struct the triangle b'h'k' as in Problem IV.; set off the given distance b'a, from b' along the line b'h,', produced if necessary; draw a'a perpendicular to xy to meet ba, drawn parallel to xy, and make ba equal to ba, a will be the horizontal projection of the point required, and a' its vertical projection.

since bb', hh, and a,a,' (figs. 20 and 22) are parallel to one another.

But BH=b'h'; therefore BA = b'a,, the given length.

Remarks. It will be seen from this proof that ba is a fourth proportional to the three lines BH, bh, and BA, the given length.

Another way of considering the problem, which gives the same construction, is to suppose Bbh, the vertical projecting plane of BH, to turn about Bb as an axis till it is parallel to the vertical plane of projection, and finding the vertical projection bh of the given line in that position; then setting off the given distance from b' along b'h,' which gives a, as the horizontal projection of the point required when the line BA is parallel to the vertical plane. The plane Bbh is then returned to its former position. In the motion of Bbh about Bb any line ba or CA at right angles to Bb will move in a horizontal plane (Theor. III. Cor. 2).

This manner of considering the solution will be easily understood from the figures.

PROBLEM VI. Figs. 19, 20, 21, 22.

Given the projections of a line, to find the angles which it makes with the planes of projection.

Let it be required to find the angle which AB makes, (1) with the horizontal plane, (2) with the vertical plane,

Construction. Draw the triangle b'a,'c' as in Problem; the angle b'a'c' will be the angle which AB makes with the horizontal plane.

Proof. Because the triangle ac is constructed equal to BAC, the angle b'a,'c' = BAC.

But since CA is parallel to ab, BAC is equal to the angle between AB and ab.

Therefore b'a'c' is the angle required.

If a right-angled triangle be constructed having a'b' for one side, and the difference of Aa and Bb' for the other, the angle between the hypotenuse and a'b' will be the angle which AB makes with the vertical plane of projection.

When the line meets a plane of projection as in figs. 21 and 22, it is generally most convenient to construct the triangle b'h'k' for determining the angle which it makes with that plane.

PROBLEM VII. Figs. 23, 24.

To determine the projections of a straight line which shall contain a given point and make given angles with the planes of projection.

Let pp' be the projections of a point P; it is required to find the projections of a straight line passing through P, making an angle a with the horizontal plane, and an angle ß with the vertical plane of projection.

Construction. Draw p'a' making an angle a with xy; with centre p and radius equal to oa' describe the circle AA1 Draw the line p'b making the angle ap'b equal to ß, and from a draw a'b perpendicular to p'b; with centre p' and radius p'b describe a circle meeting xy in a,, and draw a A1 perpendicular to xy to meet the circle AA in the point A pA, and p'a,' will be the projections of the line required. Proof. Since in the two right-angled triangles PpA, and p'oa', Pp=p'o and pA, oa, the triangles are equal in every respect.

=

Therefore the angle PA1p=p'a'o= a.

Again, the line PQ is parallel to p'a,' (Prob. III. Cor.), and therefore equal to it (Theorem XV. Cor.), and consequently equal to p'b.

Then in the two right-angled triangles PQA, and p'ba' PA1=p'a' and PQ = p'b.

Therefore the angle A,PQ= a'p'b= B.

But the angle A,PQ is equal to the angle between PA, and p'a, since PQ and p'a,' are parallel.

Therefore the inclination of PA, to the vertical plane is equal to B.

Remarks. It may be proved in a similar manner, that the three lines passing through P and the points marked 1, 2, 3 make the given angles with the planes of projection; so that, in general, there are four solutions to this problem. If ẞ were the complement of a, and therefore equal to op'a', p'b would be equal to p'o and the circle described from the