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case of plane sailing. As radius is to the distance, so is the co-sine of the course to the difference of latitude; and as radius is to the distance, so is the sine of the course to the departure.

Example. Given the course 3 points, and distance 243 miles; what is the diff. lat. and departure ?

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By inspection. In the table of triangles look under the course 3 pts. and opposite the distance 243 stands 202 the diff. lat. and 135 the departure.

· 223. Theorem.-In a right-angled plane triangle, if one side as radius be equal to the difference of latitude, and a contiguous angle be equal to the course; the side adjacent to this angle will be equal to the distance, being the secant of said angle; and the side opposite will be equal to the departure, being the tangent of said angle. Hence these solutions in the 2d case of plane sailing:

As rad. is to diff. lat. so is the secant of the course to the distance; and as rad. is to diff. lat. so is the tangent of course to the departure.

Example. Given the diff. lat. 143 and the course 34 points; what is the distance and departure?

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By inspection. Under the course 34 pts. and against the diff. lat. 143, stands the dist. 178 and the dep. 106.

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224. Theorem.In a right-angled plane triangle, if one side as radius be equal to the departure, and a contiguous angle be equal to the complement of the course, the side adjacent to this angle will be equal to the distance, being the secant of said angle, and the side opposite said angle will be equal to the difference of latitude, being the tangent of the said angle. Hence these solutions in the 3d case of plane sailing :

As rad. is to the dep. so is the co-secant of the course to the distance; and as rad. is to the dep. so is the co-tangent of the course to the diff. lat. The cosecant and the co-tangent of the course, are the same as the secant, and the tangent of the complement of the course.

Example. Given the course 2 pts. and the dep. 116; what is the distance, and the diff. lat. ?

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225. Theorem.—In a right-angled plane triangle if the hypothenuse as radius be equal to the distance, and one of the legs be equal to the difference of latitude, being the co-sine of the course, the angle formed by these sides will be equal to the course, and the side opposite this angle will be equal to the departure, being the sine of the course. Hence these solutions in the

4th case of plane sailing:

As the dist. is to rad. so is the diff. lat. to the co-sine of the course; and as rad.: dist. :: sine course dep. Example. Given the distance 212 miles and diff. lat. 187; what is the course and departure ?

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To the co-sine of course 2 pts. 9.94550

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By inspection. In the table the distance 212 with the diff. lat. 187, will give the course 2 pts. and the dep. 100 miles.

226. Theorem. In a right-angled plane triangle if the hypothenuse as radius be equal to the distance and one of the legs be equal to the departure, the angle formed by these sides will be equal to the complement of the course, and the side opposite this angle will be equal to the difference of latitude. Hence these solutions in the 5th case of plane sailing:

As dist. is to rad. so is dep. to sine of the course; and as rad. is to distance, so is the co-sine of course to diff. lat.

Example. Given the distance 104 miles and dep. 64; what is the course and diff. lat. ?

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By inspection. In the table, the distance 104 with the departure 64, give the course 38 deg. and diff. lat. 82 miles.

227. Theorem.-In a right-angled plane triangle if one of the sides as radius be equal to the difference of latitude, and another side as tangent be equal to the departure, the angle opposite the side as tangent will be equal to the course, and the side adjacent to this

angle will be equal to the distance, being the secant of said angle. Hence these solutions in the 6th case of plane sailing:

As diff. lat. is to rad. so is dep. to tangent of the the course. And as rad. is to diff. lat. so is, secant of

the course to the distance.

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Example. Given the diff. lat. 89 and dep. 36; what

is the course and distance?

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By inspection. In the table the diff. lat. 89 with the dep. 36 give the course 22 deg. and distance 96 miles.

228. Theorem.-In a right-angled plane triangle if the hypothenuse as radius be equal to the difference of longitude, and a contiguous angle be equal to the parallel of latitude sailed in-the side adjacent this angle will be equal to the meridian distance, being the cosine of said angle.-Hence this solution in the 1st case of parallel sailing:

As rad. is to the diff. lon. so is the co-sine of the lat to the meridian distance.

Example. Given the diff. lon. 140 miles and the lat.

45 deg. what is the meridian distance?

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