the line RS stands on the line PQ, the two angles PRS and

QRS are equal.

When one line standing upon another line forms two equal angles

each of these angles

is called a right angle.

Thus each of the angles

PRS and QRS is a right

angle. The lines PQ and RS are said to be perpendicular to each other, or at right angles to each other.

A

B

We shall frequently use the symbol for perpendicular. Right angles are very common and are easily recognised by the eye. In Fig. 10 ABCD represents an ordinary sheet of paper; each of the angles A, B, C, D is a right angle. Any angle is a right angle if we can fit the corner of an ordinary sheet of paper on to it; for example the angles of the cover of a book, the angles of an ordinary picture frame, the angles of a box-lid, etc. In Fig. 10, AB is perpendicular to BC, AD is perpendicular to DC, and so on.

D

Fig. 10.

Your two "set squares' are wooden triangles, and in each of these one angle is obviously a right angle. Any triangle which has one angle a right angle is called a rightangled triangle.

Now learn Definitions 11 and 12 on page 4. Note that the words obtuse and acute are Latin

words which mean blunt and sharp. Learn also Definitions 24-29 on page 6.

6. The following problems should now be practised till the student can do them from memory:

A D

C

B

Fig. 11.

Problem II.-At a given point C in a given straight line AB, to draw a line perpendicular to AB. (See Fig. 11.)

From C mark off equal portions CD and CE of any convenient length. With centres D and E, and any convenient equal radii, describe arcs of circles intersecting at F. Join CF.

Then CF is perpendicular to AB.

NOTE 1.-If the point C is very near to one end of the line it will be necessary to produce the line (i.e. lengthen it by means of the ruler) before we start the construction given above.

NOTE 2. The perpendicular CF can be more easily (though less neatly) drawn by placing the right-angled corner of a set square at C, with one edge lying along CA or CB. We can then rule CF along the other edge.

Problem III. From a given point C without a given straight line AB, to draw a line perpendicular

to AB. (See Fig. 12.)

With centre C and any convenient radius draw arcs cutting AB in D and E. With centres D and E and any convenient equal radii describe arcs cutting one another in F (on the other side of the line from C). Join CF cutting AB at G. Then CG is perpendicular to AB.

NOTE 1.-Here again it may be necessary to produce AB, before the above construction is possible.

NOTE 2.-We can place a set square so that one arm of the right angle lies along AB, and the other passes through C, and then rule CG.

NOTE 3.-The point G is called the foot of the perpendicular CG.

Problem IV.-Given a line, AB, to bisect it, or divide it into two equal parts. (See Fig. 13.)

With A as centre, and radius rather greater than half AB, describe an arc. With B as centre and an equal radius, describe

a second arc, intersecting the first one in points C and D. Join

CD, cutting AB in P, the required point.

Problem V.-To bisect a given angle, ABC. (See Fig. 14.) With B as centre, and any convenient radius, draw an arc cutting AB and BC in D and E respectively. Then with D and E in turn as centres, and a common radius a little greater than half DE, draw two arcs intersecting in F. If BF be joined it will bisect the angle ABC.

Care must be taken to join B, F, very exactly. If the angles ABF, FBC be each bisected, the whole angle ABC is divided into four equal angles.

Examples IV.

Draw again the first six triangles of Examples I.

B

Fig. 14.

1. In the first triangle bisect the base, and measure the distance of its middle point from the vertex.

2. In the second triangle bisect the vertical angle by a line cutting the base, and measure the two parts into which the base is divided.

3. In the third triangle draw a line from the vertex to the base, and measure the part of the base which lies to the left of this line.

4. In the fourth triangle draw a line 1 to the base from its right hand end to meet the left side; measure the length of this perpendicular.

5. In the fifth triangle draw a line at right angles to the base from its left extremity to meet the right side of the triangle produced; measure this perpendicular.

6. In the sixth triangle bisect the left angle by a line cutting the right side, and measure the parts into which the right side is divided.

