8. The co-tangent of either acute angle is equal to the adjacent side divided by the opposite side.

9. Either side adjacent to the right angle is equal to the hypotenuse divided by the secant of the adjacent acute angle.

10 The secant of either acute angle is equal to the hypotenuse divided by the adjacent side.

Since the angles B and P are complements of each other sec B =cosec P, sec P= cosec B; ... (9) and

11. Either side adjacent to the right angle is equal to the hypotenuse divided by the co-secant of the angle opposite that side.

12. The co-secant of either acute angle is equal to the hypotenuse divided by the side opposite that angle.

Scholium. By some authors, principles 2, 4, 6, 8, 10, and 12, have been given in the form of definitions.

Introducing radius into these formulas, by substituting for any function to the radius 1, the corresponding function to the radius R divided by R, and reducing, we have:

S. N. 5.

Given the hypotenuse and one acute angle, required the

Either side adjacent to the right angle is equal to the sine

of the opposite angle, multiplied by the hypotenuse.

. p= =h sin P.

h sin P

Introducing radius, we have, p ==

R

Applying logarithms, we have,

log plog h+ log sin P-10.

In like manner, from either formula, bh sin B, = h cos P, we find b=305.41.

or b

Given the hypotenuse and one side adjacent to the right angle, required the remaining parts.

The sine of either acute angle is equal to the opposite side divided by the hypotenuse.

Introducing radius, and multiplying by R, we have,

B 90° P-90°-60° 00′ 17′′-29° 59′ 43′′.

bh sin B, or bh cos P, .. b 55.991.

We

e can also find b as follows:

b = √ h2 — p2 = 1 (h + p) (h − p).

log b = [log (h+p) + log (h—p)].

Given one side adjacent to the right angle and one acute angle, required the remaining parts.

B 90° - P= 90° – 50° 18′ 32" 39° 41′ 28′′.

B

H