Now, if FB be refracted, so as to make with the reflecting side an angle equal to ZBY, it must (if the position of B be rightly chosen), be reflected so as to follow BC, thus making MBC = YBZ, and calling each of these angles = μ, we have the right angle NBZ made up of μ + y´ + NBK. But NBK clearly equals μ, because it is the inclination of the normals to BC and BZ, and hence y´ + 2 μ = 90°. This, therefore, forms a crucial test for the length of BC. I may only remark, that we already know the numerical value of y; and that of μ is easily found, for μ = CBA + ABM = CBA + BAM = CBA + (MAC - BAC) = CBA + ¹⁄₂ (180° - υ) - BAC. Thus knowing μ and y´, we have only to see whether
(y´ - 2 μ) - 90° = 0
We have now only to find the length of the radius AX or b X (see [fig. 73], p. 277), which will describe the reflecting surface or arc AZ b, and to determine the position of its centre X. We already know the values of y′ and y, the angles of refraction of C and b, and their difference y - y′ gives us the inclination of the rays which are to be reflected (into directions parallel to C b) at b and at A. This quantity is, of course, double the inclination of tangents to the reflecting surface AZ and b Z, and of their normals AX and b X. Again, we have the chord line
A b = AC . sin AC b sin (b CA);
and, as above,
AX b = ¹⁄₂ y - y′ = φ
And AX = b X = ρ = A b . sin ¹⁄₂ (180° - φ) sin φ = ¹⁄₂ A b . cosec ¹⁄₂ φ.
And, lastly, for the co-ordinates to X, the centre of curvature for the reflecting arc, we have
| OX | = | ρ . sin OAX |
| and OA | = | ρ . cos OAX.[69] |
[69] The angle OAX is easily found, as will be seen by referring to [fig. 73], p. 277; for, AH being horizontal by construction and AO vertical, HAO = 90°; and HAC and CAU being both known, we have