10. Graphic Demonstration of the Formula 2 sin q cos q = sin 2 g. Let BCA-q=any angle; ACD=2q; R= radius of the circle ABE. Drop a perpendicular from B to CA, BC is perpendicular to HA; therefore the rightangled triangles CBF and CAH are similar. They are equal, for the hypothenuses CB and CA are radii. They are also equal to DHC. It follows from this that the triangle DCA 2 CBF BFX FC. = = = = = From D drop a perpendicular to EC; the area of the triangle ECD DGX EC DGX R. The area of the triangle EAD=DGX EA1DGX2R=DGX R. But DEA-DEC-BFX FC-DCA DGXR-DGX R DGXR, therefore DEC-DCA. But BFR sin q; CFR cos q; DG = R sin 2 g. We have just seen that twice the triangle B CF= triangle DEC; i. e. BFX FC=1DGR, therefore R sin pXR cos q=RXR sin 2 q; hence, = 2 sin q cos q = sin 2 p. -E. HARRISON, St. Louis, Mo. 11. Note on the Trisection of an Arc.- An ellipse and concentric circle are so drawn that BR, and A B+ 2 R. Required a geometrical trisection of each of the arcs intercepted on the circumference of the circle by the ellipse. Solution.-From either of the four points of intersection, i, with a radius = B, cut A at = d, d'. Through the centre, o, draw of parallel to di, or (d'i). Then the exterior arc will be trisected at t. From t, with radius R, cut the circumference of the circle (on the same side of the ellipse's minor axis) at e. Then the interior arc will be trisected at e. The arcs may also be trisected by means of a radius A. = Proof.- Extend id to t. Angle itt and tot are equal, and arc itt twice tt...itti is trisected at t and 180°-itti is trisected by ie, which 60° — it. NOTE. = SALMON'S "Treatise on Conic Sections" gives the problem of the trisection of circular arcs, the point of trisection being determined as the intersection of the given arc with a given hyperbola. I have never met with a solution of the problem by means of the ellipse.-PLINY EARLE CHASE, Philadelphia, Pa. ON SPHERICAL ANALYSIS. By GEORGE EASTWOOD, Saxonville, Mass. [Continued from Page 190.] To find the equation of a great circle of the sphere in terms of the coordinates of its pole.- Sometimes it is more convenient to use the equation of a great circle expressed in terms of the co-ordinates of its pole, than the one expressed in terms of the intercepts of the axes. Suppose, therefore, that AB, in the annexed diagram, is the proposed great circle; that x1, 91, are the geographical coordinates of its pole or centre, and that x, y, are the geographical co-ordinates of any point P in its circumference. By inspection of the figure, π-1, π — ÿ, and, are the sides of a spherical triangle whose vertex is at Y, and whose vertical angle is x-x. Hence, by spherics, we have 1 cosπ = sin y sin y1+ cos y cos y1 cos (x1 - x), and by division and expansion, we have tan y tan yı = (cos x cos x1 + sin x sin x). But, when the axes are rectangular, we have tan y = y cos x, tan y11 cos x,; the substitution of which in tang tany, gives, yı after dividing out cos x cos x1, y y1 = — 1 — x X1, YY1 Comparing this equation with+= 1, we see that ß α Hence the equation of a great circle expressed in tangent-functions of the co-ordinates of its pole is (22) YY1 + xx1+1=0. NOTES ON PROBABILITIES. By SIMON NEWCOMB, Nautical Almanac Office, Cambridge, Mass. [Continued from Page 140.] 22. THE a priori probability of an event is its probability deduced solely from circumstances which preceded or might have preceded it. In the theory of probabilities these circumstances are regarded as causes. The a posteriori probability of an event or hypothesis is its probability deduced from events which have followed it, which events are regarded as its possible effects. Theorem. The a priori probability of an event is equal to the sum of the products obtained by multiplying the probability of each distinct cause by the probability that the event would occur on the hypothesis that the cause operates. Let m be the whole number of separate cases. Let n of these cases be favorable to the first cause, n' to the second, &c. Their probabilities will then be &c. Also, let h of the n cases, n n' n" 2 m m m h' of the n' cases, &c., be favorable to the production of the event. Its probability, on the several hypotheses of the occurrence of each separate cause, will then be h h' h" n'n'' n" , &c. Multiplying the corre sponding fractions in the two series, and taking the sum of the products we obtain h+h+h" + &c. But this is the probability h' m of the event, because h+h +h" + &c. is the whole number of cases favorable to its production. 23. Problem. Let E, E', E", &c., represent a portion of the events which may occur on a certain trial, and let their respective probabilities be p, p, p", &c. Suppose now that after I have estimated these probabilities, I am informed by a person who knows the result of the trial that one of the events E, E', E", &c., has actually occurred. What ought to be their respective probabilities in my mind after I have received this information? Let m be the whole number of possible cases; let n of these cases be favorable to E, n' to E', n" to E", &c. Then we have But when it is ascertained that one of the events E has actually occurred, the number of possible cases is reduced from m to n + n + n" + &c., those favorable to the production of some one of the events E. The probabilities of these events will therefore become that is, each of the former probabilities will be multiplied by the factor m n+n'+n" + &c. ; or, they will each be increased in the same ratio to such a degree as to make their sum equal to unity. Example. On a table are fifty small boxes, of which three each contain a gold coin, seven a silver coin, thirteen a copper coin, and twenty-seven nothing. The probability that one selected at random contains a gold coin is, silver, &c. But if by shaking the box it is found that it contains a coin, it will be known that one of the 23 boxes containing coins must have been selected. The probability that the box contains gold will then be increased to, that it contains silver to, that it contains copper to 13. 24. Problem.-To find the probability of an event or hypothesis from both a priori and a posteriori circumstances. Suppose that a possible event E, if it occur at all, must be preceded by one and one only of the causes or circumstances C, C", C", &c. Let p, p', p", &c., be the probabilities of C, C', C", &c., before it is known whether the event E has occurred. Let q be the probability of E on the hypothesis that C occurs, its probability on the hypothesis that Coccurs, &c. The a priori probabilities of the compound events CE, C'E', &c., will then be pq, p'q', &c. If, now, it is ascertained that the event E has actually occurred, it becomes certain that one of these compound events has occurred; and, by § 23, their probabilities will be increased to (1) p q p'q' Pq + p'q' + &c.' p q + p' q' + &c. &c. But, the event E being now certain, the probabilities of the causes C, C', &c., are the same as the probabilities of the compound events CE, C'E', &c.; so that the above fractions represent the probabilities of the causes or hypotheses, deduced from all the circumstances in our knowledge, which bear on the question. Example I.-A, handling a die, entertains a suspicion of that the die is so loaded that a six is as likely to be thrown as not. He throws it twice, and each time throws a six. What is the probability that his suspicion is correct? The solution is exhibited as follows: : |