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(9)

8. The co-tangent of either acute angle is equal to the adjacent side divided by the opposite side.

BH : BT :: BP

BQ, or p: 1

PH : PN :: PB:

PL, or b : 1 :: h:

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(11)

=

b

p:

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h

#1

cosec P

h cosec B

::h sec B.

sec P.

sec B:

and (12)

=

9. Either side adjacent to the right angle is equal to the hypotenuse divided by the secant of the adjacent acute angle.

sec P

10 The secant of either acute angle is equal to the hypotenuse divided by the adjacent side.

=

Since the angles B and P are complements of each other sec B: =cosec P, sec P= cosec B; ... (9) and (10) become,

h

p

h

b

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11. Either side adjacent to the right angle is equal to the hypotenuse divided by the co-secant of the angle opposite that side.

12. The co-secant of either acute angle is equal to the hypotenuse divided by the side opposite that angle.

Scholium. By some authors, principles 2, 4, 6, 8, 10, and 12, have been given in the form of definitions.

Introducing radius into these formulas, by substituting for any function to the radius 1, the corresponding function to the radius R divided by R, and reducing, we have:

S. N. 5.

(1)

(3)

(5)

(7)

(9)

(11)

(1)

p=

(2)

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(3) {

=

p=

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=

h sin P

R

h sin B

R

=

h cos B

R

h cos P

R

b tan P

R

p tan B

R

b cot B

R

p cot P

R

Rh

sec B

Rh

sec P

Rh

cosec P

Rh cosec B

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(2)

log plog
log b

log

(4)

(6)

(8)

(10)

(12)

sin P=

sin B

10+ log p

10+ log b

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Applying logarithms to these formulas, we have:

log plog hlog sin P-10.
log blog h+ log sin B-10. J

{
{

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Rp

h

Rb

h

Rp

RP.

Rp .

b

Rb

p

[ocr errors]

Rp

b

Rb

p

Rh

p

Rh

b

Rh

p

Rh

b

log h.

log h.

hlog cos B 10.

h + log cos

P10.

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1. Given

=

log p= log blog cot B
log blog plog cot

10

{

(log p
log b 10

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=

(12) { log cosec B=

10

10

[ocr errors]

p

b

[ocr errors]

p

h

365.

P 33° 12'.

10+ log cosec P 10+ log

10+ log h
10+ log h

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12:}

[ocr errors]

blog p.

log b.

10+ log p
10+ log b -- log p.

h

h

[ocr errors]

log h.

log h.

log h-log sec B.

log hlog sec P.

[ocr errors]

P-10.

B-10.

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log b.

[ocr errors]

10.

P-10.

+log h-log cosec P.
+ log h―log cosec B.S

B=90° — P — 90° - 33° 12'

log p.

log b.

65. Case I.

Given the hypotenuse and one acute angle, required the

remaining parts.

log p. log b.

B.

Requir. p.
b.

}

}

P

=

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- 56° 48'.

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Either side adjacent to the right angle is equal to the sine of the opposite angle, multiplied by the hypotenuse.

・・・ p=h sin P.

Introducing radius, we have, p =

Applying logarithms, we have,

2.56229

log h (365)
log sin P (33° 12′) 9.73843
log p
2.30072 .. p

2. Given Sh

B

3. Given

log plog h+ log sin P-10.

199.89..

In like manner, from either formula, bh sin B, or bh cos P, we find b

=

1. Given

Sh

73.26.

49° 12′ 20′′.

2195.

P-27° 38′ 50′′.

=

h
p=97.

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112.

= 305.41.

=

sin P

h sin P

R

} Requir.

} Required

66. Case II.

Given the hypotenuse and one side adjacent to the right angle, required the remaining parts.

B

p
h

P

b:

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62° 21' 10".

Requir. p=1018.512.

1944.364.

=

b

B.

b.

[ocr errors]

=

40° 47′ 40′′.

55.4625.

47.8644.

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The sine of either acute angle is equal to the opposite side divided by the hypotenuse.

... sin P

Introducing radius, and multiplying by R, we have,

Rp .

h

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Η

Applying logarithms, we have,

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b

B 90° P-90°-60° 00′ 17′′-29° 59′ 43′′.

=

2. Given

log sin P=10+ log p-log h.

log p (97) = 1.98677
log h (112)
log sin P

2.04922

9.93755

h sin B, or b

We can also find b as follows:

3. Given

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1. Given

Sh

b=Vh2-p2
− p2 = 1 ́ ́ (h + p) (h—p).

log blog (h+p) + log (h—p)].

[blocks in formation]

=

[ocr errors]
[ocr errors]
[blocks in formation]

h cos P, b=55.991.

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P 60° 00′ 17′′.

Required

B=25° 47'′ 07′′.

Required P=64° 12′ 53′′.

p=6545.

b=152.67.

P 50° 18' 32′′.

B=90° — P= 90° — 50° 18′ 32"

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67. Case III.

Given one side adjacent to the right angle and one acute

angle, required the remaining parts.

=

B.

} Requir. p.

h.

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39° 41′ 28′′.

B

H

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