SOLUTION OF TRIANGLES

Excerpts

SOLUTION OF TRIANGLES
A triangle is said to be solved when the sides and angles are calculated from the data.
SOLUTION OF TRIANGLES
Then if the side $AB$ be taken as radius, the ratio of $BC$ to $AB$ is the sine of $A$; and of $AC$ to $AB$ is its cosine;
SOLUTION OF TRIANGLES
or *the sides are proportional to the sines of the opposite angles*. This is true of all triangles.
SOLUTION OF TRIANGLES
There will always be this ambiguity in determining the angle of a triangle from its sine, unless there be something to point out whether the angle is acute or obtuse: and therefore it is generally inconvenient to use a method involving the determination of the angle from its sine.
SOLUTION OF TRIANGLES
These relations give the simplest logarithmic solution of a triangle, when two sides and the contained angle are given
SOLUTION OF TRIANGLES
The formulæ used here were discovered by William Purser of Dublin, in 1632.
SOLUTION OF TRIANGLES
There is also a simple verification of the process afforded by taking the sum of the three angles, which ought to be $180°$

Equations

SOLUTION OF TRIANGLES
a &= c \sin A
In a right triangle with hypotenuse c, the side a opposite A equals c times the sine of A.
SOLUTION OF TRIANGLES
b &= c \cos A
In a right triangle, the side b adjacent to A equals the hypotenuse c times the cosine of A.
SOLUTION OF TRIANGLES
b &= c \sin B
In a right triangle, the side b opposite B equals the hypotenuse c times the sine of B.
SOLUTION OF TRIANGLES
a &= c \cos B
In a right triangle, the side a adjacent to B equals the hypotenuse c times the cosine of B.
SOLUTION OF TRIANGLES
a &= b \tan A
In a right triangle, the side a opposite A equals the side b adjacent to A times the tangent of A.
SOLUTION OF TRIANGLES
c &= b \sec A
In a right triangle, the hypotenuse c equals the side b adjacent to A times the secant of A.
SOLUTION OF TRIANGLES
b &= a \tan B
In a right triangle, the side b opposite B equals the side a adjacent to B times the tangent of B.
SOLUTION OF TRIANGLES
c &= a \sec B
In a right triangle, the hypotenuse c equals the side a adjacent to B times the secant of B.
SOLUTION OF TRIANGLES
\log a &= \log c + \tab\log \sin A - 10
Common logarithm of side a equals the common logarithm of c plus the common logarithm of sin A, less 10, used to solve a right triangle given c and A.
SOLUTION OF TRIANGLES
\log b &= \log c + \tab\log \cos A - 10
Common logarithm of side b equals the common logarithm of c plus the common logarithm of cos A, less 10, used to solve a right triangle given c and A.
SOLUTION OF TRIANGLES
\log b &= \log a + \tab\log \tan B - 10
Common logarithm of side b equals the common logarithm of a plus the common logarithm of tan B, less 10, used to solve a right triangle given a and B.
SOLUTION OF TRIANGLES
\log c &= \log a + \tab\log \sec B - 10
Common logarithm of the hypotenuse c equals the common logarithm of a plus the common logarithm of sec B, less 10, used to solve a right triangle given a and B.
SOLUTION OF TRIANGLES
\tab\log \sin A &= 10 + \log a - \log c
Common logarithm of sin A equals 10 plus log a minus log c, used to find A when c and a are given.
SOLUTION OF TRIANGLES
B = 90° - A
In a right triangle with C = 90°, the two acute angles A and B are complementary.
SOLUTION OF TRIANGLES
AD = AB \sin B = c \sin B
The altitude AD from A to line BC equals AB times the sine of B, which equals c sin B.
SOLUTION OF TRIANGLES
AD = CA \sin C = b \sin C
The same altitude AD equals CA times the sine of C, which equals b sin C.
SOLUTION OF TRIANGLES
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
In any triangle, each side is proportional to the sine of the opposite angle, with the same constant for all three sides.
SOLUTION OF TRIANGLES
\Tab\log\sin B = (\tab\log\sin C - \log c) + \log b
The common logarithm of sin B equals the common logarithm of sin C minus log c plus log b, a logarithmic equation determining B from b, c and C.
SOLUTION OF TRIANGLES
A = 180° - (B + C)
The third angle of a triangle is 180 degrees minus the sum of the other two.
SOLUTION OF TRIANGLES
\log a = \tab\log\sin A - (\tab\log\sin C - \log c)
