8 66
THEORY AND PRACTICE OF ENGINEERING.
Book 11.
Fig.1105.
extent of the houses, &c. Lastly, to have the angles of thestreets, take them with a recipiangle, and their length andbreadth by means of a cord and square ; then draw with aprotractor the various angles, as before described, andset off the breadths and lengths till the whole is com-pleted. It must always be borne in mind that the threeinternal angles of every triangle are equal to two rightangles, and make 180°. In the right-angled triangle,all its three angles taken together make 180°, as is seenby the space of a circle described from the point of eachangle, comprised by the sides of the same angle. Thesame is the case with the obtuse-angled triangle and theacute. To have the angle of the centre of a regularfigure, divide 360° by the number of sides of the figure,the quotient will give the angle.
In the regular hexagon abedef, to have the centralangle aib , divide 360° by 6, the number of sides of thehexagon, the quotient will give 60° for the angle,where it will be seen that whatever number of sides aregular or irregular figure may have, the central anglestogether equal 360°, because they occupy a circle de-scribed from the centre of the figure, and which is dividedinto that number of degrees.
To find the Polygonal Angles of a regular Figure, sub-tract from 180° one of the central angles, the remainderwill be the polygonal angle.
In a regular hexagon subtract 60°, the central angle,from 180°, the remainder, 120°, will be the polygonalangle: multiply 120 3 by 6, the number of sides in ahexagon, and we have 720° for the six polygonalangles.
To have all the polygonal angles of an irregularfigure, it is a rule for those of the same number of sides,both regular and irregular, that the sum total of the poly-gonal angles of the regular figures is equal to that of allthe polygonal angles of the irregular figure.
If in a regular hexagon six lines are drawn from itscentre to its six polygonal angles, the hexagon will bedivided into six equal triangles; and as already stated,the three angles of a triangle are equal to 180° : we shall,
therefore, have for the angles of the six triangles 1080°, and as the six central angles ofthe hexagon are equal to 360°; if from 1080° we subtract 360°, there remains 720° forthe six polygonal angles of a regular hexagon. By the same rule it will be found that thesix polygonal angles of an irregular hexagon are together equal to 720°, which is easilyseen by drawing lines from its centre to its six polygonal angles, dividing the hexagon intosix triangles, each of which is equal to 180°; we therefore have 1080° for the anglesof the six triangles of the hexagon, and if we subtract 360° from 1080°, there remains720° for the polygonal angles which are equal to those of the regular hexagon.
To find in plans laid down, if the sum of the polygonal angles is correct, multiply180° by the number of sides of the plan; from the product subtract 360°, the remainderwill be the value of all the polygonal angles of the figure.
In taking the polygonal angles of the irregular hexagon, R c
the addition of the six polygonal angles gives 720, andif it is not known whether the sum is correct, to ascer-tain it, by the preceding rule multiply 180° by 6, thenumber of sides of the proposed hexagon; from theproduct 1080° subtract 360°, the value of all the centralangles, the remainder, 720°, will be the number of degreesof the six angles of the irregular polygon ; and since thesum of the angles of the polygon which have been takenagree exactly with this, it proves that they have been cor-rectly taken.
To find whether the Polygonal Angles have each separately been correctly taken : by thesecond of the preceding rules it may be found if the polygon is regular ; but if not, forexample, in the angle D E F, take on the ground each angle of the complement, by meansof a cord stretched in the line of the angle of the polygon required, and if, by adding toge-ther, the value of these two angles is equal to 180°, it may be concluded that the angle is
Fig. 1106.
Fig. 1107.