Imágenes de páginas
PDF
EPUB
[merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small]

(5) D2= R2 +

=

2 Rr cos (C-B)
sin 4

[ocr errors]

4 Rr' cos B cos C, by (2) and (3).

sin A

Substituting (4) in (1), and reducing by (d) and (b), we have

[ocr errors][merged small]

.. (6) D' VR2+2 Rr'.

Art. 111, (2).

.. (7) D'V R2 + 2 Rr".
D" =

。。。 (8) D′′"=V R2 + 2 Rr"".

3. Given p 100, A required D', D", D".

[blocks in formation]

=

118. Examples.

1. The three sides of a triangle are 21, 23, 26; required the distances from the center of the circumscribed circle to the centers of the three escribed circles. Ans. 25.19, 26.64, 29.73.

2. The angles of a triangle are 56', 60', 64°, the greatest side is 25; required the distances from the center of the circumscribed circle to the centers of the three escribed circles. Ans. 26.96, 27.80, 28.65.

R2 + 2 Rr'.

65°,

55°,
B 60°, C
Ans. 37.10, 38.55, 40.01.

=

119. Problem.

To find the distance between the centers of the inscribed and escribed circles.

[blocks in formation]
[merged small][merged small][ocr errors]

a

Z p(p-a)

bc

b

Substituting the values of r', r, and sin A, we have

(2) D1

≥ p (≥ p -- b)

ac

B

с

p(p-c)
ab

[ocr errors][merged small][merged small]

120. Examples.

1. The three sides of a triangle are 30, 50, 60; required the distances between the centers of the inscribed and escribed circles. Ans. 31.05, 56.69, 87.83.

2. The sides of a triangle are 500, 600, 700; required the sides of the triangle formed by joining the centers of the inscribed and circumscribed circles and the center of the escribed circle, tangent to the sides 600 and 700 produced. Ans. 540.06, 104.58, 624.58.

121. Miscellaneous Exercises.

1. Prove that sin 15°

tan 15° 2

21 2

√ 3+1

- V 3,

cosec =

Ans. sin 75°

√3-1
"
21 2

cot 15° 2+ √ 3,

212

[ocr errors]
[ocr errors]

=

13

1

2. Find the sine and co-sine of 75°.

√3 1
21 2
cos 15°, and cos 75° sin 15'?

V3+1
21 2

cos 15°

[ocr errors]
[ocr errors]

cos 75°

[ocr errors]

3. Why is sin 75°

4. How may the values of tangent, co-tangent, secant, and co-secant of 75° be found from the values of the sine and co-sine?

5. Find the functions of 150'.

,cos 150°

Ans. sin 150' 6. Given sin a + cos a = 12, to find a. Ans. 45°, or 45°+ 360°; or, in general,

√ 3+1

21 2

sec 15' =

10. Prove that ABC is isosceles if cos A

V 3
2

π

4

7. Given sin 2 a = cos a, to find a.

π

Ans. + 2 πη, or ğ π + 2 πη.

6

11. Prove that the sum of the diameters scribed and circumscribed circles of any angle ABC is

a cot Ab cot B + c cot C.

[ocr errors]

8. Prove that the sum of the tangents of the three angles of a plane triangle is equal to their product.

=

+ 2 n.

9. Prove that the sum of the co-tangents of one-half the angles of a plane triangle is equal to their product. sin B 2 sin. C

of the in-
plane tri-

[ocr errors][ocr errors][merged small]

12. If b is the base of the triangle ABC, p, the perpendicular to the base from the vertex of the opposite angle, and s, the sum of the sides a and c, prove that 2 bp (8+b) (8b)

tan B

tan B

13. If b is the base of the triangle ABC, p, the perpendicular to the base from the vertex of the opposite angle, and d, the difference of the sides a and c, prove that

=

[blocks in formation]

14. If a, b, and c be the sides of the triangle ABC, s, the sum of the sides a and c, and r, the radius of the inscribed circle, prove that

tan B

... (2) cos 1':

=

[ocr errors]

2 r

122. Computation of Natural Functions.

π

Dividing the length of the semi-circumference to the radius 1, which is 3.141592653589793... by 1080, the number of minutes in 180', the quotient, which is .0002908882..., will be the length of the arc 1', and will differ insensibly from its sine.

(1) sin 1'=.0002908882.

b

-

.9999999577.

Adding (a) and (c), then (b) and (d), articles 89, 91, and transposing,

V 1 sin2 1':

(3) sin (a+b)= 2 sin a cos b- sin (ab).

(4)

cos (a + b) = 2 cos a cos b cos (a - b).

If in (3) and (4) b=1, a = 1, 2, 3..., in succession, we have

[merged small][merged small][merged small][ocr errors]

=

2 cos 1' cos 1'

2 cos 1' cos 2'.

=

[ocr errors]
[ocr errors]
[ocr errors]

[ocr errors]

- =

To facilitate computation, for 2 cos 1' use its equal, 2.0000000846. Then we have

sin a

cos a

[blocks in formation]

sin 2'
sin 3' 2 sin 2' .0000000846 sin 2′-

cos 0.9999998308.
.9999996193.

cos l' =

2 sin 1'.0000000846 sin 1'- sin 0.
sin 1'.

1.9999999154,

[ocr errors]

After finding the sines and co-sines, the tangents and co-tangents can be calculated from the formulas:

(5) tan a

(6) cot a

It is not necessary to carry the computation beyond 45°, since sin a = cos (90° — a), etc.

cos a

sin a

The logarithmic functions can be found from the corresponding natural functions by the method of article 60.

SPHERICAL TRIGONOMETRY.

123. Definition and Remarks.

Spherical Trigonometry is that branch of Trigonometry which treats of the solution of spherical triangles.

If any three of the six parts of a spherical triangle are given, the remaining parts can be computed.

The radius of the sphere is taken equal to 1, and

[merged small][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors]
« AnteriorContinuar »