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SOLUTION OF PLANE RIGHT TEIANGLES. 51<br />
CHAPTER VI.<br />
SOLUTION OF PLANE RIGHT TRIANGLES.<br />
106. IN order to solve a plane right triangle, two parts in addition<br />
to the right angle must be given, one of which must be a side,<br />
The solution is effected directly by means of our Mg. 15.<br />
definitions of sine, &c., which are expressed by<br />
the equations (1). As three of the six functions<br />
are only the reciprocals of the other three, we<br />
shall base the solutions upon the foUowino- three ;<br />
(Fig. 15):<br />
, a , b , a<br />
svn A = — cos A = — tan A = -r<br />
0 c 0<br />
Since each of these equations expresses a relation between three<br />
parts-^-an angle and two sides—it follows that in order to apply them,<br />
or in order to solve the triangle trigonometrically, there must be<br />
given two of these parts ; and that of the three parts considered,<br />
one must be an angle while the other two are sides. Thus, if an<br />
angle and side are given, the third part sought must be a side; but<br />
if two sides are given, the third part sought must be an angle.<br />
In every instance the choice of the proper equation will be determined<br />
by the precept,—employ that trigonometric function of the<br />
angle which is equal to the ratio of the two sides considered.<br />
107. CASE. I. G-iven the hypotenuse and one angle, or c and A.<br />
To find a. We consider a, c and A; and since the ratio of a and<br />
c is given by the sine, we have<br />
sin A = — whence a = c sin A (195)<br />
0<br />
To find b. Considering b, c and A, we have the ratio of b and c<br />
expressed by the cosine, or<br />
b<br />
cos A = — whence b = c cos A (19^)<br />
e<br />
To find B. We have 5 = 90° - A.
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