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Since it has been shewn in the corollaries to Proposition Tenth of Book Fourth, that the angle subtended at the centre by the greater segment of the radius divided medially is 36°, half the greater segment of the radius divided medially is the sine of 18°; now if x represent the greater segment, and radius be 1, by Proposition Eleventh of Book Second, the following equation is obtained :

ma = 1 – X, or x? + x = 1;

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The first ten of the following exercises in Trigonometry, may be solved by means of the values found above, and the rules formerly established, without the aid of trigonometrical tables.

EXERCISES. 1. The hypotenuse of a right-angled triangle is 200, and one of its acute angles is 30°; find the two sides.

Ans. 100 and 100x3 = 173.205. 2. The base of a right-angled triangle is 80, and the vertical angle is 30°; find the hypotenuse and the perpendicular.

Ans. Hyp. = 160, perp.= 80/3.

3. The base of a right-angled triangle is 30, and each of its acute angles is 45°; find the hypotenuse and perpendicular. .

Ans. Hyp. = 3002, and the perp. = 30. 4. The base of a triangle is 150, and the angles at the base are 60° and 75°; find the third angle and the remaining sides. d Ans. The third angle = 45°, and the sides are 7506 and

75(W 3 + 1). 5. The vertical angle of an isosceles triangle is 30°, and the base is 200; find the length of the equal sides.

Ans. The sides = 100(16 +x2). 6. In the triangle of last question find the length of the perpendicular from the vertex on the base, and the perpendicular from the extremity of the base on one of the equal sides.

Ans. 100(2 + 13) and 50(V6 +12). 7. The three sides of a triangle are 7, 713, and 14; find the three angles. . . . . . . Ans. 30°, 60°, and 90.

8. One angle of a triangle is 45°, and the side opposite to it is 80, another side is 4006; find the remaining angles, and the third side. . Ans. 60°, 75°, and the third side 40(3 + 1).

9. A tower stands on a horizontal plane, the angle of elevation of its top above the plane at a certain point was 45°, and 120 feet further from it the same angle was 30°; find the height of the tower. . . . . . . Ans. 60(N3 + 1).

10. One side of a triangle is 96 feet, and the angle at the base adjacent to it is 75°; find the perpendicular from the vertex on the base and the distance of the foot of the perpendicular from the angle.

Ans. Perp. = 24(V6 +12), and dist. = 24(16 – N2). 11. If in any triangle a perpendicular be drawn from the vertex upon the base, the segments of the base have the same ratio as the tangents of the parts into which the vertical angle is divided.

12. The base of a triangle is to the sum of its two sides, as the cosine of half the sum of the angles at the base to the cosine of half their difference.

13. The base of a triangle is to the difference of its sides, as the sine of half the sum of the angles at the base to the sine of half their difference.

14. The base of a triangle is to the difference of its segments, as the sine of the vertical angle to the sine of the difference of the angles at the base.

15. Half the perimeter of a triangle is to its excess above the base, as the cotangent of half either of the angles at the base to the tangent of half the other angle. b.16. The excess of half the perimeter of a triangle above the less side is to its excess above the greater, as the tangent of half the greater angle at the base to the tangent of half the less.

17. In a right-angled triangle, radius is to the sine of double one of the acute angles, as the square of half the hypotenuse to the area of the triangle. 18 18. Radius is to the tangent of half the vertical angle of a triangle, as the rectangle under half the perimeter and its excess above the base to the area of a triangle.

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APPENDIX

PROPOSITION (A). THEOREM.

[To be substituted for I. 7 and 8 of Elements.] If two triangles have three sides of the one respectively equal to the three sides of the other, they are equal in every respect.

Given ABC, DEF two triangles, having the side AC.E. DF, CB = FE, and AB = DE ; to prove that they are every way equal.

(Const.) For let the triangle ABC be applied to DEF, so that the point A may coincide with D, and the line AB with DE,

then shall the point B coincide with E; and let the sides

AC, CB take respectively the positions DG and GE

and join

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(Dem.) Then because FD and DG are each equal to AC, FD is equal to DG, therefore (I. 5) the angle DGF is equal to the angle DFG; and because FE and EG are each equal to CB, FE is equal to EG, therefore the angle EGF is equal to the angle EFD. But the angle DGF was proved equal to the angle DFG, therefore the whole angle DGE is equal to the whole angle DFE. Again, the angle DGE is the angle ACB in a different position, therefore AC and CB and the contained angle ACB, are equal to DF and FE and the contained angle DFE; hence (I. 4) the triangles are equal in every respect.

PROPOSITION XXVI. If this proposition had been transposed, and placed after Propo, sition 32, its demonstration might have been given in a more simple manner, and as no use is made of it until it is applied in demonstrating the 34th Proposition, the logical sequence would not be affected by such transposition. It might then be demonstrated as follows:

If two triangles have two angles of the one equal to two angles of the other, and a side of the one equal to a corresponding side of the other; the triangles are equal in every respect.

Given two triangles ABC and DEF, which have the two angles ABC and ACB of the one, respectively equal to DEF and DFE of the other, and consequently(I. 32, Cor.3) the third angle BAC of the one equal to the third angle EDF of the other; also let BC be given equal to EF; it is required to prove that the triangles are equal in every respect.

(Dem.) Conceive the triangle ABC to be applied to the triangle DEF, so that the point B may be on E, and the side BC on EF, the point C will coincide with the point F, because BC is equal to EF; and since the angle at B is equal to the angle at E, the side BA will fall on ED, also since the angle at C is equal to the angle at F, the side CA will fall on FD; therefore the point A will coincide with the point D, and since B coincides with E, and A with D, the side AB is equal to DE; and since the point C coincides with F, and A with D, the side AC is equal to the side DF; the triangles thus coinciding are equal in every respect, and have their sides equal that are opposite to the equal angles, and the area of the one is equal to the area of the other.

PROPOSITION XXXII. There are some interesting ways in which this proposition may be demonstrated, which are not generally known; two of these are here given, they are as follow, and depend on the following principle, which may be demonstrated by the 13th Proposition. If a line be turned round a point or points in it. till its ends be reversed, it has been turned through two right angles.

Let ABC be any triangle, and suppose a rod longer than any of its sides to be laid along BA, suppose it to be turned round the point B till it coincide with BC, it will then have changed its

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