Take BAD from BAC,

ACD;

find one of the other sides AD. the remainder, DAC is the angle included between two known sides AD, AC; from which the angles ADC and ACD may be found, by chap. iii. case 2. The angle CAP = 180° (APC + ACD). Also, BCP BCA and PBC ABC + PBA ABC + sup. ADC. Hence, the three required distances are found by these proportions. As sin APC: AC:: sin PAC: PC, and :: sin PCA: PA; and lastly, as sin BPC: BC :: sin BCP : BP. The results of the computation are, PA = 709-33, PC = 1042-66, PB 934 yards.

*** The computation of problems of this kind, however, may be a little shortened by means of the following

Put AC

General Investigation.

a, BC= b, APC = P, BPCP, ACB = C, and let there be taken for unknown quantities PAC = x, PBCy. The triangles PAC and PBC give

sin APC: sin CAP:: AC: CP

and sin BPC: sin CBP :: BC: CP;

This expression being separated into two parts, we have

Hence, a being thus determined, we get y from the equation y = R z, and cr from either of the expressions above given.

Note. It will be a useful exercise for the student, to work out the computation by both these methods. The comparison of the results will serve to give him confidence in the deductions from the analytical investigations.

EXAMPLE XVIII.

It is required to find the distances from Edystone light-house to Plymouth, Start Point, and the Lizard, respectively, from the following data:

}

(Plymouth to Lizard....607

(Plymouth to miles.

Start Point to Plymouth 20

bears from Edystone rock

North.

W.S.W.

E. by N.

Ans. From Edystone to Lizard .....53-04 miles to Plymouth ..14-333 ditto to Start Point.. 17.36 ditto. Remark. The general problem of which the last two examples are particular cases, was originally proposed by Richard Townley, Esq. and solved by Mr. John Collins, in the Philosophical Transactions, N° 69, A. D. 1671. See New Abridgement, vol. i. p. 563.

The six cases there considered are, 1. When the station is out of the triangle made by the objects, but in one of its sides produced. 2. When the station is in one of the sides of the triangle. 3. When the three objects lie in a right line. 4. When the station is not within the triangle formed by the objects. 5. When the station is within that triangle. 6. When, by reason of the relation of the sides and angles, the points c and d (see the preceding diagram) fall so near together as to make the continuation to P of doubtful accuracy. To these, later writers on trigonometry have added another case, viz. 7. When the point c falls between the line AB and the station P.

It will be advantageous to the student, to modify the construction and computation to suit all these cases.

CHAPTER VI.

Spherical Trigonometry.

SECTION I.

Fundamental Principles, General Properties, and

Formula.

1. THREE planes, AOC, AOB, BOC, all of which pass

through the centre o of a sphere, intersect the surface of that sphere in portions of great circles which form a spherical triangle ABC. Thus also is constituted the spherical pyramid or tetraedron which has for its base the triangle ABC, and for its vertex the centre o of the sphere. The angle A

C

a

B

and 14°, and the two equal measured distances 84 feet. Required the height of the object.

Ans. 53.964 feet.

EXAMPLE XVI.

Let the distances be AB = 100 yards, BC = 400 yards, and the angles, at A = 5° 24', at B = = 6° 27', at c8° 36'. What is the height of the object? Ans. 44-46 yards.

=

EXAMPLE XVII.

From a convenient station P, where could be seen three objects A, B, and C, whose distances from each other were known (viz. AB = 800, ac = 600, BC 400 yards), I took the horizontal angles APC = 33° 45′, BPC

22° 30'. It is hence required to determine the respective distances of my station from each object.

Here it will be necessary, as preparatory to the computation, to describe the manner of

C

B

Construction. Draw the given triangle ABC from any convenient scale. From the point A draw a line AD to make with AB an angle equal to 22° 30′ and from в a line BD to make an angle DBA = 33° 45′. Let a circle be described to pass through their intersection D, and through the points A and B. Through c and D draw a right line to meet the circle again in P: so shall P be the point required. For, drawing PA, PB, the angle APD is evidently ABD, since it stands on the same arc AD: and for a like reason BPD BAD. So that P is the point where the angles have the assigned value.

Manner of computation. In the triangle ABC where the sides are known, find the angles. In the triangle ABD, where all the angles are known, and the side AB,

ACD;

find one of the other sides AD. Take BAD from BAC, the remainder, DAC is the angle included between two known sides AD, AC; from which the angles ADC and ACD may be found, by chap. iii. case 2. The angle CAP = 180° — (APC + ACD). Also, BCP = BCA — and PBC = ABC + PBA = ABC + sup. ADC. Hence, the three required distances are found by these proportions. As sin APC: AC:: sin PAC: PC, and :: sin PCA: PA; and lastly, as sin BPC: BC :: sin BCP: BP. The results

of the computation are, PA 709-33, PC PB 934 yards.

=

1042-66,

***The computation of problems of this kind, however, may be a little shortened by means of the following

Put AC

General Investigation.

a, BC= b, APCP, BPC = P', ACB = c, and let there be taken for unknown quantities PAC = x, PBCy. The triangles PAC and PBC give

sin APC: sin CAP:: AC: CP

and sin BPC: sin CBP :: BC: CP;

y= Rx, and consequently,

a sin p' sin x b sin P (sin R cos x cos R sin x) Dividing by sin x there results,

= 0.

COS X

a sin p ́

b sin P (sin R

sin z

COS R) = 0.