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

63 sion; show that its area = (2 a – b) (36 – 2 a).

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 determined 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

a

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a

Also, sin A, or log a

log c = L sin A 10. . log c = 10 + log a L sin A ......... Hence the three elements, A, B, C, are determined.

(3).

a

CASE 2. When the hypothenuse and a side are given.
Let a be the given side.
We have sin A =

or L sin A

10 log a – log c.
.. L sin A
10 + log a

log C........... ...(1), and B = 90° A

.(2), also 62 ca a (c + a) (c – a);

::. log b = 1 {log (c + a) + log (c – a)} .....(3). CASE 3. When an acute angle and a side are given Let A, a be the given angle and side. Then B 90° - A....

(1), 6 also = tan B, or logo L tan B 10 + log a.....(2),

a

a

= 90°

a

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and = sin A, or log a log c = L sin A - 10,

с or log c = 10 + log a L sin A. CASE 4. When the hypothenuse and an acute angle are given. Let A be the given acute angle. We have B A .........

..(1). Also = sin A, or log a = log c + L sin A - 10...(2).

C

6 And = cos A, or log b = log c + L cos A

10....(3). It is evident, from Art. 30, that when the angles only of a triangle are known, we can determine the ratio only of the three sides of the triangle to each other. Ex. 1. Given A

23° 41', a = 35, solve the triangle.
This is an example of Case 3.
We have B 90° - 23° 41' = 66° 19'.
Again, logo

L tan B – 10 – log a
L tan 66° 19'. 10 - log 35
10.3579092 – 10 + 1.5440680
1.9019772

- log 79.795
...6

79.795.

Also log c = 10 + log a – L sin A

= 10 + log 35 - L sin 23° 41'
= 10 + 1:5440680 – 9.6038817

= 1.9401863 log 87.134

..C = 87.134. Ex. 2. Given a =

214,6

317, solve the triangle. This falls under Case 1. We have L tan A 10 + log a

log 6 10 + log 214 – log 317 = 10 + 2-3304138 – log 2.5010593

9.8293545, Next lower in tables is 9.8292599 L tan 34° 1'; ...d

946. Also, by tables, D 2724, d

946 x 60" And x 6011 =

= 21" nearly
D

2724
.:. L tan A = L tan 34° 1' 21",

34° 1' 21".
Hence B = 90 34° 1' 21" 45° 58' 39".
And similarly may c be determined.

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or A

Ex. VII.

a.

1. Given a =

32, A = 63° 45', find b.
Log 32 = 1:5051500, L cot 63° 45' = 9•6929750,

Log 15780 = 4.1981070, log 15781 = 4.1901345. 2. Given c = 151, A = 37° 42', find

Log 151 = 2.1789769, L sin 37° 42 = 9.7864157,

Log 92340 = 4.9653899, tab. diff. = 47. 3. Given a = 60, c =

65, find b, A.
•3010300, log 3 4771213,
Log 65 = 1.8129134, L sin 67° 22! = 9.9651953,

L sin 67° 23! = 9.9652480.
4. Given a = 6 84, find A, C.
Log 73

1.8633229, L tan 40° 59! 9.9389079,

1.9242793, L tan 41° 9.9391631, L sin 40° 59' = 9.8167975, L sin 41° = 9.8169429, Log 111.288

2.0464479.

Log 2

73,

Log 84

5. Given B 71° 41' 10", c =

24, find 6. Log 24

1:3802112, L cos 18° 18' 9.9774609, L cos 18° 19' = 9.9774191, log 2278.4 = 3-3576300, Log 2278.5

3:3576490.
6. Given a =

293, c = 751, find b.
Log 1044 = 3:0187005, log 458 = 2.6608655,

Log 691.49 = 2.8397830.
7. Given a = 12, A = 30°, find b, c.
8. Given c =
10, B

find

a,

6. 9. Given a = 17, c =

34, find A, b. 10. Given a = 5, 6 5 /3, find A. 11. Given a =

15°, find the length of the perpendicular from C on AB.

12. CD is the perpendicular from C on AB, and DE the perpendicular from Don BC. Given B = 60°, a = 20, find DE

75°,

.

=

28, B

CHAPTER IX.

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SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 38. Given the three sides of a triangle, to find the remaining parts.

62 + ca a? We have, Art. 31, cos A =

from which A

2 bc may be determined ; and from similar formulæ we may find B and C. These formulæ are not however adapted to logarithmic computation. We shall therefore find it generally advisable to use the formulæ of Art. 34.

A We have, sin 2

bc A

8(8 - a)
2

bc
A (s b) (8 c)
tan
2

8(8 - a)

с

(8 ) (

COS

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From either of these formulæ we can determine A, and from similar formulæ determine the other angles.

39. Given one side and two angles, to find the remaining parts.

Of course the third angle is at once known. Let a be the given side.

a sin B We have, Art. 30, 6

sin A

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C

b + c

both of which formulæ are adapted to logarithmic computation.

40. Given two sides and the included angle, to find the remaining parts. Let b, c be the given sides, and A the included angle.

t We have, Art. 36, tan : B C) cot 1 A.

This formula is adapted to logarithmic computation, and determines } (B – C). We know also ! (B + C), for it is the complement of A. , are easily .

6 sin A Then, Art. 30, a =

which determines a.

sin B' NOTE.—When the two given sides are equal, the solution may be effected more easily by drawing a perpendicular from the given angle upon the opposite side, and so bisecting it. By drawing a figure, it is easily seen that, in this case, B = C = 90° – A, and a = 2 cos B.

41. Given two sides and an angle opposite to one of them, to find the remaining parts.

Let a, b, B be the given elements.
Then we have, Art. 30, sin A *sin

sin B.

a (1.) Let the value of sin B be unity.

5 We then have sin A = 1 sin 90°, and : A = 90°. Hence the triangle is right-angled at A.

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