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Remarks on Angles, Triangles, &c. 1. If from a point D in a right line AB, one or more right lines be drawn on the same side of it, the angles thus formed at the point D will be together equal to two right angles, or 180°; thus ADE + EDB = two right angles, or 180°: also ADC+CDE + EDB = two right angles, or 180°. Fig. 35.
2. Since the angles thus formed at the point D, on the other side of AB would also be equal to two right angles, the sum of all the angles formed about a point is equal to four right angles or 360°.
3. If two right lines cut one another, the opposite angles will be equal: thus AEC = BED and AED= CEB.
4. The sum of the three angles of a plane triangle is equal to two right angles, or 180°.
5. If the sum of two angles of a triangle be subtracted from 180°, the remainder will be the third angle.
6. If one angle of a triangle be subtracted from 180°, the remainder will be the sum of the other two angles.
7. In right angled triangles, if one of the acute angles be subtracted from 90°, the remainder will be the other acute angle.
8. The angles at the base of an isosceles triangle are equal to one another.
9. If one side of a triangle be produced, the external angle will be equal to the sum of the two internal and opposite angles: thus the external angle CBD, of the triangle ABC, is equal to the sum of the internal and opposite angles A and C. Fig. 37.
10. The angle at the centre of a circle is double of the angle at the circumference, upon the same base, that is, upon the same part of the circumference: thus the angle BEC is double of the angle BAC. Fig. 38.
11. The angles in the same segment of a cirele are equal to one another: thus the angle BAD is equal to the angle BED; also the angle BCD is equal to the angle BFD. Fig. 39.
12. The angle in a semicircle is a right angle; thus the angle ECF, Fig. 45, is a right angle.
13. This mark 'placed on the sides or in the angles of a triangle, indicates that they are given; and this mark placed in the same way, indicates that they are required.
PRACTICAL RULES FOR SOLVING ALL THE CASES OF
The angles and one side of any plane triangle being given,
to find the other sides.
As the sine of the angle opposite the given side,
* DEMONSTRATION. Let ABC, Fig. 40, be any plane triangle; take
Note 1.-The proportions in trigonometry are worked by logarithms; thus, from the sum of the logarithms of the second and third terms, subtract the logarithm of the first term, and the remainder will be the logarithm of the fourth term.
2. .The logarithmic sine of a right angle or 90° is 10.00000, being the same as the logarithm of the radius.
1. In the triangle ABC, there are given the angle A= 32° 15', the angle B=114° 24', and consequently the angle C= 33° 21', and the side AB = 98;* required the sides AC and BC.
By Construction, Fig. 41. Make AB equal to 98 by a scale of equal parts, and draw AC, making the angle A=32° 15'; also make the angle B= 114° 24' and produce BC, AC, till they meet in C, then is ABC the triangle required; and AC, measured by the same scale of equal parts, is 162, and BC is 95.
will be the sines of the angles A and B to the equal radii AC and BF. Now the triangles BDC and BEF being similar, we have CD:FE:: BC:BF:or AC; that is sin. A: sin. B::BC: AC. In like manner it is proved, that sin. A: sin. C:: BC: AB. When one of the angles is obtuse, the demonstration is the same. Hence it appears, that in any plane triangle, the sides are to one another as the sines of their opposite angles.
* This 98 may express so many feet, or yards, &c., and the other sides will be of the same denomination as the given.
By Gunter's Scale. Extend the compasses, on the line of sines, from. 33° 21' to 65° 36, the supplement of the angle B; that extent will reach, on the line of numbers, from 98 to 162, the side AC.
Extend the compasses from 33° 21' to 32° 15' on the line of sines ; that extent will reach, on the line of num. bers, from 98 to 95, the side BC.
2. In the right angled triangle ABC, are given the hypothenuse AC = 480, and the angle A = 539 8. To find the base AB and perpendicular BC.
From 90° subtract the angle A = 53° 8'; the remainder 36° 52' will be the angle C. The angle B, being a right angle is 90°
By Construction, Fig. 42. This may be constructed as in the preceding example; or otherwise thus,
Draw the line AB of any length, and draw AC making the angle A = 53° 8'; make AC = 480 by a scale of equal parts, and from C draw CB perpendicular to AB, then ABC is the triangle required. AB, measured by the same scale of equal parts, will be 288, and BC will be 384.