Page images
PDF
EPUB
[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][subsumed][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

CB

AAKC

BC

B

&

and

In this Synopfis

FC is CG.

Thus is the Oblique Angled Triangle refolved into 2 Right-angled ones, in either of which the Hypothenufe is known, and also the Bafe.

fignifies two Lines or Angles added to

gether, and-fignifies Subtraction, or the difference of two

Lines or Angles.

And thus have you the feven Cafes of Right-angled, and five of Oblique-angled plain Triangles fufficiently demonftrated and explained. I fhall next proceed to fome Problems Trigonometrical.

С НА Р. V.

Problems Trigonometrical.

Problem I. One Side of an Oblique angled plain Triangle, the Angle oppofite to that Side, and the Sum of the other two Sides being given; to find the other two Sides, and the Angles feverally.

[blocks in formation]

be given.

[blocks in formation]

Extend the Side BA to D, making AD equal to AC, and draw the Line DC; fo fhall you have two other Oblique angled Triangles BDC and ADC. In the Triangle ABC is given the Angle BAC 110° 30′, and confequently in the Triangle ACD you have given CAD 69° 30', it being the Complement of the other to 180°, (by Def. 16. Chap. 1.) alfo the Triangle ADC is Equicrural, by Construction; and therefore the Angles at the Bafe ADC and ACD are equal (by Def. 19. Chap. 1.) viz. each of them is equal to half the given Angle 55° 15. Now in the Triangle BCD there is given,

1. BC 532

2. BD637

3. The Angle BDC 55° 15′

From whence you may find the Angle DCB (by Cafe 2. Chap. 4.)

[blocks in formation]

From which fubtract the Angle ACD 55° 15', and the Remainder is 45° 04' for the ACB; and the Angle ABC is 24° 26', and is found by fubtracting the Sum of BCD 100° 19' and D 55° 15' from 180°. The Sides AB and AC may be found thus (by Cafe I. Chap. 4.)

[merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small]

By Scale and Compaffes.

9.6166164

2.3709404

For the first Proportion, extend from 532 to 637 in the Line of Numbers, that extent will reach from 55° 15' to 79° 40′ in the Sines; but because the Angle fought is Obtufe (as appears by the Figure) therefore the Complement of 79° 40′ to 180° is 100

19' the Angle required. For the fecond Proportion, extend from 69° 30' (the Complement of 110° 30' to 180°) to 45° 04' in the Sines, that extent will reach** from 532 to 402.08. For the third Proportion, extend from 69° 30' to 24° 26' in the Sines; that extent will reach from 532 to 234.93 in the Line of Numbers.

By the Sliding-Rule.

For the first Proportion, fet 532 on B to 637 on A; then against 55° 15' on S, you will find 79° 40′ on SS. For the fecond Proportion, fet 69° 30′ on SS, to 45° 04' on S, then against 532 on A, is 402.08 on B. For the third Proportion, fet 69° 30' on SS, to 24° 26' on S, then against 532 on A is 234.93 on B.

Problem II. One fide of an oblique angled Plain Triangle, the Angle oppofite thereto, and the Dif ference of the other two Sides being given, to find the other Angles, and the two Sides feverally.

In the Triangle ABC, let BC 250.2, and the Difference of the other two Sides 106, B

and the Angle

106

D

250-2

BAC (opposite to BC) 96° 50' be given.

C

Make AD equal to AC, and draw CD, conftitu ting the Triangle ACD Equicrural; and the Angle DAC being 96° 50', the Complement thereof to 180° is 83° 10' for the two Angles ADC and ACD; which being equal one to the other, therefore each of them is half 83°10', that is 41° 35'; and by drawing the LineCD, there is also another Triangle made, wherein is given the Side BC 250.2, and BD 106 equal to

E

the

50 the Difference of the two Sides AB and AC; and there is alfo given the Angle BDC 138° 25', equal to the Complement of ADC, 41° 35', to 180°. From these things given, you may find the Angles ACB and ABC, and alfo the Sides AB and AC, feverally.

[blocks in formation]

To which if ACD 41° 35' be added, the Sum is 57° 54' 52" for the whole Angle ACB. And if the Angle A be added to it, and the Sum fubtracted from 180, there will remain 25° 15′ 08′′ for the Angle ABC. Find the fides AB and AC as in the last Question, thus.

As the s.<BAC 96° 50' (or Ar. Com. 0.0030960

its Comp. 83°10')

Is to the Side BC 250.2

(or}

So is the s. ACB 57° 54′ 52"

To the Side AB 213.5

Again,

2.3982873

9.9280146

2.3293979

As the s. of the BAC 96° 50' Ar. Com. 0.0030960

Is to the Side BC 250.2

So is the s.< ABC 25° 15′ 8′′

To the Side AC 107.5

By Scale and Compaffes.

2.3982873

9.6300247

2.0314080

For the first Proportion, extend from 250.2 to 106 in the Line of Numbers; that extent will reach from 41° 35' to 16° 20' in the Line of Sines. For the fe= cond Proportion, extend from 83° 10' to 57° 55' in

the

« PreviousContinue »