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152. In the formulæ of the two preceding articles, it is convenient to abbreviate by writing

a+b+c=2s, so that a + b-c= 28 – 2c, and so on. Thus

1 ab sin C = 1{8 (8 -. a) (8 b) (8 c)}, С

c)

С (8 a) (8 —b)
ab
2

ab
С
tan
2

8 ()
Since } ab sin C = area of triangle = S (say),

:: S = \{8 (s-a) (8 - b) (8 -c)}.
153. To prove the sum-and-difference formula :-

tan} (A - B)

tan ] (A + B) Assume that A is greater than B; and therefore a greater than b.

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А.

F

B

E

With centre C and radius CA describe a semi-circle cutting BC in D and BC produced in E. :: angle ADE at circumference = 1, angle ACE at centre

=} (A + B),
:: angle DAB=ADE - ABD = } (A + B) B = 1 (4 B).
Also angle DAE in a semi-circle is a right-angle.
Draw DF parallel to AE, or at right-angles to DA.

tan 1 (A B) tan DAF DF AE DF
tan (A + B) tan ADE AD AD AE

BD

by similar triangles,
BE
BC - DC - 6
BC+CE

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a

a +6°

154. The student should observe that each of the identities connecting the sides and angles of a triangle (except the double-cosine formula) involves four out of the six elements A, B, C, a, b, c, two of the four being sides.

Thus by means of these identities we can find all the elements of a triangle when three, including a side, are given. [Compare Art. 29.]

In fact, since A + B + C = 180°, we have no more specific information about any particular triangle, when 3 angles are given, than when 2 angles are given. Hence a side must always be one of the elements given.

155. The formulæ above given may be thus classified.
Class I. Involving three angles.

A + B + C = 180°.
Class II. Involving three sides and two angles.

a=b cos C + c cos B.
Class III. Involving two sides and the two opposite angles.

a sin B = b sin A, A - B

A + B
tan

tan
2
a + b

2 Class IV. Involving two sides, the included angle and another angle.

c (cot C + cot B) = a cosec B,

c sin B
tan c

a-c cos B
Class V. Involving three sides and an angle.

Co = a+ b2 – 2ab cos C,
{8 (s a) (8 6) (s – c)} = { ab sin C,
С

(8 - a) (8-6))

s sin

ab

a-6

=

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J. T.

7

b-c

с

sin B

b-C

156. The substitution of d sin A, d sin B, d sin C for a, b, c respectively is often useful in the working of examples on Triangular formulæ. b

tan 3 (BẠC) Example 1. Given

show that sin C

b+c

tan 4(B+C) Put b=d sin B, c=d sin C. Then

d sin B-d sin C sin B-sin C b+cd sin B+ d sin C sin B+sin C

2 sin } (B-C) cos } (B+C)_ tan } (B-C)

2 sin } (B+C) cos }(B - Ctan (B+C)* Example 2. Show that tan B : tan C=a2 + b2 – 2 : q2 62+c?,

tan B sin B cos Cb cos C 2ab cos C a2 +62 - 02
tan Csin C cos B

ccos B

2ac cos B a? + c2 - 62 Example 3. Show that 8 cos A cos B cos C is never greater than 1. We have 2 cos A cos B=cos (A + B)+cos (A - B).

Keeping A+B (and, therefore, C) constant, this has its greatest value, when cos (A - B)=1, i.e. when A=B. Hence the given expression has its greatest value, when A=B=C=60°, i.e. when

8 cos A cos B cos C=1.

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EXAMPLES VI. B.

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If A, B, C are the angles of a triangle, prove the following statements :

1. sin (A + B + C) = cos } (A + B + C) = 0.
2, - cos (A + B+C) = sin } (A + B + C) = 1.

sin A - sin B С А B 3.

= tan tan sin A + sin B 2

2 4. tan A + tan B=sin. sec A. sec B. 5. tan A + tan B + tan C = tan A . tan B. tan C. 6. sin A + sin B – sin C == 4 sin } 1. sin } B.cos C. 7. cos A + cos B + cos C = 4 sin } A.sin 1 B. sin } C + 1. 8. cos A + cos B cos C = 4 cos į A.cos 1 B.sin 1C - 1. g. sin 2A + sin 2B + sin 2C = 4 sin A. sin B. sin C.

