and from B intersect the arc at C; through C draw the line A C; then will CAB be the angle required. Or by the following rule:

Problem 13.

The measure of an angle is an arc of a circle contained between two lines which form that angle, as CB; the angular point being the centre of the circle as at A, and is estimated by the number of degrees contained in that arc.

Thus from the scale of chords (which is generally engraved on the ivory protractor marked CH) take in the compass 60 degrees; and from the angle at A, Fig. 13, Plate 1, describe the arc B C; then with the compass take the distance from B to C, and apply it to the scale of chords, which will give the measure or number of degrees and minutes contained in the angle.

Problem 14.

To measure an angle by the protractor.

Place the centre of the protractor at the point or angle A and the edge along the line A B; extend the lines sufficiently long to read where it cuts the outer edge of the protractor, and the number of degrees and minutes it reads from B to C will be the measure of the angle required, equal to 29° 5'.

Problem 15.

To bisect or divide an angle into two equal parts, Fig. 14. From A, with any radius, describe the arc bc; then from b and c, as centres, describe arcs intersecting each other at d; draw the line d A, which will divide the angle as required.

Problem 16.

To trisect or divide a right angle into three equal parts, Fig. 15.

From A, with any radius, describe the arc BC; with the same radius from B and C, describe other arcs intersecting at d and e; then draw the lines Ad and Ae, which will divide the angle as required.

Problem 17.

Another method: To trisect a given angle A B C less than a right angle, Fig. 16.

With any radius describe a circle; draw the line ABD; from the points A and C as centres, with the radius A C, describe arcs intersecting each other at E; from E draw the line Ea D, cutting the circle at a; with the distance C a mark off b, then will the angle A B C be trisected.

Problem 18.

When two lines intersect each other as at E, Fig. 16a, the opposite angles are equal to each other: viz. the angle A E C is equal to the angle D E B; and the angle A E D is equal to the angle C E B.

Note. This problem is the main principle on which large surveys are conducted, for when any of the two lines are connected by another line, as, for instance, from A to C, or from B to D, the whole becomes fixed, and all other lines intersecting them must, when plotted, agree with the number or distance measured; examples of which will be hereafter given.

TRIANGLES.

Problem 19.

To describe a right angle triangle, the base and perpendicular being given, Fig. 17, Plate 2.

At the point or angle A, erect a perpendicular, and set off the required height; also the given distance on the base from A to B; then draw the hypothenuse from C to D will be the triangle required.

Note.-In any right angle triangle, the square of the hypothenuse is equal to the sum of the squares of the other two sides.

Problem 20.

To describe an equilateral triangle, Fig. 18.

From the points A and B, with the radius A B, describe arcs intersecting each other at C; draw the lines C A and C B, then will A B C be the triangle required.

Problem 21.

To describe an isosceles triangle, Fig. 18, Plate 2.

From the points A and B, with the radius C B, describe arcs intersecting each other at C; draw the lines CA and C B, then will A B C be the triangle required.

Problem 22.

To describe a scalene triangle, the three sides being given, Fig. 19.

Set off the given length of the base A B; from A, with the distance A C, describe an arc; and from B, with the distance B C, intersect the arc at C, and draw the lines A C, and C B will be the triangle required.

Note.-The longest side of any triangle is opposite the greatest angle, and the shortest side will be opposite the smallest angle.

In any triangle the sum of all three angles are equal to two right angles, or 180 degrees.

The sum of two angles in any triangle taken from 180 degrees, leaves the measure of the third angle.

All the outer angles of any triangle added together are equal to four right angles.

Every triangle has six parts-viz. three sides and three angles.

Problem 23.

Fig. 17. Two sides of a right angled triangle given to find the third.

Rule 1. To the square of the base add the square of the perpendicular; the square root of that sum will be the hypothenuse.

Rule 2. Multiply the sum of the hypothenuse and one side by their difference, and the square root of that product will be the other side.-See Euclid, Prob. 47, book i.

Example 1. Given the base A B = 320 and perpendicular B C = 246, required the length of the hypothenuse A C.

Ans. 3202 + 2462 = √162916 = 403 the hypothenuse

Example 2. Given the base A B = 320 and the hypothenuse BC= 403, required the length of the perpendicular B C.

Ans. 403 + 320 × 83 = √60009245 the perpendicular

See also Trigonometry, Case 1.

PARALLELOGRAMS.

Problem 24.

To describe a quadrilateral or square on a given base, Fig. 20, Plate 2.

From the given length A and B, draw perpendiculars; on which set off the length equal to A B; then CD will be equal to A B, and all the angles right angles.

Another method: From A and B as centres, describe the arcs AC and BD, cutting each other in m; bisect Am or Bm in n; with the radius m n, at m as a centre, describe arcs cutting A C and BD; then draw the lines A D, D C, C D will be the square required.

Note.-If diagonals are drawn from the opposite angles of the square and the rhomboid, they will be at right angles to each other.

Problem 25.

To describe a rectangle whose base and perpendicular are given, Fig. 21.

From the given base raise perpendiculars at A and B; on which set off the given lengths of the sides A D and B C; then CD will be equal to A B, and all the angles right angles.

Problem 26.

To describe a rhombus on a given side, Fig. 22.

Mark off the length of the given side A B; from B, with the radius A B, describe the arc DC; from A, with the same radius, intersect the arc at D; then with the same radius from D intersect the arc at C; draw lines through the several points, and the rhombus will be complete. The opposite angles will be equal to one another, two obtuse and two acute.

Problem 27.

To describe a rhomboid whose sides and angles are given, Fig. 23.

The length of AB = 400; the length of AD = 250; the

angle D A D = 60°. Set off the line A B to the given length; by the protractor from the point A set off the given angle; draw the line BC parallel to AD, and mark off the given length of the sides A D and B C; then draw the line D C equal and parallel to A B; the angles are the same as the rhombus.

Problem 28.

To construct a trapezoid whose base and perpendiculars are given, Fig. 24, Plate 2.

From A and B raise perpendiculars, and mark off the given lengths of the sides AD and BC; draw the line D C, which gives the trapezoid; the angles A and B are right angles; the angle C is obtuse, and the angle D acute.

Problem 29.

To construct a trapezium, whose base A B and all the sides are given, Fig. 25.

Draw a line, and set off A B equal to the given length; from A, with the given length of A C, describe an arc; then from B and length of B C intersect that arc at C; draw the lines A C and C B; in like manner from A and B draw arcs equal to the given lengths of the sides A D and D B, intersecting each other at D; draw the lines A D and D B, the trapezium required.

Problem 30.

To find the side of a square that shall be any number of times the area of a given square, Fig. 26.

Let ABCD be the given square, then will the diagonal BD be the side of the square A E F G, double in area to the given square ABCD; and if the diagonal be drawn from B to G, it will be the side of the square A H K L, three times the area of the square A B C D; and the diagonal B L will be equal to the side of the square, four times the area of the square A B C D