Chapter Five Dihedral Angels.docx

chapter five dihedral angels, perpendicular planes and polyhedral angel

1. dihedral angels

56. definitions. a stright line liying in a plane divides it into two halfplanes. A dihedral angel is a geometric figure formed by two half-planes issuing from a common stright line (fig.119). The two half-planes forming a dihedral angel are its faces.

parallel to one another (Theorem 1 in 50). each face of the dihedral angle intersects the planes along parallel lines (Theorem in 51) and angles with parallel arms having the same direction are equal (Theorem in 54). 57. equal Dihedral Angles. Two dihedral angles are called equal if they are congruent (see the definition in 38) of two unequal dihedral angles 𝛼AB𝛽 and xxxxxxx the angle 𝛼AB𝛽 is the greater if in making the face xx coincide with 𝛼 (Fig. 121) the face 𝛽1 takes a position between 𝛼 and 𝛽 Theorem. 1. If two diledral angles are equal their linear angles also are equal

ts edge.

2. If two dihedral angles are unequal their linear angles are unequal and a greater linear angle corresponds to the greater dihedral angle

If the two faces of a dihedral angel are in the same plane the angel is a developed dihedral angel.

Given: (1) xxxxxxx; their linear angles NMP nnd N1M1PI and the planes of the linear angles y andy2;

We shall denote a dihedral angel whit faces 𝛼 and 𝛽 and an edge AB or a by 𝛼a𝛽 or 𝛼AB𝛽

(2) xxxxxxx; their linear angles NMP and N2M2P2 and the planes of the linear angles y and Y2 (Fig. 115) To prove that: (1) LNMP LNIMPI and (2) LNMP

The stright line from which the faces of the dihedral angel issue is

The linear angel of a dihedral angel is that formed by two perpendiculars raised from any point in the edge and lying in the faces (fig.120) Obviously the plane 𝛼 of the linear angle is perpendicular to the edge since two straight lines (the arms of the lineal angles) which lie in the plane are perpendicular to the edge. The value of a linear angle is independent of the position of its vertex in the edge. Indeed, all the planes perpendicular to the edge are

proof. 1. Place the dihedral angle xxxxxxx into the dihedral angle xxxx so that the edge xxx coincided with the edge AB, the point M1 coincided with M and the face xx with the face xx. the straight lines M1N1 and MN will coincide as they are perpendicular to the line AB at the point M in the plane xx. The two given dihedral angles being equal, the face xx will coincide with the face xx. The straight lines M1P1 and MP will coincide as they are perpendicular to the straight line AB at M in the plane xx Thus the angle N1MIPI will coincide with the angle NMP i.e.
2. Place as before the dihedral angle xxxxxx into the dihedral angle xxxxx (this is feasible, since xxxxxxxxxx) and assume that the face xx will be positioned between the faces xxxxxxx at xxxxxx. In this case the arm M2P2 of the angle xxxxxxx, will take the position MP3, i e, will be between MN and MP and conscquently
angles whose linear angles are 1' and 1'' respectively. In other words a dihedral angle is measured by its linear angle. Example. An isosceles right-angled triangle ABC is bent along its height CD so that the planes ACD and BCD make a right angle. Find the angle between line segments AB and CB. Solution. Since AD L CD and BD L CD. the angle ADB is the linear angle of the dihedral angle ACDB and therefore
61. straight Line Lying in one of Mutually perpendicular Planes. Theorem. If two planes are mutually perpendicular then any straight line lying in one of them and perpendicular to their line of intersection is perpendicular to the other plane. Given: xxx, CD is the line of intersection nr the planes and a straight line AB which is in the plane B is perpendicular to CD (Fig, 124). To prove that AB L a. Proofdraw in the plane x a straight line BM L CD. We have now at point B in the edge of the diledral angle xxx two perpendiculars to the edge which lie the faces xxxxxx. i.e.
and B onto the line of intersection of the planes are equal to x and y respectively and the distance between the feet of the perpendiculars is c. Find the lengths of the segment AB and its projections on the given planes. Solution. MN is the line of inersection of the planes P and Q, points C and D are the feet of perpendiculars dropped Fig. 125 from the points A and B respectively onto the line MN (Fig. 125). Since AC lies in the plane P and AC L MN, the line AC L Q (see 61). A similar reasoning shows that BD L P. It follows that AD is the projection of AB on the plane P and BC is the projection of AB on the plane Q. The problem is thus reduced to finding the lengths of the segments AD, AB and BC. In the right-angle triangel xxxxxxxxxxx and CD are known. Therefore xxxxxxxxxx . In the triangle ADB whose angle xxxxxxxxx Ill. Area of Projection of Plane Figure 63. Definition of Projection. Theorem. Through a given straight line not perpendicular to a plane x one and only one plane B can be drawn such that it is perpendicular to x. A perpendicular to the plane x through any point in the given straight line lies in the plane B. Fig. 136 Fig. 127 Consider a straight line p which is not perpendicular to a plane x(Fig.126) and in general is outside this plane. Take a point A in the line p and from it drop a Perpendicular AM onto the plane x. Draw a plane B through p and AM. By the theorem in 60 the plane B L x.

We shall prove now that B is the only plane perpendicular to x. It follows from the theorem in 60 that any plane drawn thought the line p and perpendicular to the plane x will also past throught the line AM, i.e. coincide with the plane B. A plane B which passes throught a stright line p and is perpendicular to a plane x is a plane which projects the stright line p on the plane x or shoerter, a projecting plane. This plane intersect the plane x along a stright line q (fig.126) which is the projection of the line p. In general the projection of a geometric figure on a plane is a figure formed by the totality of projections of all the points in the figure on ths plane called the projection plane. In particular, as shown above, the projection of a stright line m on a plane is a stright line through the projections of any two points in m; this agrees whit the definitionog the projection of an inclined line, state in 43. It follows from the theorem in 53 that projection of the same line segment on two parallel planes are equal. Indeed, in fig 127 (on p, 127) xxxxx is equal to xxx where xxx is the projction of AB on the plane x and xxx is the projection of AB on the plane B. The lenght of the projection of a line segmant is equal to the lenght of the segment multipled by the sodine of the angel made by the segment wht the projection plane. Indeed, consider xxx, the projection og the segment AB on the plane x (fig.127). draw on the projection plane throught the point A a stright line AC xxxxx. From the right-angel triangel ABC one finds xxxxxx. Since xxxxx we have xxxxxx. 64. area of projection of plane polygon. Theorem. The area of a projection of a plane polygon on a plane is equal to the area of the

polygon being projected multiplied by the cosine of the angel between the plane of the polygon and the projection plane.