10.

1.633469

18.259637

9.893106

It is almost needless to add, that the arithmetical complement may be taken from the tables by inspection.

EXAMPLE 4. In a right-angled plane triangle, given the hypothenuse 100, and one of the acute angles-49°, required the base, perpendicular and remaining angle. Answer, Base 65.607, per

pendicular 75.47, and remaining angle 41°.

EXAMPLE 5. Given the base of a right-angled plane triangle 70, and one of the acute angles-50°, required the perpendicular, hypothenuse and remaining angle. Answer, perpendicular 83.42, hypothenuse 108.90, and angle 40°.

EXAMPLE 6. In a right-angled plane triangle given the hypothenuse 832 and base 768, required the perpendicular and acute angles. Answer, perpendicular 320, angles 67° 23', and 22° 37'.

With regard to oblique angled triangles, or those which have no right-angle, the solution of them may be reduced to the four following cases:

CASE I.

39. Given any two angles B, A, and a side BC, to find the other sides BA, AC. Solution. Construct the proportion

sin A: BC :: sin B: AC=

B

BC sin B

:: sin C: AB, or:: sin (A+B)

sin A

: AB, (because sin C-sin (A+B).)

If BA had been the side given, we should have had

sin (A+B) BA:: sin B: AC:: sin A: BC.

Examples.

In the oblique angled triangle ABC, given the angle A-88°, the angle B-36o, and the side BC=56; required the other two sides. BC x sin B From the above proportion we have AC=! sin A

EXAMPLE 2. Given the angle A=41° 13′ 22", the angle C= 71° 19′ 5′′, and the side BC=55; required the other two sides.

Answer AB=79.063
AC=77.041

EXAMPLE 3. Given the angle B=78° 57', the angle C-47° 34', and the side BC=184; required the sides AC and ÃB.

Answer, AC=224.7, AB=169.

CASE II.

40. Given two sides, and an angle opposite to one of the given sides, to find the third side and the other two angles, supposing that we know of what species they are.

B

Solution. Suppose that the sides AB, AC are given, and the angle B; we shall first obtain the angle C by this proportion AC: sin B::AB: sin C, which will give the third angle. And we may then obtain the side BC by the proportion sin B: AC:: sin A: BC. But to find the side BC at once, draw

the perpendicular AD on the side BC; call AB, a;* AC, b; sin B, s; cos B, c; and we shall have

B D

R: AB (a):: sin B (s); AD= :: cos B (c): BD=

as

R

R2

ac

R

EXAMPLE. In the oblique angled triangle ABC (see the preceding figure) given AB=50, AC=40, and the angle B=329; required the side BC, and the angles A and C.

log sin C

9.821120

Whence the angle C-41° 29", and angle A-106° 31'.

BC AB cos B+√AC2¬AB2 (sin B), and by substituting the values of AB, AC, sin B, cos B, we shall find BC=72.368, as before.

EXAMPLE 2. In the oblique angled triangle ABC let AB=120 feet, AC=42 feet, and the angle B=57° 27; what are the other angles and the third side?

EXAMPLE 3. In the plane triangle ABC, let BC=345 feet AC=232 feet, and the angle B=37° 20′, what are the other ar gles and the third side, supposing the triangle to be acute-angled, and what, if obtuse-angled?

Answer, the angle A-64° 21' 1", or 115° 35' 59"

C-78 15 59, or 27 41 1 and the side AB=374.559, or 174.073.

N. B. The angle found by this method is ambiguous when th given angle is acute, and at the same time the side opposite the given angle is less than the other given side.

CASE III.

41. Given two sides and the included angle, to find the other two angles and the remaining side.

Solution. Let the angle A and the two sides AB, AC be given, and the angles B and C, and the side BC be required.

First we have,

sin B sin C, or by composition

B

AC AB

sin B+ sin C

AC+AB: AC-AB : : sin B+sin C ; sin B-sin C ::

:

:1

sin B-sin C

AC + AB ; AC—AB:: tang (B+C) : tang— (B—C)

And since B+C=180°—A, we have

1

tang (B+C)=tang (90° —— A)= cot_— A.

2

And consequently since we know B-C and B+C, it is easy to obtain the angles B and C; and we shall then have sin B: AC:: sin A: BC, which will give the third side BC.

This third case may also be solved in the following manner :
Draw the perpendicular BF upon the side

AC, and let AB=a, AC=b, sin A-s, cos
A=c, we shall have

'ac R

ac

B

F

Therefore FC-b- But in the right-angled triangle BFC,

we have

ac

R

FC (b −): BF ($) :: R: tang C =

R

R

aRs

b R-ac

If the angle A is obtuse, C becomes negative, and we have

'+ (b − a )2 } = √ (a2 + b2 _ 2 abc); or √

R

(a2 + b2 + 2_abc), if the angle A is obtuse

R

42. From the proportion

I

AC+CB: AC—CB:: tang (B+C) : tang (B—C)

2

it follows that in any triangle the sum of any two sides is to their difference, as the tangent of half the sum of the two angles opposite these sides, is to the tangent of half the difference of these same angles.

Let then AC-a, AB=d, (see preceding figure) the angle B=B, the angle C=C; we shall have

tang (B+C):tang (B-C)::a+d:a-d::R:

Therefore if represents the tangent of any arc U, we shall d

R tang U-RR

have

R:

tang U+R

or by (22)

R: tang (U—45°) : : tang-1 (B+C): tang (B—C)

2

Hence if we construct this proportion; the less side AB is to the greater AC:: R: tang arc U, and if we subtract 459 from this angle, radius will be to tangent of the remainder, as tang (B+C)

1

: tang (B–C).

2

1 2

EXAMPLE 1. In the oblique angled triangle ABC (preceding figure) given AB=320, AC=562, and the angle A=1288 4; required the side BC.

cotA=tang (B+C.

But AC+AB=882, AC-AB-242, and cot A-tang (

2

=tang 25° 58'.

Therefore tang (B–C) — tang 25° 58′ × 242

802

B-C.
2

Whence the half difference of the angles B and C is 7° 36', and by a well known theorem 25°58′+7° 36′=33° 34' will be the greater angle, and 25° 58—7° 36=18° 22=the less angle.

By the proportion sin B: AC:: sin A: BC, we shall then find the side BC-800.

As the second method is not adapted to logarithmic calculation, we shall not examplify it.

arith.co.log AB (320) = 17.494850

log tang arc U = 10.244586

Whence arc U=60° 20′, and U-45°-15° 20′