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10. cos 2A + cos 2B + cos 2C =- 4 cos A.cos B. cos C - 1. 11. cos” } A + cos” ] B - cos} C = 2 cos } A.cos } B. sin } C.

+ 1 COS A

С A - B 12.

cot tan cos B + COS A 2.

2 13. tan 1 B. tan} C + tan} C.tan 1 A + tan 1 A. tan; B=1. 14. sin? A = cosB + cosa C + 2 cos A.cos B. cos C. 15. cos? A + cos2 B + 2 cos A cos B cos C

- sino A + sin? B - 2 sin A sin B cos C. 16. 8 sin } A. sin } B. sin } C is never greater than 1. In any triangle ABC prove the following statements :

a sin C 17. tan A

6 - a cosĆ:

18. b (tan B + tan C) = a tan B sec C.
19. a cos A + b cos B + c cos C = 2a sin B sin C.
20.

(a + b) cos C + (b + c) cos A +(c + a) cos B = a +b+c. 21.

a (6 cos C c cos B) = 62 c?. 22. a sino B = b (cos A cos B + cos C). 23. a+ b2 + c= 2ab cos C + 2bc cos A + 2ca cos B. 24. ab sin? C =c (a cos B cos C + b cos C cos A + c cos A cos B). 25. 4S = a (26 sin B cos A + a sin 2B). 26. 26 (1 - sin B cos A cosec C) = a sin 2B cosec C. 27. 2 cos C (a sin A 6 sin B) = c (sin 2B – sin 2A). 28. bc cos A + ca cos B + 2ab cos C = a + b2. 29. b sin 2A 2a (sin C cos B sin A). 30. S (a + b + c)=abc (sin B cos' } A + sin A cos” } B). 31. (a - b) cos } C =c sin } (A – B). 32. a” cos B +62 cos C + c cos A

2ab sino 1 A + 2bc sin’ } B + 2ca sin } C. 33. (62 - c) cos A + (co – a) cos B + (a62) cos C

=a cos A (6 c) + b cos B (c- a) + c cos C (a - b). 34. 8 (2ab + 2bc + 2ca a62 c^) = 2abc (cos | A + cosa 1 B + cosa 1 C).

+

=

tan A a2 + c - 62 35.

tan B 72 + co – a?' 36. tan A (62 +ca – aạ) = 4S. 37. a sec A - 6 sec B -- sec C (6 sec A - a sec B). 38. Ss=abc (sin A + sin B + sin C).

(8 - a) sin A (8-b) sin’ | B

s ] bį (8 - c) sino LC S52 39.

6

sabc 40. a sin B + b sin C + csin A А B+C

B-C

B A+C A

( + cos
2

2
2

2 If C is a right angle, prove the following statements :-(41—50) 41. tan B= cot A.

42. tan A + tan B=sec A sec B. 43. C+a : b=b:c-a. 44. tan 2 A + tan 2B = 0.

a

C

= COS

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C COS

+ a cos

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+ COS

C COS

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a

sec 2B

+

+

a2 + 72

b 47.

48.

cosec 2B
a62

26 2a
2 sin A sin B
49. tan2B

50. 4S2 = abc cos A cos B. sino A- sin’B 51. If any one of the above equations (41–50) holds, examine in each case whether C is a right angle.

52. If, in any triangle ABC, d, e, f are the distances of the angles from the middle points of the opposite sides,

4 (do + e + f2) = 3 (a+ b + c^). 53. If l, m, n are the perpendiculars from A, B, C on the opposite sides,

2 (l cos A + m cos B+n cos C)=a sin A + b sin B +c sin C.

54. If BC be bisected in D and produced to E, cot A is the Arithmetic Mean between cot DAC and cot ACE.

35. If D be the middle point of BC, H the point where the bisector of A cuts BC, L the foot of the perpendicular from A on BC, then

4DH.DL = (6-0)

=

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