Plane trigonometry with practical applications, by Leonard E. Dickson.

116

TRIGONOMETRY

[Art. 88

We shall see that formulas (1) enable us to solve a triangle if
we are given a side and two angles, or two sides and the angle
opposite to one of them. A suitable check formula is furnished by
any one of the formulas (5), (6), (7) of Art. 88.
88. Law of tangents. Given two sides a, b and the included
angle C, we cannot solve the triangle by use of the law of sines,
since each formula (1) contains two unknowns. We proceed to
prove a formula suitable for this purpose.
Let ABC be a triangle in which the
~ B side BC = a is greater than the side
AC = b (Fig. 72). With center C and
radius CB describe a semicircle meeting
A C produced in D and E. Join B with D
E C bA D and E. Then
FIG. 72 EA = EC + b = a + b,
AD = CD-b = a-b.
Since Z CBD = Z CDB and Z CBD + Z CDB =180- C =
A + B, we have
ZADB = Z CDB= (A + B),
ABD = Z CBD- Z B = (A +B)-B = (A- B).
Since Z EBD is inscribed in a semicircle, it is a right angle. Hence,
by Art. 12,
ZAEB = 90~- ZADB=90~- ~(A+B), sin AEB=cos (A+B),
ZEBA = 90~- Z ABD = 90 — (A - ), sin EBA = cos1 (A-B).
Applying the law of sines to triangle ABD, we get
a-b AD sin ABD sin '(A-B)
c - AB sin ADB sinI (A+B)
Similarly, applying the law of sines to triangle EAB, we get
a+b EA sin EBA cos -(A-B)
(3) c AB sinAEB cos
c - AB - sin AEB - cos '(A +B)

Hence by division of (2) by (3),