WILLIAM F. BRADBURY, A. M.,

HOPKINS MASTER IN THE CAMBRIDGE HIGH SCHOOL; AUTHOR OF A TREATISE ON
TRIGONOMETRY AND SURVEYING, OF AN ELEMENTARY GEOMETRY AND
TRIGONOMETRY, AND OF AN ELEMENTARY ALGEBRA.

BOSTON:
THOMPSON, BIGELOW, AND BROWN,

25 & 29 CORNHILL.

1872.

USED WITH UNEXAMPLED SUCCESS IN THE BEST SCHOOLS AND ACADEMIES OF THE COUNTRY.

EATON'S PRIMARY ARITHMETIC.

EATON'S INTELLECTUAL ARITHMETIC.

EATON'S COMMON SCHOOL ARITHMETIC.

EATON'S HIGH SCHOOL ARITHMETIC.

EATON'S ELEMENTS OF ARITHMETIC.

EATON'S GRAMMAR SCHOOL ARITHMETIC.

BRADBURY'S EATON'S ELEMENTARY ALGEBRA.

BRADBURY'S ELEMENTARY GEOMETRY.

BRADBURY'S ELEMENTARY TRIGONOMETRY.

BRADBURY'S GEOMETRY AND TRIGONOMETRY, in one volume.
BRADBURY'S TRIGONOMETRY AND SURVEYING.

KEYS OF SOLUTIONS TO COMMON SCHOOL AND HIGH
SCHOOL ARITHMETICS, TO ELEMENTARY ALGEBRA, GEOM-
ETRY, AND TRIGONOMETRY, AND TRIGONOMETRY AND SUR-
VEYING, for the use of Teachers.

Entered according to Act of Congress, in the year 1872,

BY WILLIAM F. BRADBURY,

in the Office of the Librarian of Congress, at Washington.

UNIVERSITY PRESS: WELCH, BIGELOW, & Co.,
CAMBRIDGE.

KEY

ΤΟ

BRADBURY'S ELEMENTARY GEOMETRY.

BOOK I.

PRACTICAL QUESTIONS.

1. 1st. They do, unless parallel (1). 2d. They do (1).

2. It does not.

3. Complements, 63°, 39°, -1°, -63 (5). Supplements, 167°, 97°, 83°, —37° (6).

4. 180°(15°+27° +99°) = 39° (7).

(11°+53° + 74° + 19° + 117°) = 86° (9).

16. Not necessarily. (See Fig., II. 22.)

17. 1st. Yes. 2d. Not necessarily.

18. 1st. Yes. 2d. Not necessarily.

19. 1st. Three; the two not parallel and one of the parallel sides. 2d. Three.

20. 180°(5 — 2)5108°; 180°(62) ÷ 6 = 120°; 180°(82) 8 = 135°; 180°(102) 10 = 144°; ÷ 180°(122) 12 150° (67). ÷ =

21. (813) ÷ 2 = 10.5 (66).

22. Opposite angle 120° (62); each of the other angles 180° 120°: = 60°.

68. (Fig. Art. 12.) ABC GEH (12); therefore also of ABC.

EXERCISES.

As ABC and DEG. Produce D E.
DEG is supplement of GEH, and

69. (Fig. Art. 26.) AB+BC > AC (Axiom 9)

70. Produce AD till it cuts BC at E. AB+BE> A E and DE + EC > DC (Axiom 9); adding these two inequalities, we have

AB+BE+DE+EC>AE+DC

but BE+EC BC and AE-AD+DE .. subtracting DE from both sides,

AB+ B C > A D + D C

71. (Fig. 70.) ADC > DEC; but DEC>A BE (39) ; much more then is A D C > B.

and BDC is equiangular and therefore equilateral (46).

73. Let ABC be equilateral, A D C and AEC isosceles. As all the triangles are isosceles, a perpendicular bisecting AC will pass through D, B, and E (53); that is, D, B, and E are in the same straight line. Draw DE; DE bisects the angles at E and B (43).

E

C

60°

180° (30°

30°) = 120°

2d. As DBC is an exterior angle of the triangle BEC, and

74. DBA=BAC and E BC — BCA D(17). Hence,

DBA+ABC+EBC=A+ABC+C

but (7)

DBA+ABC+E BC= two right angles 4 .. A + ABC +C=two right angles.

B

E

C