COS α = cos b cos c + sin b sin c cos A

In the same way, we may deduce,

cos b = cos a cos c + sin a sin c cos B

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

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

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

a = 180° - A',

b = 180°

B', c = 180° - C',

A 180° - a', B = 180° — b′, C 180°. c'.

Substituting these values in Equation (3), of the preceding article, and recollecting that,

cos (180°-A') = cos A', sin (180° - B')

we have,

= sin B', &c.,

cos A' cos B' cos C' - sin B' sin C' cos a';

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

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

81. From Equation (3), Art. 79, we deduce,

cos a

cos A =

cos b cos c sin b sin c

(1.)

If we add this equation, member by member, to the num ber 1, and recollect that 1+cos A, in the first member, is equal to 2 cos24 (Art. 66), and reduce, we have,

cos acos (b+c) = 2 sin (a + b + c) sin (b + c − a),

Equation (2) becomes, after dividing both members by 2,