Page 109
PROP. VII. The Latitude of the Place and the Sun's Declination being given, to find at what Hour the Sun will be upon the true East or West Points.
UPON the Projection there are two Right-angled Sphe∣ricall Triangles, by either of which this Proposition may be solved. The one is the Triangle Z P o, made by the Intersections of Z o, an Arch of the Prime Verticall, P o, an Arch of an Hour-Circle, and Z P, an Arch of the Meridian. In which Triangle there is given Z P, the Perpendicular, the Complement of the Latitude of the Place 38 degr. 30 min. and the Hypotenuse P o, the Complement of the Sun's De∣clination 70 degr. to find the Angle at the Perpendicular Z P o, which you may doe by the 14. Case of Right-angled Sphericall Triangles.
The other Triangle is o C A, right-angled at C, and is con∣stituted of o C, an Arch of an Hour-Circle, C A, an Arch of the Aequinoctial, and o A, an Arch of the Prime Verticall. In which Triangle you have given, (1.) the Perpendicular, O C, the Sun's Declination; (2.) the Angle at the Base, C A o, the Latitude 51 degr. 30 min. to find the Base C A. Thus having the Perpendicular and the Angle at the Base, you may find the Base C A as followeth, this being
As the Tangent of the Latitude 51 degr. 30 min. is to the Tangent of the Sun's Declination 20 degr.
So is the Radius 90 degr. to the Co-sine of the Hour from Noon.