TRANSIT WITH SOLAR ATTACHMENT. The circles shown in the cut are intended to represent in miniature circles supposed to be drawn upon the concave surface of the heavens. 16. sin A † (5+ √7) = 0.95572; etc. 17. cos A = (√31-1) = 0.57097; sin A = (√31 + 1) = 0.82097; etc. 18. a 12.3. 19. b 1.54. 20. a = 9. 21. b = 68. 22. c = 40. 24. Construct a rt. ▲ with legs equal to 3 and 2 respectively; then construct a similar ▲ with hypotenuse equal to 6. In like manner, 25, 26, 27, may be solved. 28. a = 1.5 miles; b = 2 miles. 31. 400,000 miles. 30. a=0.342, b = 0.940; a = 1.368, b = 3.760. 32. 142.926 yards. EXERCISE III. 5. Through A (Fig. 3) draw a tangent, and take AT=3; the angle AOT is the required angle. 6. From 0 (Fig. 3) as a centre, with a radius = 2, describe an arc cutting at T the tangent drawn through B; the angle AOT is the required angle. 7. In Fig. 3, take OM, and erect MP OA and intersecting the circumference at P; the angle POM is the required angle. 8. Since sin x = cos x, OM = PM (Fig. 3), and x = 45°; hence, construct x = 45°. 9. Construct a rt. ▲ with one leg = twice the other; the angle opposite the longer leg is the required angle. 10. Divide OA (Fig. 3) into four equal parts; at the first point of division from O erect a perpendicular to meet the circumference at some point P. Join OP; the angle AOP is the required angle. Leg adjacent to A = nc, leg opposite to A = mc. 21. r sin x. 22. 4. tan Acot A = cot (90° — A); hence, A = 90° — A and A = 45°. 11. 10°. cot A=0.29167, sec A=3.5714. 11 4. sin A=0.96, tan A=3.4285, cot A=0.75, sec A=1.6667, csc A=1.25. 10. cos A= √1— m2, tan A= cot A=1, csc A= √2. csc A=1.11. csc A=√3· sec A= √2. √1-m2, cot A== V1 — m2. m 1 - m2 |