The angle or arc corresponding to the logarithmic tangent 9.497200 is 17° 26' 33". The angle or are corresponding to the logarithmic cosine 9.792477 is 51° 40′ 30′′. EXAMPLES. 1. Required the logarithmic sine of 28° 42'. Ans. 9.681443. 2. Required the logarithmic cosine of 59° 33' 47". Ans. 9.704657. 3. Required the logarithmic cotangent of 127° 2'. Ans. 9.877640. 4. Required the logarithmic sine of 81° 20' Ans. 10.207523. 6. Required the logarithmic tangent of 74° 21' 20". Ans. 10.552778. 7. Required the logarithmic cosecant of 102° 24′ 41′′. Ans. 10.010270. 8. Required the logarithmic tangent of 1° 59′ 51′′.8. Ans. 8.542587. 9. Required the angle of the logarithmic sine 9.999969. Ans. 89° 19'. 10. Required the arc of the logarithmic tangent 9.645270. Ans. 23° 50′ 17′′. 11. Required the angle of the logarithmic cosine 9.598075. Ans. 66° 39'. 12. Required the angle of the logarithmic cotangent 10.301470. Ans. 26° 32′ 31′′. 13. Required the arc of the logarithmic sine 9.893410. Ans. 51° 28′ 40′′. 14. Required the angle of the logarithmic cosine 9.421157. Ans. 105° 17′ 29′′. 15. Required the arc of the logarithmic tangent 9.692125. Ans. 26° 12′ 20′′. 16. Required the angle of the logarithmic cotangent 9.421901. Ans. 75° 12' 6". BOOK III. SOLUTION OF PLANE TRIANGLES. 108. THE SOLUTION OF TRIANGLES is the process by which, when the values of a sufficient number of their elements are given, the values of the remaining elements are computed. The elements of every triangle are the three sides and the three angles. Three of these elements must be given, one of which must be a side, in order to solve a plane triangle. The solution of plane triangles depends upon the following FUNDAMENTAL PROPOSITIONS. 109. In a right-angled triangle, the side opposite to an acute angle is equal to the product of the hypothenuse into the sine of the angle; and the side adjacent to an acute angle is equal to the product of the hypothenuse into the cosine of the angle. Let ABC be a triangle having a right angle at C; then, by (1), h 110. In a right-angled triangle, the side opposite to an acute angle is equal to the product of the other side into the tangent of the angle; and the side adjacent to an acute angle is equal to the product of the other side into the cotangent of the angle. 111. In any plane triangle, the sides are proportional to the sines of the opposite angles. B Let A B C be any triangle, in which the sides opposite the angles A, B, C, respectively, are denoted by a, b, and c. From one of the angles, as B, draw BD perpendicular to the opposite side A C, and denote the line BD by p. Then the A right-angled triangles, CBD, A B D, give, by (85), D The angle was acute, but had it been angle, the results would have been the same. therefore, applies in every case. obtuse, or a right The proposition, 112. In any plane triangle, the sum of any two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference. For, by (90), ab sin A: sin B; whence (Geom., Prop. XII. Bk. II.), a+b: b:: sin ▲ + sin B: sin A — sin B, which may also be written, a+b: a b :: tan 1 (4+ B) : tan † (4 — B). (95) 113. In any triangle, the square of any side is equal to the sum of the squares of the two other sides, diminished by twice the rectangle of these sides multiplied by the cosine of the included angle. Let ABC be any plane triangle, in which the sides opposite the angles A, B, C, respectively, are denoted by a, b, c. Draw BD from one of the angles, B, perpendicular to the opposite side, AC. Then, if A is acute, we have by Geometry (Prop. XII. Bk. IV.), B A D b a2 = b2 + c2 2b AD; a but from the right-angled triangle ABD, by (86), we have AD c cos A; C therefore, a2 = b2 + c3 — 2 be cos A. (96) B When the angle A is obtuse, the point D will fall on the other side of A, and we have by Geometry (Prop. XIII. Bk. IV.), a2 = b2 + c2 + 26 AD. But since BAD is now the supplement of BAC, by Art. 77 we have ᎪᎠ c cos BAD-c cos BAC―c cos A. Substituting this value of AD, we have as before, When A is a right angle and a the hypothenuse, cos is zero (30), and (96) becomes a2 = b2 + c2, and thus the formula (96) is true, whatever the angle A may be. In like manner we have 114. The cosine of any angle of a plane triangle is equal to the fraction whose numerator is the sum of the squares of the containing sides, diminished by the square of the opposite side, and whose denominator is twice the product of the containing sides. For, by (96), a2 = b2 + c2 — 2 be cos A, 115. By these formulæ the angles of a triangle can be found when the sides are given, but they cannot be conveniently applied in computation by logarithms. We then subtract both members of formula (99) from 1, and |