CASE IV.

If the three angles of the spherical triangle DEF be given to find the sides; we take the supplements of the given angles, we have the three sides of the supplemental or polar triangle ABC; and, the angles of the triangle being found by case 1, their supplements will be the sides required in the triangle DEF.

CASE V.

If the side DE and the adjacent angles at D and E, be given to find the remaining parts of the triangle DEF; by taking the supplements of the given parts we have the two sides AB, AC and the contained angle BAC, to find the remaining parts of the triangle ABC by case 2.

CASE VI.

If the side DF and the two angles at D and E be given in the triangle DEF to find the remaining parts; we have in the triangle ABC the two sides AB, AC, and the angle ABC opposite to on of them, to find the remaining side by case 3.

CHAPTER

CHAPTER IV.

CONSTRUCTION OF THE CONIC SECTIONS,

PROBLEM I,

To construct a conic surface.

1. Draw the ground line AB, in the horizontal plane; takę any point p' for the centre of the circular base of the cone, and with the radius of the base describe about p' as a centre, the circle co'n' for the base of the cone. From P' let fall on

AB, the perpendicular P'K, in which produced take KP" equal to the axis of the cone; that is to the distance between the vertex of the cone, and the centre p' of the base; and r" will be the vertical projection of the vertex of the cone.

Because the axis of the cone is at right angles to the base, it is evident that the horizontal projection of the axis is simply the point p'; and KP" at right angles to the ground line AB is its vertical projection: and therefore, P' and P" are the horizontal and vertical projections of the vertex of the cone.

2. To

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2. To project the slant side of the cone, take any point e^ in the circumference of the base, through which and the centre e' draw 'P'R' a diameter of the base; also, from a draw QʻL at right angles to the ground line AB; and join P′′L; then will r'Q' and P′′L be the horizontal projections of the slant side of the cone which passes through the point a'.

This construction of the slant side is evident, because L being the vertical projection of the horizontal point q'; therefore P and Q are the horizontal projections of two points of the slant side, and P", the corresponding vertical projections; and consequently, P'Q', PL are the required projections of the slant side passing through a'.

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If we make a similar construction for the slant side passing through R' we have the construction of the two slant sides of the cone, in which the curve surface of the cone is intersected by a vertical plane passing through the axis of the cone., Thus PL and P"м are the vertical projections of the opposite slant sides passing through the extremities of the diameter Q'R' of the base, and the opposite radii r'a', P'R' are the corresponding horizontal projections.

3. To find the vertical projection of any point of the surface corresponding to any given horizontal projection. Let s' be any given horizontal projection of a point of the curve surface of the cone; draw the radius r's'a', and having constructed the slant side by its projections P'q', P′′L, draw s'ns" at right angles to the ground line AB, meeting the vertical projection P'L in s", and s" will be the vertical projection of that point of the conic surface which has s' for its horizontal projection.

In the preceding construction we have considered only that part of the whole conic surface which is between the vertex and base; but as the conic surface may be extended indefinitely downwards below the base, and upwards above the vertex, it is plain that the horizontal projection R'r'Q' of the opposite slant sides as well as the vertical projections P′′K, P′′м, should be produced indefinitely both ways; that is, n'q' towards F' and '; and P′′K, P′′м towards E", R"; H′′, H′′, D′′.

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Now, the vertical projections R"E", D"H" being both in a vertical plane passing through F's', if we produce s's" to meet D"H" in T", we shall have T" for the vertical projection of the point in which a perpendicular to the horizontal plane at s' meets the slant side which passes through the point R of the base this perpendicular therefore meets the conic surface in two points, of which s" and r" are the vertical projections, the horizontal projections being coincident in the point s'. Also, the vertical ordinates of these two points being

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Ns" and NT", it is evident that the part of the perpendicular at s' which is projected into s"r", falls without the conic surface; the remaining parts of it falling within the upper and lower divisions of the conic surface.

If we produce Q'x to meet D"A" in q", the point q” will be the vertical projection of the point in which the perpendicular from o' to the horizontal plane, meets the upper division of the conic surface. In like manner, if R'м be produced to

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R", we have the vertical projection of the point in which the
perpendicular at ' meets the slant side of the cone passing
through a': and, because мв" and κq” are equal, as is evident
from the construction, it follows that R" and Q" are the vertical
projections of two points diametrically opposite in a circular
section of the upper conic surface parallel to the base.

If we take any point F' in 'Q' produced, and draw D'F', OE'
at right angles to the ground line AB, it is plain that D" and E"
are the vertical projections of the points in the upper and
lower divisions of the conic surface through which a straight
line passes, that is, perpendicular to the horizontal plane at F',
so that oF', OD" are the horizontal and vertical ordinates of the
point of intersection in the upper division, and DF', OE" in the
lower. Also that part of the perpendicular at F', that is re-
presented by D"E", falls without the conic surface; and the re-
maining parts above D" and below E" fall within the upper and
lower divisions of the conic surface.

PROBLEM II.

To find the point in which a given plane is cut by the given slant
side of a cone.

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