CHAPTER III.

OF RIGHT TRIANGLES.

35. Fundamental relations. Let OCN be a right triangle of which a and b are the sides which contain the right angle, c the hypothenuse, a and 6 the angles opposite a and b respectively.

If we take ON as a radius and draw the arc NX from the centre O, the side NC will, by definition, represent the sine of XON, and OC its cosine, when the radius is ON. That is,

NO
ON = sin a;
ΟΝ

We may show in the same way, by taking N as the centre and NO as the radius,

Ос
ON

= sin ß;

NC
ON

= cos B.

(2)

We might also have deduced these equations from (1), because B = 90°

α,

whence

=

sin ß
sin (90° — a) = cos α;
cos B = cos (90° — a) = sin a.

Again, by taking OC as a radius, we find

Putting NC = a, OC = b, ON = c, the equations (1), (2), and (3) give the relations

We may express the same relations in terms of 6, using the complementary functions, as follows:

a = c cos B = b cot ß;

bc sin ß = a tan ẞ;

c = a sec ẞ

=

b cosec B.

(5)

These relations may be summed up in the following general theorems:

I. The hypothenuse of any right triangle is equal to a side into the secant of its adjacent angle or the cosecant of its opposite angle.

II. A side is equal to the hypothenuse into the sine of the opposite angle or the cosine of the adjacent angle.

III. One side is equal to the other side into the tangent of the angle adjacent to that other side or the cotangent of the angle adjacent to itself.

EXERCISE.

Show by the above equations how each side will be expressed in terms of the others when a 30° and when α = 45°, using the values of sin a, etc., already found—namely,

and show how the results agree with those of elementary geometry.

36. Examples and exercises in expression. In the accompanying figure OQN, ONP, and OXN

It is now required to express all the other lines in terms of OX and trigonometric functions of a and B.

1. By the same process express OQ, QN, OX, NX, NP, and XP in terms of ON and trigonometric functions.

2. Express the same quantities in terms of NP.

3. Express NX separately in terms of OX and XP, and by multiplying the two values prove the geometric theorem that NX is a mean proportional between OX and XP.

4. In a right triangle the sides which contain the right angle are a and b, (a > b), and ♂ is the difference of the angles at the base (hypothenuse). Express the length of the perpendicular from the vertex upon the base in two ways, and the lengths of the segments into which it divides the base each in one way. The expressions are all to be in terms of a, b and S.

Ans. (in part). One expression for the perpendicular is

pa sin (45° – † 6).

Solution of Right Triangles.

Since in a right triangle one angle-the right angle-is given, only two other independent parts are required to solve the triangle. These two parts may be any two of the sides or one side and one angle. What parts soever are given, the remaining parts may be found by the equations (4) and (5). The following are all the essentially different cases.

37. CASE I. Given the two sides a and b adjacent to the

Therefore the quotient of the two sides gives the tangent of the angle opposite the dividend side.

From the tangent the angle a is itself found by the trigonometric tables; then sec a or cos a; then the hypothenuse c from the equation

c = b sec a =

Example. Given a 9 metres, 6 12 metres, to find the remaining parts of the triangle.

Solution by numbers and measurement.

tan α = = 0.75.

On the tangent line XN (§22) measure a distance from X equal to 0.75 of the radius OX; join the end of the distance to O, and measure the angle XON which the joining line makes with OX. This angle will be a. The length of the joining line divided by OX will be the secant, which multiplied by b = 12 will give the hypothenuse c.

* It is recommended that in commencing this subject the student first solve a few of the problems by his own process, and without the use of any tables but those he may construct for himself by measurement as described in § 18.

The first two exercises are made purposely simple, that they may be performed by measurement.

38. CASE II. Given the hypothenuse and one side. Solution. From the equation

which may be used to find a when a and c are given. Then the remaining side is found by the equation

=

b = c cos α.

Example. Given a 13, c = 20, to find the remaining parts. Solution by numbers and measurement.

larly to OB, take its ratio to OX, and multiply it by c= 5. This will give the side b.