3. When one of the circles is a diameter of the primitive, and the other is inclined to the latter. Let AFB and AIB be the two circles, and FAI the given angle. Draw the radius AC of the circle AIB, and AH perpendicular to AFB, and the angle HAC measures the given angle. Or, find P and E the poles of the circles; draw AE, AP, then GE measures the given angle, which is 50°. 4. When both the circles are inclined to the primitive. Let ABD, A'BD', be the two circles, and ABA' the given angle. Find C, C', the centres of the circles, then the A two radii drawn from these to B, will contain an angle CBC' equal to that at B. Or find P, P', the poles of the circles, and lines drawn from B through these points, will intercept on the primitive an arc which measures the given angle. The angle, in this instance, is 32°. с 'P P B F D 301. PROBLEM VII.-Through a given point in a given projected great circle, to describe the projection of another great circle, cutting the former at a given angle. Let ABCD be the primitive, and Z the given angle. 1. When the given circle is the primitive. Let A be the given point; draw the perpendicular diameters AC, BD; make angle EAF = Z=32° suppose; and from F as a centre, with a radius FA, describe the circle AGC; it is the required projection, and angle GAD = 32°. When the angle is a right angle, the diameter AC is evidently the required projection. B E Ꮐ D 2. When the given projected circle is a diameter of the primitive. Let BD be the given projection, and F the given point. Find GH the locus of all the great circles passing through F; draw FL perpendicular to BD, and FH, making an angle LFH = Z= 46°, for instance; from the centre H, with the B radius HF, describe the circle IFK ; it is the required projection, and angle DFK = 46°. F L E G F H If the angle be a right angle, G is the centre, and AFC the required projection, for angle LFG a right angle. Or, since the required circle is, in this case, perpendicular to BFD, it must pass through its poles A and C. Hence, the circle AFC, passing through the three points A, F, C, is the required projection. 3. When the given circle is inclined to the primitive. B Let AFC be the given circle, and F the given point in it. Find EG the locus of the centres of all the great circles passing through F. Draw FH a radius of the given circle, and draw FG, making the angle GFHZ= 23° suppose; from the centre G, with I the radius GF, describe IFE; and it is the required projection, and angle IFC23°. A E K H D G D When the angle Z is a right angle, draw from F a line perpendicular to FH, and it will cut EG in the centre of the required circle. Or since, in this case, the required projection must pass through the pole of AFC, find its pole, and describe the projection of a great circle passing through this pole and the point F (296), and it will be the required circle. 302. PROBLEM VIII.-Through a given point in the plane of the primitive, to describe the projection of a great circle cutting that of another great circle at a given angle. Let AKB be the given circle, Z the given angle, and C the given point in the plane of the primitive AMB. A Find F the pole of AKB, and about it describe a small circle IGN, at a distance from its pole equal to the measure of angle Z 44°, for example. About the given point C, as a pole, describe a great circle LHM, intersecting the small circle in L and G. About either of these points, as G, for a pole, describe a great circle DCE, and it is the required projection. For the circle DCE must pass through C, since C is at the distance of a quadrant from G, a point of the circle LGM. Also, the distance between F and G, the poles of AKB and DCE, is the measure of the given angle, and hence the inclination of the circles is equal to that angle = 44°. B K M E Schol. 1.-Let an arc of a great circle FCK be described through F and C; then, FK and CH being quadrants, FHCK. Now, FH must not exceed FN, the measure of the angle, otherwise the circle LHM would not meet IGN, and the problem would be impossible. But CK=FH; therefore, the distance of the given point from the given circle must not exceed the measure of the angle. Schol. 2.-If the point C were in the centre of the primitive, the circle LGM would coincide with the primitive. If C were in the circumference of the primitive, the circle LGM would be a diameter perpendicular to that passing through C. 303. PROBLEM IX.-To describe the projection of a great circle that shall cut the primitive and a given great circle at given angles. Let ADB be the primitive, AEB the given circle, and X, Y, the given angles, which the required circle makes respectively with these circles, and let these angles be respectively 47° and 45°. About F, the pole of the primitive, describe a small circle at a distance of 47°, the measure of angle X, and about G, the pole of AEB, describe another small circle at a distance of 45°, the measure of angle Y. Then, from either of the points of intersection H, I, as I for a pole, describe the great circle CED, and it is the required circle. For the distances of its pole I from F and G, the poles of the given circles, are equal to the measures of the angles X and Y; and therefore the inclinations of CED to the given circles are equal to these angles, that is, angle ACE = 47°, and AEC = 45°. B H Schol. When any of the angles exceeds a right angle, the distance of the small circle from its pole is greater than a quadrant. The same small circle will be determined by finding the more remote pole, that is, the projection of the pole nearest to the projecting point, and then describing a small circle about it, at a distance equal to the supplement of the measure of the angle. STEREOGRAPHIC PROJECTION OF THE CASES PROJECTION OF THE CASES OF RIGHT-ANGLED E H 304. Case I.-Given the hypotenuse AC 64°, and the angle C 46°, to construct the triangle, and to measure its other parts. K Let ECFD be the primitive; draw the circle CAD, making angle C = 46° (301); about C, as a pole, describe the small circle IAH at a distance 64° from C (298); then through A draw the diameter BK; and ABC is the given triangle. Measure the sides AB, BC, and angle A (299 and 300); and it will be found that AB = 40° 17', BC= 54° 55', and A = 65° 35′. B F and 305. Case II.-Given the hypotenuse AC. 70° 24', the side BC 65° 10', to construct the triangle. C Make the arc BC = 65° 10′, and describe the small circle IAH at a distance from its pole C equal to 70° 24′ (298); draw the diameter BAG, and then through A and C describe the great circle CAD; and ABC is the required tri- E angle. G Measure the side AB, and angles A and C, as in the preceding problem. Angle C39° 42', A = 74° 26', and AB=37°. AC 70° 24', A = 74° 26', = C = 39° 42'. 306. Case III.—Given the side AB 37°, and BC 65° 10′, to construct the triangle. E Make BC 65° 10′; draw the diameter BAK; and about G, as a pole, describe the small circle AIH at a distance from G the complement of AB 53° (298), then is AB37°; through A and C describe the great circle CAD (296); and ABC is the required triangle. Measure AC, and angle A and C, K as before. and Draw a diameter BL, and find its pole P (297); about the pole P describe the small circle KI'I at a distance from P of 32° 30'; and about G, the pole of the primitive, describe E a small circle I'IQ at a distance from it 73° 36', the supplement of angle C (298); and about I, the intersection of these small circles, describe = M D A A B I 307. Case IV.-Given angle A 32° 30', and C 106° 24', to construct the triangle. B K F B F |