By a like process, formula (2) reduces to

92. If, in formula I, we advance one letter, we have

(II)

And dividing tanA by tan B, and reducing, we obtain

Multiplying tan A by tan B, and reducing, we obtain

By division and composition, and by 88 30, 38, this be

Proceeding in a similar way with formula II, we obtain

93. In the spherical triangle ABC, suppose CD drawn perpendicularly to AB, then, by the formulas for right spherical triangles,

Remark. If (A+B)>180°, then (a+b)>180°, and if (A+B)<

180°, then (a+b)< 180°.

94. All cases of oblique-angled triangles may be solved by applying one or more of the formulas I, II, III, IV, V, VI, VII, as shown in the following cases.

CASES

(1.) Given three sides, to find the angles. Apply formula I. Check: apply V or VI.

(2.) Given three angles, to find the sides.

Apply formula II. Check: apply III or IV. (3.) Given two sides and the included angle.

Apply V and VI, and VII. Check: apply III or IV.

(4.) Given two angles and included side.

Apply III and IV, and VII. Check: apply V or VI. (5.) Given two angles and an opposite side. Apply VII, V, and III. Check: apply IV.

97. (1.) Two sides and an angle opposite one of them are the given parts.

If the side opposite the given angle differs from 90° more than the other given side, the given angle and the side opposite being either both less or both greater than 90°, there are two solutions.

(2.) Two angles and a side opposite one of them are the given parts. If the angle opposite the given side differs from 90° more than the other given angle, the given side and the angle opposite being either both less or both greater than 90°, there are two solutions.

Remark.-There is no solution if, in either of the formulas,

the numerator of the fraction is greater than the denominator.