7. On the measurement of angles.-For the purpose of measuring angles, we divide a right angle into 90 equal parts, and we call each part a degree. A degree is obviously a very small angle.

In order to state the size of an angle, we state how many degrees it contains; obviously half a right angle will contain 45 degrees, two-fifths of a right angle will contain of 90 or 36 degrees, and so on. We use the symbol to denote degrees, so that 23° means an angle containing 23 degrees. In Fig. 7, GPA is a right angle, and it is divided into six equal parts. Hence each of these parts contains 15 degrees. It follows that the angle

BPD 30°,

The protractor (Fig. 15) is an instrument for measuring the number of degrees in an angle. The curved edge of the protractor is divided into 180 equal parts by small marks (only 36 of them are shown in the diagram). If we draw a line from the point marked 90 to the point marked 0 (in the centre of the

straight edge) this line forms a right angle with OA and another

165

150

135

120

6

T

with OB. If we join the point to each of the marks on the curved edge we shall divide each of these right angles into 90 equal parts, i.e. into degrees; so that if we draw a line from the point marked 15 to the point marked 0, then this line forms with OA an angle of 15°; and if we draw a line from the point marked 105 to the point marked 0, this line forms with OA an angle of 105°, and so on.

180

B

Fig. 15.

If we join any point Q on the curved edge of the protractor to the point 0, it is obvious that the two angles QOA and QOB together contain 180° or two right angles. Hence if one straight line stands upon another straight line the two angles so formed are together equal to two right angles.

To measure an angle, place the protractor with O at the vertex of the angle and OA along one arm of the angle. If the other arm of the angle passes through the point marked 75 on the protractor the measurement of the angle is 75°. If the other arm of the angle does not reach to the curved edge of the protractor, it must be produced.

Examples V.

Draw again the triangles of Examples I. and measure the three angles of each triangle.

In the course of these measurements note the following points :(i) The three angles of a triangle when added together make up 180°. (ii) The largest angle of a triangle is always opposite to its largest side, and the smallest angle is always opposite to the smallest side.

(iii) If two sides of a triangle are equal, the angles opposite to these two sides are also equal.

(iv) If the three sides of a triangle are equal, its three angles are equal and each measures 60°.

These four results must be committed to memory.

8. On the construction of angles.-To construct an angle (say of 135) given one arm and the angular point;-place the protractor so that O is on the given angular point and OA lies along the given arm. Then make a pencil mark on the paper against the mark 135 on the protractor, and join this pencil mark to the given angular point.

LGCA

=

Certain angles can be constructed accurately without using a protractor. Thus by Problem II. we can construct an angle of 90° at the point C (Fig. 11). If in the same figure we bisect the angle FCB by a line CG, then GCB = 45°, 135°. By bisecting an angle of 45° we can obtain an angle of 2210, and so on. Again we can construct an equilateral triangle by Problem I., and we then obtain an angle of 60° (see Examples V., note iv.). Bisecting an angle of 60° we obtain an angle of 30°, and so on.

9. On the construction of triangles.

Problem VI.-To construct a triangle given two sides and the included angle.

Construct a triangle having an angle of 72° and the sides containing that angle of lengths 2 and 3 inches respectively. Construct HAE 72° (Fig 16). From AE mark off AB = 2"; from AH mark off AC=3"; join BC. ABC is the required triangle.

H

=

Problem VII.-To construct a triangle given two angles and the adjacent side.

Construct a triangle whose base is 2 inches, having angles of 40° and 55° respectively at the extremities of the base.

Draw BC of length 2 inches (Fig. 17). Using the protractor, make an angle of 40° at B (viz. CBH), and an angle at 55° at C (viz. BCK). If BH and CK intersect at A, ABC is the required triangle.

Problem VIII.-To construct a triangle given two angles and the side opposite to one of them.

Construct a triangle having two angles of 30° and 85°, and a side of 2.3 inches opposite to the second angle.

Since the three angles of any A are together equal to 180°, we obtain the third angle by adding the two given angles and subtracting the result from 180°. Thus the third angle is 65°.