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6)

ac

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or, 2 cosa

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or, cos?

a

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The form of this result gives us also-
B (8 a) (8 - c) C

a) (8 Sin

sin 2

2

ab
Again, Art. 18, we have
A

12 + c
2 cos? 1 = cos A =
2

2 bc
А

72 + c a> (b + c) - a
1+
2
2 bc

2 bc
(a + b + c)(b + c a) 28.2 (8 a);
2 bc

2 bc
A 8 (8 - a)

2 bc
Α. 8 (8
.'. COS

(2) 2

bc The form of this result gives us also B 8 (8

6)

С 8 (8 82

2 Now, (1) ^ (2), we have

A
Sin
2

6) (8 c) 8(8
bc

bc
Cos

2 tan (8 6) (8

:(3). 2

8(8 a) The form of this result gives us also— B a) (8

C tan

2

2 8(8 b)

s(s – c)
35. In any triangle ABC, a = b cos + c cos B.
Using the figures of Art. 30, we have~
(1.) When C is acute-

BD AB cos B, and DC AC cos C.
... BD + DC AC cos C + AB cos B, or

b cos C + c cos B,

COS

COS

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ab

ac

8

a

or

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c) tan

a =

BD =

or a

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a

or, sin A

(2.) When C is obtuse

AB cos B, and DC

AC cos ACD = AC cos (180° — C) = - AC cos C. .:. BD - DC = AC cos C + AB cos B,

b cos C + c cos B, as before. The form of this result gives us also

7

= a cos C + c cos A, c = a cos B + 6 cos A. Cor. 1. Hence sin (B + C) sin B cos C + cos B sin C.

6 For we have, 1

cos C + cos B, or, by Art. 30,

a sin B

sin C cos C +

cos B, sin A

sin A

sin B cos C + sin C cos B. But sin A sin (180° – A) sin (B + C). ... Sin (B + C) sin B cos C + cos B sin C. COR 2. Hence also sin (B - C) sin B cos C

cos B sin C. For, since this result has been proved for any angles of a triangle, it will be also true for a triangle which contains an angle supplementary to B; that is, it will be true, if we put 180° B for B. We then have Sin (180° – B + C) = sin (180° B) cos C

+ cos (180° - B) sin C. But sin (180° – B + C) sin (180° - B - C) sin (B-C),

sin (180° - B) = sin B, cos (180° – B) Hence, sin (B - C) = sin B cos C cos B cos C. NOTE.-The results of Cor. 1 and Cor. 2 have been proved only for angles less than two right angles. We shall see in Vol. II, they are generally true. 36. In any triangle ABC

6 Tan A (B – C)

cot} A.

b + c We have, Art. 30,

Sin B 6 sin B - sin c 6
Sin C c sin B + sin C b + c

cos B.

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or

.....(1)

.

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....

= sin

sin

2 cos

B + C

B C and sin C sin

2

2

, or, by Cor. 2, Art.35, B + C B С

B + C. B - C
COS

COS
sin

(3).
2
2

2

2

B + c B-C
(2)
(3), then sin B sin C

2 cos
2

2

B + c B-C and (2) + (3), then sin B + sin C = 2 sin COS

2

2 B + c B-C

sin sin B - sin C

2

2 sin B + sin c

B + C B 2 sin

COS 2

2 B + c B С

B - C
tan
= cot (90°

tan
2
2

2
A B - C
tan tan

.(4). 2 2

A B-C 6 (4) = (1), then tan

tan 2

2 B-C

6 .. tan

cot : A. Q.E.D. 2

b + c

= cot

b + c

с

Ex. VI.
1. If a = 5, C = 30°, sin A }, find c.
2. Find a, having given b = 12, c = 15, A 60°.

Α.
3. Find tan when a =

6, 6 7,0

= 8. 2' 14. What is the area of a triangle whose sides are 48, 52, 20?

5. Given two sides of a triangle to be 18 and 24, and the included angle 45', find the area,

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6. Two of the sides of a triangle are as 2:1, and the included angle is 60°, find the other angles. 7. In any triangle show thatb cos C

Na? - 4 bc cos B cos C.

sin B sin c 8. Show that the area of a triangle = } a:.

sin A 9. An object is observed from two stations 100 yards apart, and the angles subtended by the distance between the object and either station are 45° and 60° respectively. Find the distance of the object from each station.

10. An observation is made from a point known to be distant 120 and 230 yards respectively from two trees, and the angle which the trees subtend is found to be 120°. Find the distance between the trees. 11. If a sino 0 + b cosa o : m,

1 1 1 1 b sin 0 + a cosa o

6 a tan o 12. If a', b, c be the sides of the triangle formed by joining the feet of the perpendiculars from the angles A, B, C of the triangle ABC upon the opposite sides, then

al b! ci a” + 72 + c2

a? 72 ca 2 abc 13. A perpendicular AD is drawn from the angle A of a triangle, meeting the opposite side BC and D; and from D a perpendicular is drawn to AC, meeting it in E. Show that DE b sin C cos C. 14. Show that the length of AD in the last example

bc sin A + ac sin B + ab sin

En, } then

+ a

to

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n2

+

+

3 a

15. Show that V8 (8 a) (s 6) (8 c) = jab, when the triangle is right-angled at C. 16. Show that (a + b + c)2 sin A sin B

(sin A + sin B + sin C) ab. 17. Show that in any triangle

cosa A + cos2 B + cos O + 2 cos A cos B cos C = 1.

18. The sides of a triangle ABC are in arithmetical progres

b3 sion; show that its area = (2 a ) (36 2a).

2

CHAPTER VIII.

SOLUTION OF RIGHT-ANGLED TRIANGLES.

37. A triangle can always be determined when

any

three elements, with the exception of the three angles, are given. In the latter case we have only the same data as when two angles are given, for the third can always be found by subtracting the sum of the other two from two right angles (Euc. I., 32).

Hence a right-angled triangle can always be det when any two elements, other than the two acute angles, are given besides the right angle. And when one of the acute angles is given, the other may be obtained by subtracting it from a right angle.

We have the following cases :

CASE 1, When the two sides containing the right angle are given.

A We shall take C as the right angle
in every case.
Now tan A

Ő
or, L tan A

10 log 06 - log 0; or, L tan A = 10 + log a - log b...(1).

This determines A, and we then have B 90° - A........

..(2) B4 Also, sin A, or log a

log'c = I sin A - 10. .. logo 10 + log a L sin A ........

(3) Hence the three elements, A, B, C, are determined.

a

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