the given 4, gives the

opposite the greater of the

two given sides; and, subtracted from it, gives the ‹

opposite the less.

Or,

b~c

-cot

b+c

플

Tan (B~C) =+ Cot & A.

L tan (B~C) = € L (b + c) + 1 (b~c) + L cot

A — 10.

144

L

Then, (90°- A) + 1⁄2 (B ~ C) = opp. greater side; and (90° — ¦ A) — Į (B ~ C) = <opp. less side. If the values of b, c, in this case, should be given in logarithms, instead of the natural numbers, as is sometimes the case in astronomical calculations, the following formulæ will be found more convenient in practice than that given above.

Let € 1 greater side + L less side = L tan ø; in which case arc (9) is always less than 45°.

Then,

Ltan (B~C) = L cota + L tan (45° — p) — 10. Where (90° — A) + § (B~C)

= opp. greater side,

= < opp. less side, as

Or, either of the two 2 may be found by the following formulæ :

V. Given two sides and their included angle, to find the other side.

RULE.

Find the other two 4 by case iv; and then the remaining side by case 1.

Or, the side may be found by the following formula:

c = √ (a~b)2+4ab sin2¿c=± √√ (a+b)a sina‡c+(a~b)*cos2 ‡ c.

But these do not admit of convenient logarithmic expressions.

If the A be isosceles, or have a = b, thè rule will give

2 a sin c

C=

, or LcLa + L sin

r

9.6989700.

VI. Given the three sides, to find either of the angles.

RULE I.

As the longest side, taken as a base : sum of the other two sides :: difference of those sides : difference of the segments of the base, made by a perpendicular from the opposite <.

Then, this difference added to the base, gives

the greater segment, or that next the greater side; and subtracted from it, gives the less.

And as the perpendicular divides the A into two right-angled A', in each of which the hypothenuse and a leg are known, the angles may be found by case II. of right-angled triangles.

tan A=

¤ L ‡s + ¤ L ( ÷ s − a) + L (1⁄2 s − b ) + L (‡s—c)

2

Where s denotes the sum of the three sides a+b+c; andza is always acute.

Or, the latter of these two formulæ may be ex-, pressed in words at length, as follows:

Add together the log. of the sum of the three sides, and the log. of the difference between this sum and the side opposite the sought, and find the complement of their sum, by subtracting the index from 19, and the rest of the figures from 9, as usual.

Then to this complement add the logarithms of the differences between the said sum and each of the other two sides, and the result, divided by 2, will give the tangent of the required angle.

To the formula above given, we may also add the following:

Sin ÷ a = r√ √ (‡ s — 6) × (‡s—c); cos↓a=r√ (₤s (s-a)

L sin, A =

L COSA =

b c

Or,

bc

€ Lb + €LC+L (7 s − b ) + L († 5 — c)

2

The former of which, or that for the sine, must be

used when a differs considerably from 90°, and the A 2 latter when it is near 90°.

It may likewise be observed, that when the A is isosceles, or has b = c, we shall have

ra

COSA

Sin + A=2; cos; A = 26 √(2b+a) × (26 − a)

26

Or, L sin A € Lb + La .301300, and L cos

ठ्ठ

--

=

{ A = { € L And if the

2

+ 1⁄2 {L (2 b + a) + 1 (2 b − a)}

sides

A be equilateral, having each of its

=

a, we shall have area A a2/3; or L area 2 La

.3634993.

We may here also subjoin the two following formulæ :

MISCELLANEOUS EXAMPLES.

1. How many inches does an angle of 1" subtend at the distance of six miles; supposing the object to be perpendicular to the horizon?

Ans. 17 inch nearly.

2. The hypothenuse of a right-angled triangle being 13 feet 10 inches, and the base 9 feet 7 inches, it is required to find the perpendicular.

Ans. 9 feet 11 inches.

3. The hypothenuse of a right-angled triangle bebeing 6395 feet, and one of the acute angles 39°50′38′′, it is required to find the two legs.

Ans. 4097.262 and 4910.035

4. The hypothenuse of a right-angled triangle being 19630040, and one of the legs 19630000, it is required to find the two acute angles.

Ans. 6′ 56′′ and 89° 53′ 3′′

5. If the base of an oblique-angled plane triangle be 60, and the other two sides 30 and 40, what is its perpendicular altitude? Ans. 17.77561

6. If the base of a plane triangle be 60, and the other two sides 30 and 40, what are the lengths of the segments of the base, made by a line bisecting the vertical angle? Ans. 34.28571 and 25.71428

7. Supposing one angle of a plane triangle to be 139° 54′, and the two sides about that angle in the ratio of 5 to 9, it is required to find the other two angles. Ans. 26° 0′ 10′′ and 14° 5′ 49′′ 8. The sides of a plane triangle being 14373, 13172, and 11845, it is required to find the three angles.

Ans. 69° 54′ 10′′, 59° 23′ 19′′, and 50° 42′ 31′′ 9. If the three angles of a plane triangle be 104° 42′ 15′′, 49° 24′ 5′′, and 25° 53′ 40′′, and the side opposite the greatest angle 476.75 yards, what are the other two sides?

Ans. 215.2533 and 374.2469