(25) To construct an angle of 75° at the point A on the straight line AB.

Still taking the figure (19) for finding a right angle at A, bisect the angle CAF by describing arcs with centres C and F and distance AB to meet in M.

Then the angle MAB = the sum of CAB and CAM = 75°, being the sum of 60° and 15°.

(26) To construct an angle of 105° at the point A, on the straight line AB.

With the same figure as in the case of 45° and 75° (24, 25) bisect the angle FAD by a similar construction to that used for 75°. Then the angle NAB the sum of the angles 90° and 15° = 105°.

=

In the same way by taking arcs subtending 15° at the centre and adding them on or taking them from the arcs subtending angles already known we can find any angles from o° to 180°, differing by 15°, at the centre of the circle.

Other angles are found by dividing arcs by trial, or by use of the protractor.

(27) To describe a right-angled triangle CAB where the lengths of the hypotenuse AB and one side CB are given.

[The hypotenuse is the side opposite the right angle.]

Draw AB and bisect it at O. Make the semicircle ACB. Mark off the given distance CB with centre B. Join CA, CB. (28) To describe a rectangle ACBD where the lengths of the diagonal and one side are given.

Draw the diagonal AB and bisect it in O.

Describe the circle ACBD with centre O. Mark off distances BC and AD each equal to the other given side. Join AC, CB, BD, DA.

(29) To make an angle BAC at the point A in the straight line AC equal to a given angle DEF.

With centres A and E and any the same distance make arcs BC and DF. Measure the arc BC = the arc DF. Join AB. Then the angle BAC

= DEF.

(30) To draw a tangent to a given circle at a given point B on it.

Draw the diameter AB, and CD at right angles to it at B. CD is the tangent required.

(31) To draw tangents to a given circle from a given external point A.

Find O the centre of the given circle. Join OA and draw a circle with OA as diameter [that is bisect OA at M and draw a circle with centre M and distance MA] cutting the given circle at B and C.

AB and AC are tangents to the given circle.

[If one tangent only is required draw AB or AC.]

(32) To describe a segment of a circle AEB to contain an angle equal to a given angle, on a given straight line AB. Draw a straight line CD bisecting AB at right angles. Make the angle BAQ :

=

the given angle.

Draw AO perpendicular to AQ meeting CD in O. A segment described with centre O and distance equal to OA contains an angle equal to BAQ.

(33) To cut off from a given circle a segment ABC to contain an angle equal to a given angle.

Draw a tangent at a point A. Make the angle DAC the given angle, then ABC is the segment required.

(34) To find the centre of a circle of which a part ABC is given.

Bisect any two arcs AB, BC which form part of the given segment. The two bisectors meet at the centre.

[An arc is bisected exactly in the same way as a straight line.]

(35) Two straight lines DE, CF are given in position, it is required to find without producing them the straight line which would bisect the angle between them.

Draw from a point A in DE, AG parallel to CF.

Bisect the angle DAG by AB, and bisect AB at right angles by HK. Then HK is the straight line required.

(36) Two straight lines DE, CF are given in position, it is required to find a straight line which passes through their intersection, and a given point P, without producing them.

Draw any two parallels LM, AB. Join AP, BP and draw LQ, MQ parallel to AP and BP intersecting in Q. PQ is the straight line required.