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54. Eliminate 0 and $ from the equations
x cos 0 + y sin 0 = a, b sin (8 + )= a sin ,
Ans. x2 + y2 = a? +
62 55. Eliminate 0 from the equations
sec” 0 cos20
56. Eliminate from the equations
(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
22 + y2
62 58. Eliminate 6 and from the equations
aʼcos? 0 — b2 cosa d = c, a cos 0 + b cos $=r,
(2 — c) 59. Eliminate from the equations
n sin 0 — m cos 0 = 2 m sin ,
Ans. (n sin 0 + m cos 0)2 = 2 m (m + n). 60. Eliminate a from the equations
tan (a-B)= y tan (+B),
Ans. 2 + 4xy = 2(x + y) cos2 B.
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.
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:
b a=csin A=ccos B=b tan A= b cot B
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:
* 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.
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.
62 — a?
3 ab? — a3 7. tan 2 A
8. sin 3 A = 62 - a?
4. sin 2 A = 20$
95. Law of Sines. - In any triangle the sides are proportional to the sines of the opposite angles.
Let ABC be any triangle. Draw
We have, then, in both figures
Similarly, by drawing a perpendicular from A or B to the opposite side, we may prove that
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
a = b +- 2c. AD. But AD=bcos A.
с .. a=b? +- 2 bc cos A.
In an obtuse-angled triangle (see second figure) we have (Geom., Book III., Prop. 27)
BC = AC + AB* + 2 AB X AD,
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.
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.
a+b sin A + sin B
sin A sin B
tan (A - B)
btc tan }(B+C) Similarly
b-C tan }(B – C)
( 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)
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.
b=ccos A ta cos C,
a=bcos C+ccos B.