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EXAMPLE 3.
In the plane triangle ABC,
AC 120

LA 57°27'00"

poles. Given BC 112

Ans. LB 64 34 21

AB 112:65 poles. Required the other parts.

R 2C 57° 58' 39"; }

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PROBLEM III.

Given the three sides ; to find the angles.

In any plane triangle, having let fall a pependicular from the greatest angle to the opposite side or base, dividing it into two segments, and the whole triangle into two right-angled triangles; it will be

;
As the base, or sum of the segments
Is to the sum of the other two sides
So is the difference of those sides
To the difference of the segments of the base.*

:

*

Then

*

26

B

* .

DEMONSTRATION. From one end B of the least side AB, of the triangle ABC,

a centre, and radius AB, describe a circle cutting the other two sides in E and F; produce CB to the circle at G, and let fall the perpendicular BD. Then is GB=B=AB, and (by 3 III. Eucl.) AD= DE, and consequently EC=CD-DA the difference of the s£gments, FC=GB-BA the difference of the sides, and GC=

CB

Then half the difference, being added to and subtracted from half their sum, will give the greater and less segment.

Hence, in each of the right-angled triangles, will be known two sides, and the angle opposite to one of them ; and consequently the other angles will be found by the first problem.

NOTE. In the above proportions, if half the difference of the sides be taken for the third term, then the fourth term will be half the difference of the segments. Which will commonly be more convenient to use than the whole difference.

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Geometrically.
1. Draw AB equal to 345, by a scale of equal parts.

2. With

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CB+BA the sum of the sides. But (by Cor. to 36 III. Eucl.) the rectangle CAXCE=CGX CF, or CA : CG :: CF : CE, that is, AC : CB+BA :: CB-BA : CD-DA.

Q. E. D. And that the half-sum of two quantities, increased and diminished by their half-difference, gives the greater and less quantities respectively, was proved in the last problem.

2. With the centres A and B, and radii 174'07 and 232, taken from the same scale, describe arcs intersecting in C.

3. Draw AC and BC, and it is done.

01

· Then, by measuring the angles, they appear to be nearly of the following dimensions, viz. ZA=370, ZB= 27', and C=1150*

Arithmetically

Having let fall the perpendicular CP, it will be, As AB =345 : BCTCA=406'07 :: BC-CA=57*93 406.07 X 57093

=68.18BP-PA. 345

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Then, in the triangle APC, right-angled at P,

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Whence the ZA=37 20', the B=27° 204', and the 2C=115° 36'.

Instrumentally.

In the first proportion, extend from 345 to 406, on the line of numbers ; that extent will reach, upon the same line, 'from 58 to 68.2, the difference of the segments of the base.

In the second proportion, extend from 174 to 138-on the numbers ; that will reach, on the sines, from 90%

; to 52° :

In the third proportion, extend from 232 to 206, and that extent will reach from 90° to 63o.

EXAMPLE EXAMPLE 2.

In the plane triangle ABC,
SAB 162 -

ZA 53° 07' 481
Given

Ans.

ZB

go oo oo
BC 216;

ZC 36 52 12
Required the angles.

AC 270

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EXAMPLE 3.

In the plane triangle ABC,
SA
AB 112'65

ZA 57° 27' oo"
Given
AC 120

Ans.

ZB 64 34 21 BC 112 ;

LC 57 58 39 Required the angles.

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EXAMPLE 4:

.-13

In the plane triangle ABC,
SAB 53

ZA 121°14
Given

Ans. ZB

29 23
BC 92'36 ;

ZC 29 23
Required the angles.

AC 53

1

1

Note 1. These three problems include all the casus of plane triangles, as well right-angled as oblique-angled. There are some other theorems, suited to some particular forms of triangles, which are often more expeditious in practice than the preceding general methods. One of which, as the case, for which it serves, so often occurs, is here given.

PROBLEM,

VOL. II.

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