eft fide be 70° 30'; required the other two angles and the third fide?

THEOREM, Fig. 33. The half difference of any two quantities added to their half fum, produceth the greater quantity; and their half difference being ta ken from their half fum leaves the leffer quantity.

Let AB and BC reprefent two quantities, whofe half fum is AF or FC, and half difference FB. It is manifeft, that FB, the half difference, added to AF the half fum, is equal to AB, the greater quantity; and FB, the half difference, being taken from FC, the half fum, leaves BC the leffer quantity.

CASE 3. Fig. 34. Two fides of an oblique angled triangle AB and BC, with the angle ABC contained between them, being given, to find the other angles at A and C, and the third fide AC.

If the two given fides are equal between themselves, fubtract the given angle from 180°, and half of the remainder is either of the other angles; and find the third fide by Case 1.

If the given fides are unequal, the angles may be found by this Theorem.

In any plain triangle ABC,

As the fum of any two fides BC+BA
Is to their difference

BC-BA

So is the tangent of half the sum of

the angles oppofite to those fides

To the tangent of half their difference

Take

Take BE and BF, each equal to AB; join FA and AE; and through C draw CD parallel to AE: Then F, A, E, are points in the circumference of a circle whofe center is B, and FAE, FDC are right angles; FC is the fum of the fides AB and BC, and EC their difference.

The angle BAE is equal to the angle AEB, and their fum is equal to the fum of the two angles BAC and ACB; because the angle ABF is equal to either of these fums; therefore, the angle AEB, or its equal FCD, is the half fum of the angles BAC, ACB; and the angle EAC, or its equal ACD, is their half difference; because, this added to BAE, their half fum, is equal to BAC, the greater of the two.

Making CD the radius, FD is the tangent of the angle FCD, the half fum, and DA is the tangent of ACD, the half difference of the angles BAC and ACB; the triangles AFE, DFC are fimilar; there fore,

As FC: CE::FD: DA; that is,

As FC, the fum of AB and BC,

Is to EC their difference,

So is FD, the tangent of half the fum of the angles
BAC and ACB,

To DA, the tangent of half their difference.

The half difference added to the half fum, gives the greater angle, and fubtracted gives the leffer.

EXAM. I. In the oblique angled triangle ABC, there is given the fide AB=317, the fide BC=429, with

the contained angle at B=48° 40'; required the angles at A and C, and the fide AC?

To tan. A-C= 18° 22′ From the halffum 65° 40′ Subt. the halfdiff. 18 22

2

The leffer ang.C=47° 18′

Having found the angles, the fide AC is found by Cafe 1. thus, as Sine C: Sine B :: BA: AC=323.8.

EXAM. 2. Fig. 32. Suppose CDE a triangular field; the fide DE being measured, is 564 links of a furveying chain, and EC-368 links; alfo the angle at E being measured with an instrument, is 54° 20′; what are the angles at C and D, and what is the length of the fide CD?

3. Suppose the distance of the earth from the fun is 81000000 miles, and the distance of the moon from the earth is 240000 miles; and that the angle contained by two right lines drawn from the center of the earth to the centers of the fun and moon is 89° 6′; what is the diftance of the moon from the fun?

CASE 4. When the three fides AB, BC, CA, of any oblique angled plain triangle ABC are given; to find the angles at A, B, and C.

If any two of the given fides are equal, the triangle being ifoceles, its angles may be found thus. Fig. 35. Suppofe AB is equal to AC, let fall the perpendicular AE, and the point E will fall in the middle of the fide BC; therefore, by Case 2. of right angled triangles, as AB: BË :: Rad. : Sine BAE, whose comple. ment is the angle at B or C; and its double is the angle BAC.

When the fides are unequal, Fig. 36. let fall the perpendicular AE upon the base, or longest fide BC; and, about the center A, with a radius equal to the shortest fide AC, defcribe the circle FDG. Produce BA to G; then BG is the fum of the fides BA and AC, and BF their difference; alfo BE and EC are the fegments of the bafe, and BD their difference. The rectangle under CB and BD is equal to the rectangle under GB and BF, by the property of the circle; therefore, as CB: BG: BF: BD; that is,

As the bafe BC

Is to (BG) the fum of the other two fides BA+AC,
So is (BF) the difference of thefe fides BA-AC
To (BD) the difference of the fegments of the base
BE-EC.

Now, take the half of this difference, and also the half of BC, their fum is BE the greater fegment, and their difference is EC the leffer fegment. Then, in the right angled triangle ABE, there is given AB and

BE; to find the angles, which is done by Cafe 2. of right angled triangles, viz.

:

As AB BE: Rad.: Sine BAE, whofe complement is the angle at B. For the fame reafon, in the right angled triangle AEC, as AC: CE :: Rad. : Sine EAC, whose complement is the angle at C.

BAE+EAC the angle BAC.

And

EXAM. in the oblique angled triangle ABC, there are given the three fides, viz AB 264, BC=384, and AC 196; to find all the angles at A, B, and C.

(2.) As AB: BE:: Rad: Sine BAE=61° 49′ whofe

complement 28° 11′ is the angle at B.

(3.) As AC: CE:: Rad.: Sine EAC 50° 31', whose complement 39° 29'the angle at C, and

BAC=112° 20′.

EXAM. 2. Fig. 37. Let D and E reprefent two ports in the fame latitude, whofe distance is 480 miles; and suppose a ship from D fails 300 miles between north and eaft, and another fails from E 240 miles between

north