Imágenes de páginas
PDF
EPUB
[ocr errors]

54. Eliminate 0 and $ from the equations

x cos 0 + y sin 0 = a, b sin (8 + )= a sin ,
3 cos (@ +20)- y sin (0+2)=a.

Ans. x2 + y2 = a? +

a’ya.

62 55. Eliminate 0 from the equations

sec” 0 cos20
sec @ + cosao
=sec0 + cos20.

2c2

1. y

[ocr errors]

a

25

y2

a2

62

56. Eliminate from the equations

+ a2

=1.

(a + b) tan (@- )=(a - b) tan ( + 0),

a cos 2+ bcos 20 =c. Ans. b=c+a? — 2 ac cos 2 d. 57. Eliminate from the equations

cos? sin? 0 1
x sino
y cos 0 = V a2 + y?,

+
a 62

22 + y2

22 Ans.

ya

62 58. Eliminate 6 and from the equations

aʼcos? 0 b2 cosa d = c, a cos 0 + b cos $=r,
a tan0 = b tan d.

47a
al

47-62 Ans.

(2 — c) 59. Eliminate from the equations

n sin 0 m cos 0 = 2 m sin ,
n sin 20 m cos 20= n.

Ans. (n sin 0 + m cos 0)2 = 2 m (m + n). 60. Eliminate a from the equations

tan (a-B)= y tan (+B),
(20 - y) cos 2 a + (+ y) cos 2 B= 2.

Ans. 2 + 4xy = 2(x + y) cos2 B.

[ocr errors]

(+]=o

=

= 2.

CHAPTER VI.

RELATIONS BETWEEN THE SIDES OF A TRIANGLE

AND THE FUNCTIONS OF ITS ANGLES.

93. Formulæ. — In this chapter we shall deduce formulæ which express

certain relations between the sides of a triangle and the functions of its angles. These relations will be applied in the next chapter to the solution of triangles. One of the principal objects of Trigonometry, as its name implies (Art. 1), is to establish certain relations between the sides and angles of triangles, so that when some of these are known the rest may be determined.

RIGHT TRIANGLES,

[ocr errors]

94. Let ABC be a triangle, right-angled at C. Denote the angles of the triangle by the let

B ters A, B, C, and the lengths of the sides respectively opposite these angles, by the letters a, b, c.

Then we have (Art. 14) the following relations:

А

b a=csin A=ccos B=b tan A= b cot B

(1)

:

b=csin B =cCOS A = a tan B= a cotA (2) c=bsec A = a sec B= b cosec B =

= a cosec A .

(3) which may be expressed in the following general theorems:

[ocr errors]

* The student must remember that a, b, c, are numbers expressing the lengths of the sides in terms of some unit of length, such as a foot or a mile. The unit may be whatever we please, but must be the same for all the sides.

I. In a right triangle each side is equal to the product of the hypotenuse into the sine of the opposite angle or the cosine of the adjacent angle.

II. In a right triangle each side is equal to the product of the other side into the tangent of the angle adjacent to that other side, or the cotangent of the angle adjacent to itself.

III. In a right triangle the hypotenuse is equal to the product of a side into the secant of its adjacent angle, or the cosecant of its opposite angle.

EXAMPLES.

In a right triangle ABC, in which C is a right angle, prove the following: 1. tan B = cot A + cos C. 2. sin 2 A sin 2 B.

ab 3. cos 2 A + cos2B=0.

2 =

c
b
5. cosec 2 B

62 a?
26 2 a
6. cos 2 A=

c2
2 ab

3 ab? a3 7. tan 2 A

8. sin 3 A = 62 - a?

CP

4. sin 2 A = 20$

=

a

+

[ocr errors]
[ocr errors]

OBLIQUE TRIANGLES.

95. Law of Sines. - In any triangle the sides are proportional to the sines of the opposite angles.

Let ABC be any triangle. Draw
CD perpendicular to AB.

We have, then, in both figures
CD casin B=b sin A. (Art. 94)
... a sin B = b sin A.

A
6
sin A sin B

A

a

[ocr errors]

с

B

a

Similarly, by drawing a perpendicular from A or B to the opposite side, we may prove that

A

с

B

D

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

96. Law of Cosines. In any triangle the square of any side is equal to the sum of the squares of the other two sides minus twice the product of these sides and the cosine of the included angle.

In an acute-angled triangle (see
first figure) we have (Geom., Book
III., Prop. 26)
BC2 = AC + AB’ – 2 AB X AD,

a = b +- 2c. AD. But AD=bcos A.

с .. a=b? +- 2 bc cos A.

b

or

A

=

a

[ocr errors]

D

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

In an obtuse-angled triangle (see second figure) we have (Geom., Book III., Prop. 27)

BC = AC + AB* + 2 AB X AD,
a=b2 + 2 + 2 c. AD.

or

=

But AD= b cos CAD=- bcos A.

.a = b +- 2 bc cos A. Similarly, 62 = c + a2 — 2 ca cos B,

c= a + b2 — 2 ab cos C.

NOTE. — When one equation in the solution of triangles has been obtained, the other two may generally be obtained by advancing the letters so that a becomes b, b becomes c, and c becomes a; the order is abc, bca, cab. It is obvious that the formulæ thus obtained are true, since the naming of the sides makes no difference, provided the right order is maintained.

[ocr errors]

[ocr errors]

97. Law of Tangents. - In any triangle the sum of any two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference.

By Art. 95, a:b= sin A:sin B.
By composition and division,

a+b sin A + sin B

sin A sin B
tan }(A + B), by (13) of Art. 61 (1)

tan (A - B)

btc tan }(B+C) Similarly

b-C tan }(B – C)
cta tan }(C + A)

( tan (C - A) Since tan (A + B)= tan (90° – +C)= cot C, the result in (1) may be written cot C

(4) tan }(A -- B)

(2) (3)

C-a

=

[ocr errors]

a

b

and similar expressions for (2) and (3).

98. To show that in any triangle c= a cos B + b cos A.

In an acute-angled triangle (first figure of Art. 96) we have

c= DB + DA

= a cos B + b cos A. In an obtuse-angled triangle (second figure of Art. 96) we have

c= DB – DA

= a cos B - b cos CAD.

..c= a cos B + bcos A.

Similarly,

b=ccos A ta cos C,

a=bcos C+ccos B.

« AnteriorContinuar »