4 Cos sin sin

and similar expressions for sin (S – B) and sin (S

C).

-

Take the colunar triangle, whose angles are π - В, π - С, A, and sides

[It is evident that the numerator of the fraction is of the arc joining the middle points of the sides - b, triangle; therefore, &c. (Art. 26, and Ex. 4.)]

33. In any triangle

equal to the cosine c, of the colunar

37. If a spherical triangle is equal and similar to its polar triangle,

sec a sec b sec c + tan b tan c.

Give a particular example of such a triangle.

(Science and Art Exam. Papers.)

38. Prove that by a closer approximation, employing the notation used in Legendre's Theorem (Art. 29),

39. Deduce the following formula for a + b in terms of the opposite angles

(See Todhunter's Plane Trig., p. 238, Art. 299.)

40. From any three points on a sphere are drawn to each of three points on a great circle arcs, a1, b1, c1; a2, b2, c2; aз, bз, c3; making angles with the great circle, respectively, A1, A2, A3, etc. relation

.; prove the determinant

sin ai cos i
sin bi cos B1 sin b2 cos B2

sin a2 cos 42

sin a3 cos A3

sin b3 cos B3

= 0.

sin ci cos C1 sin c2 cos C2 sin c3 cos C3

41. If a and b are the sides, and e the base of a triangle, prove that if be the angle which B, the bisector of the base, makes with the base,

[Find tana, and tan2b, and multiply the results].

42. Find the locus of the vertex, having given the base and

(1) Ratio of cotangents of the base angles.

(2) Sum or difference of cotangents.

(3) cot A+ m cot B.

(Cf. p. 58, Ex. 10.) (Educational Times, April, 1885.)

CHAPTER IV.

RIGHT-ANGLED TRIANGLES.

SECTION I.

44. Geometrical Deduction of Formulæ connecting the Parts of a Right-angled Triangle.— The formulæ necessary for the solution of right-angled triangles have been deduced in Chapter III. as particular cases of the general formula

cos a = cos b cos c + sin b sin c cos A.

We, however, add the following geometrical method of obtaining them, as it is both simple and instructive :—

Let ABC (fig. 24) be a triangle right-angled at C', and let O be the centre of the sphere on which it is described. Since C is a right angle, the planes OCA and OCB are perpendicular to each other (Art. 6). Take P any point on the radius OA, and draw PR perpendicular to OC, and RQ perpendicular to OB. Join PQ. Since the planes. COA and COB are at right angles, the angle PRQ is right, and, as in Art. 32, the angle OQP is also right. Hence we have

From R draw RP' perpendicular to OA, and draw P'Q' also perpendicular to OA, in the plane AOB, and meeting OB in Q. Then (Art. 6) the angle Q'P'R is equal to the angle A. Join Q'R.

Now PQ P'Q" + PP = PR+ RQ2; (Art. 32.) therefore

PQPR RQ PP. But PR PP'2 = RP2.