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α=

1800.

6. Given b = 2000.

B=111° 15'.

75. Case III.

Given two sides and their included angle, required the

remaining parts.

Req.

Let ABC be a triangle, and let the sides opposite the angles A, B, C, be denoted, respectively, by a, b, C. Let a and b, and their included angle C, be given, and the remaining parts, A, B, and c, required.

A+ B

A 57° 0′ 50′′.

=

C11° 44' 10".
C= 436.49.

E

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The sum of the angles A and B is found from the formula,

180° - C.

With C as a center, and b, the shorter of the two given sides, as a radius, describe a circumference cutting a in D, a produced in E, and c in H. Draw AE, AD, CH, and DF parallel to AE. The angle DAE is a right angle, since it is inscribed in a semi-circle; hence, its alternate angle, ADF, is also a right angle.

HF B

The angle ACE being exterior to the triangle ABC, is equal to A + B. But ACE having its vertex at the center, is measured by the intercepted are AE. The inscribed angle ADE is measured by one-half the arc AE; hence, ADE } ACE = {(A + B).

CHCA, since they are radii of the same circle; hence, the angle CHA A. The angle CHA being exterior to the triangle CHB is equal to HCB+B; hence,

HCB+B= A. .'. HCB= A — B.

But HCB, having its vertex at the center, is measured by the intercepted arc DH; and DAF, being an inscribed angle, is measured by one-half the arc DH; hence, DAF = } HCB = 1 (A — B).

In the right triangles ADE and ADF we have

AD tan

(A+B).

AD tan DAF AD tan (A — B).

From the similar triangles, ABE and FBD, we have

BE BD: AE: DF.

AE AD tan ADE

DF

=

=

Since CECA, BE BC+ CA=a+b.
Since CDCA, BD

=

=

BC-CA-a-b.

Substituting the values of BE, BD, AE, and DF in the above proportion, and omitting the common factor AD in the second couplet, we have

tan (A — B) =

a+ba−b :: tan 1⁄2(A+B) : tan 1⁄2 (A— B).

Hence, In any plane triangle, the sum of the sides including an angle is to their difference as the tangent of half the sum of the other two angles is to the tangent of half their difference.

We find from the proportion, the equation

We have now found A = { (A + B) + (A−B), ¦

(a - b) tan 1 (A+B).
a + b

... log tan (A——B) = log (a —b) + log tan 1⁄2 (A+B) a. c. log (a + b) --- 10.

(A+B) and 1⁄2 (A — B).

B={ (A + B) — } (A —- B).

sin A sin C:: a: c, '. C=

a sin C

sin A

.. log clog a + log sin C+ a. c. log sin A-10.

S. N. 6.

1. Given

...

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2. Given

b

C=68° 25'.

}

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Req.

a+bab tan †(A+B) : tan †(A — B).

::

tan (AB)=

... log tan (A—B) = log (a —b) + log tan (A+B) +a. c. log (a+b) — 10.

log a (37.56)

log sin C (68° 25′)

a. c. log sin A (74° 7′ 25′′)

log c

A.

B.

(ab) tan (A+B)

a + b

log (a-b) (13.81)

1.14019

log tan (A+B) (55° 47′ 30′′) = 10.16761 a. c. log (a+b) (61.31)

= 8.21247

log tan (A-B)

9.52027

(A-B)= 18° 19′ 55′′.

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sin A sin C:: ac, .'. C=

=

A

=

..

A = 1⁄2 (A + B) + 1⁄2 (A — B) = 74° 7′ 25′′.
7'

B = ↓ (A+B) — (A-B)

37° 27′ 35′′.

a sin C

sin A

log clog a log sin C+ a. c. log sin A-10.

[ocr errors]

a =

996.63.

b = 712.83.

C 72° 29′ 48′′.

с

1.57473
9.96843

0.01689

1.56005, ... c=36.312.

A

Req. B

B

66° 30′ 37′′.

40° 59′ 35′′.

C = 1036.35.

[blocks in formation]

77. Case IV.

Given the three sides of a triangle, required the angles.

Let ABC be a triangle, take the longest side for the base, and draw the perpendicular p from the vertex B to the base.

Ն

Denote the segments of the base by s and s' respect

ively.

A

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[ocr errors]

B

(a+c) (ac)
8 +8

P

[ocr errors]

a

Then, (1) c2-s'2 — p2, and (2) a2-s2= p2.

... (3) c2—s′2—a2—s2, .'. (4) s2—s′2—a2—c2.

.'. (5) (8+8′) (8 −− s'′) = (a + c) (a−c).

·

.'. (6) s + s′ : a + c : : a -C: S - s'.

Hence, The sum of the segments of the base is to the sum of the other sides as the difference of those sides is to the difference of the segments.

(6) gives (7) 8—8′ =
s S

... (8) log (s—s′) = log (a+c) -+ log (a−c)
a. c. log (88) 10.

с

In case the sides of the triangle are small, find s S' from (7); otherwise, it will be more convenient to employ (8).

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Having s+s' and s-s', we find s and s' thus,

(9) 8 1

=

= 1⁄2 (8 + 8′ ) + 1⁄2 (8 —s′), (10) s'=1(s + s′ ) — ¦ (s—8′).

(11) cos A (12) cos C C Introducing R, reducing, and applying logarithms, (13) log cos A= 10+ log s'-log c. (14) log cos C 10+ log slog a.

From which we find A and C.

1. Given b

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Then, (15) B=180°- (A+ C).

a 125.

Cos C

78. Examples.

150.

100.

A.

Req. B.
C.

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a

B

s + s': a + c :: a

(a+c) (a−c)
8+8

225 × 25
150

8 = 12 (8 + 8') + 4 (8-8)=75+18.75=93.75.

8' (88) — (8-8)75-18.75-56.25.
— 11 1

COS A

, or introducing R, cos A=

с

.. log cos A 10+ log slog c.

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37.5.

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S

=, or introducing R, cos C

=

a

Rs'

·. A = 55° 46′ 18".

R$

a

с

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