Using the same letters as before to designate the three sides and the
subtending angles of oblique-angled triangles, we have the following
cases:—
(1.) Given one side and two angles, as _a_, _a_°, and _b_°, to find _b_,
_c_, and _c_°.
In the first place, _c_° = 180° − (_a_° + _b_°); also we note that, as
the sides are proportional to the sines of the opposite angles, _b_ =
(_a_ sine _b_°)/(sine _a_°) and _c_ = (_a_ sine _c_°)/(sine _a_°).
Taking as an example, _a_ = 45, _a_° = 57°, and _b_° = 63°, we have _c_°
= 180 − (57 + 63) = 60°. To find _b_ and _c_, set _a_° on S to _a_ on A,
and read off on A above 63° and 60° the values of _b_ (= 47·8) and _c_
(= 46·4) respectively.
(2.) Given _a_, _b_, and _a_°, to find _b_°, _c_°, and _c_.
In this case the angle _a_° on S is placed under the length of side _a_
on A and under _b_ on A is found the angle _b_° on S. The angle _c_° =
180 − (_a_° + _b_°), whence the length _c_ can be read off on A over
_c_° on S.
(3.) Given the sides and the included angle, to find the other side and
the remaining angles.
If, for example, there are given _a_ = 65, _b_ = 42, and the included
angle _c_° = 55°, we have (_a_ + _b_) ∶ (_a_ − _b_) = tan. (_a_° +
_b_°)/(2) ∶ tan. (_a_° − _b_°)/(2). Then, since _a_° + _b_° = 180° − 55°
= 125°, it follows that (_a_° + _b_°)/(2) = (125°)/(2) = 62° 30′.
By the rule for tangents of angles greater than 45°, we find tan. 62°
30′ = 1·92. Inserting in the above proportion the values thus found, we
have 107 ∶ 23 = 1·92 ∶ tan. (_a_° − _b_°)/(2). From this it is found
that the value of the tangent is 0·412, and placing the slide with all
indices coinciding, it is seen that this value on D corresponds to an
angle of 22° 25′. Therefore, since (_a_° + _b_°)/(2) = 62° 30′, and
(_a_° − _b_°)/(2) = 22° 25′, it follows that _a_° = 84° 55′, and _b_° =
40° 5′. Finally, to determine the side _c_, we have _c_ = (_a_ sin
_c_°)/(sin _a_°) as before.