Then by Propofition 3, fay;

As the Sine of half the Sum of An. is to the Sine of half their Differ. fo is Tan. of half the includ. Side

85 00=0.0016558 45 00=9.8494850 18 00 9.5117760

to Tang. of half the Dif. of Sides 12 59=9.3629168

Then again by Propofition 4.

Co. Ar.

As Co-fine of half Sum of Ang. 85 00 = 1.0597030

is to the Co-fine of half their Dif. 45 00

fo the Tan. of half the incl. Side 18 00

9.8494850 9.5117760

to Tan. of half Sum of two Sides 69 33110.4209640

The two Sides being thus found, the other Angle C is had as in the common manner, by Theorem 35.

1. To find the Angle C, by Theorem 35.

The Angle Cmuft firft be found, as in the above Variety; then find the half Sums, and half Differences of the Sides D B and B C, and of the Angles D and C, as in the fecond Variety of Cafe ; Then Jay by Propofition 2,

As the Sine of Dif. of the Angle 6 299.9472515 is to the Sine of half their Sum 33 30=9.7420603 fo is Tang. of half Dif. of Sides 10 07

=9.2514612

=9.9407730

to Tang. of half the includ. Side 41 061

=

as required.

Which doubled, gives the Side DC 822 12,

3. To find the Angle B..

SAT

This is the fame with Variety 2, of Cafe ; as the fecond of this, was the fame with the third of that Cafe; and therefore need not be again repeated.

Then by Propofition 1, fay

D

doing 2 sdt slot vd

As Sine of half the Sum of Sides 59 06 is to the Sine of half their Differ. 23 064 fo Co-tan. of half contained An. 20 00

to Tan. of half the Dif. of Ang. 51 29

Co. Ar.

0.0664547

9.5937206

Then by Propofition 2, fay;

As Coffine of half Sum of Sides 59 06

10.4389341

10.0991094

Co. Ar. 0.2894940

is to Co-fine of half their Dif. 23 061 ± 9.9636856 fo is Co-Tan. of halfiinclu. An. 20 000

the Tangent of half the

10.4389341

Sum of the other two Ang. 78 3972=10.6921137

Then, to the half Sum of the Angles

78 3013

add half the Difference of the Angles 51 29+

the Sum, is the greater Angle B =

130 00

the Differ. is the leffer Angle C = 27 01 2

The

The Side B C is found, by Theorem 35, fince all the other Parts are known.

This may be done two ways; firft by Theorem 39, already fufficiently exemplified in the laft Chapter, Cafe 5. The fecond Way is by finding the Segments of the Bafe D C, made by a perpendicular Arch falling thereon fubtending the Angle fought, by Propofition 5 Thus having taken the half Sum of the two Sides D B and B C, viz. 46° 07', and alfo the half Difference thereof, 10° 07', as likewife half the Base DC, 41° 06'; fay thus ;

As the Tangent of half the Bafe 41 061 is to Tang. of half Sum of Sides 46 07 fo is the Tang. of half their Dif. 10 07

to the Tang. of half Differ. of the Segments of the Bafe

Then from the whole Bafe - 82 124 DC take the Difference of the Segments 24 00

there will remain

= VC

the half of which is the leffer Segm. 29 06 =DS = SV

24 001=VC

the Sum is the greater Segment 53 061=SC

Then fay,

As the Tangent of the Side BD=36 00 is to the Tang. of the Seg. DS=29 06

fo is Radius

9.8612610 9.7455376

10.0000000

} D=40 00

9.8842766

to the Co-fine of the

Angle fought

And any other of the Angles may be found in the

fame Manner.

The only Way for folving this Cafe, is by Converfion of Triangles, as taught in Theorem 13, and exemplified in this Cafe of the laft Chapter; and fo in the foregoing Scheme, the Reciprocal Triangle AEG, hath its three Sides equal to the Angles of the Triangle DB C, the Complement of B being taken; for AED 40° 00'; and E G=C=27° 01'; and laftly, AG = Complement of B, viz. 50° ool;

Hence