Vertical Angle upon the Bafe, being given, thence to find the Bafe and the Sides.

I S

1155 the Sum of the Sides.

2d

495 the Difference of the Segments.

3 P300 the Perpendicular.

4 a

the leaft Segment.

the leaft Side.

d2a the Bafe.

7s2e

the Difference of the Sides:

10 ✔aa+pp=e II | dd+2da=ss-25e 2se-ss-dd-2da Suppofe 13 2x=ss-dd

12

Then 14 2se = 2x·
x-da

1425 15e= S

-2da

2e: d

S

17 X ss 18 xx

=aa+pp

2xda + ddaa = ssaa +sspp

18+ Ig ssaa ddaa+2xda xx―sspp

The Area of any Oblique-angled plain Triangle, the Difference of the Sides, and the Difference of the Segments of the Base, being gi ven, thence to find the Bafe, &c.

19 X-4dd 20

{

bba' -2ddba3+d+aa

=4A1+

bba-2ddba+d+aa= 16 AAdd + dda+ 2ddba3 + ddbba2

dda++ d'a2 bba

a-2ba+bba

ddbba 16 AAdd

16 A Add

bb dd

16AADD

24+ dd 25 aa dd + √16 AAdd

aadd√16 ·+÷d4

bb. dd

PROBLEM XV.

There is an Oblique-angled plain Triangle, wherein a Perpendicular is let fall from the Vertical Angle upon the Bafe; the least Side and the Bafe are given; and the Rectangle of the Difference of the Sides into the leaft Side is equal to the Square of the Difference of the Segments of the Bale: 'Tis requir'd to find the Segments of the Bafe, &c.

The Value of a, in this Equation, may be found as in the Examples Page 238, viz. by putting rea, &c. as in thɔ Examples you will find a = 37,55502, &c.

PROBLEM XVI.

The three Chords or Subtenfes of three Arches compleating a Semicircle being each given, thence to find the Diameter of that Circle. That is,

Any Trapezium being inferib'd in a Semicircle, if one of its Sides be the Diameter, and the other three Sides be given, thence to find the Diameter or fourth Side.

U u 2

Let

This Equation being folv'd, as in Example 2, Page 240, you will find a 8,05581, &c.

PBOBLEM XVII.

In any Right-angled Triangle, the Area and the Sum of the Hypothenufe, when added to either Side, being given, thence to find the Sides, &c.

The Value of a, in this Equation, may be found as in the third Example, Page 241; that is, by making r+e=a, &c. ̈ ̈ it will be found that a

60.

PROBLEM XVIII.

There is an Oblique-angled plain Triangle, wherein a Perpendicular is let fall from the Vertical Angle upon the Bafe; the Sum of each Segment of the Bafe, when added to its adjacent or next Side, and the Area of the Triangle, are given, to find the Perpendicular and each Side.

Let

{

|y+b=x=1500

2 /e u=3=1500} Quære y, b, e, and u

3 A the Area = 141750

And 4 a the Perpendicular fought.
Then 5y+ex — a = A