The common logarithm of side a equals log sin A minus (log sin C minus log c), giving the third side once the angles are known.
SOLUTION OF TRIANGLES
\tab\log\sin B_1 = \log b + (\tab\log\sin C - \log c)
In the ambiguous case, the acute angle B_1 is determined by its common logarithm of sine, which equals log b plus (log sin C minus log c).
SOLUTION OF TRIANGLES
B_2 = 180° - B_1
The second possible value of B in the ambiguous case is the supplement of the acute value B_1. The source text prints 'B_2 = 180° - B' in one place, which reads as a typo for B_1; flagged, not silently corrected in the record.
SOLUTION OF TRIANGLES
A_1 = 180° - (B_1 + C)
With the acute value B_1, the third angle A_1 is 180 degrees minus B_1 and C.
SOLUTION OF TRIANGLES
A_2 = 180° - (B_2 + C)
With the obtuse value B_2, the third angle A_2 is 180 degrees minus B_2 and C.
SOLUTION OF TRIANGLES
\log a_1 = \tab\log\sin A_1 - (\tab\log\sin C - \log c)
Common logarithm of the third side a_1 for the first solution of the ambiguous case.
SOLUTION OF TRIANGLES
\log a_2 = \tab\log\sin A_2 - (\tab\log\sin C - \log c)
Common logarithm of the third side a_2 for the second solution of the ambiguous case.
SOLUTION OF TRIANGLES
DCB = A + B = 2BED
In the construction with C as centre, the angle DCB equals A + B and is twice the angle BED.
SOLUTION OF TRIANGLES
A = ECF + CFA = ECF + B
Angle A decomposes into ECF plus CFA, and CFA equals B, so A equals ECF plus B.
SOLUTION OF TRIANGLES
ECF = A - B = 2FBE
The angle ECF equals A minus B, and is twice the angle FBE.
SOLUTION OF TRIANGLES
BED = \dfrac{1}{2}(A + B)
The angle BED is half the sum of A and B.
SOLUTION OF TRIANGLES
FBE = \dfrac{1}{2}(A - B)
The angle FBE is half the difference of A and B.
SOLUTION OF TRIANGLES
\frac{a+b}{\cos\dfrac{1}{2}(A - B)} = \frac{c}{\cos\dfrac{1}{2}(A + B)}
The sum of two sides a+b over the cosine of half the difference of the opposite angles equals c over the cosine of half their sum; the relation used for the two-sides-and-included-angle solution.
SOLUTION OF TRIANGLES
\frac{a-b}{\sin\dfrac{1}{2}(A - B)} = \frac{c}{\sin\dfrac{1}{2}(A + B)}
The difference a-b over the sine of half the difference of the opposite angles equals c over the sine of half their sum.
SOLUTION OF TRIANGLES
\frac{\tan\dfrac{1}{2}(A - B)}{\tan\dfrac{1}{2}(A + B)} = \frac{a - b}{a + b}
The ratio of the tangents of half the difference and half the sum of two angles equals (a-b)/(a+b); the tangent formula for two sides and the included angle.
SOLUTION OF TRIANGLES
A + B = 180° - C
The two angles A and B together make 180 degrees minus C.
SOLUTION OF TRIANGLES
\tab\log\tan \frac{1}{2} (A - B) = \tab\log\tan \frac{1}{2} (A + B) \\ + \log (a - b) - \log (a + b)
The common logarithm of tan of half the difference of A and B equals the common logarithm of tan of half their sum, plus log(a-b) minus log(a+b); determines the acute angle (A-B)/2.
SOLUTION OF TRIANGLES
A &= \frac{1}{2} (A + B) + \frac{1}{2} (A - B)
Angle A is half the sum of A and B plus half their difference.
SOLUTION OF TRIANGLES
B &= \frac{1}{2} (A + B) - \frac{1}{2} (A - B)
Angle B is half the sum of A and B minus half their difference.
SOLUTION OF TRIANGLES
\log c = \tab\log\sin C - (\tab\log\sin A - \log a)
The common logarithm of c equals log sin C minus (log sin A minus log a), an alternative route to the third side.
SOLUTION OF TRIANGLES
\log c = \tab\log\cos \frac{1}{2} (A + B) - \tab\log\cos \frac{1}{2} (A - B) + \log (a + b)
The common logarithm of the third side c equals log cos of half the sum minus log cos of half the difference, plus log(a+b).
SOLUTION OF TRIANGLES
\log c = \tab\log\sin \frac{1}{2} (A + B) - \tab\log\sin \frac{1}{2} (A - B) + \log (a - b)
The common logarithm of the third side c equals log sin of half the sum minus log sin of half the difference, plus log(a-b).
This equation is in OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE (OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE)
SOLUTION OF TRIANGLES
a + b + c = 2s
The sum of the three sides is twice s, which is defined as half the perimeter.
SOLUTION OF TRIANGLES
r = (s - a) \tan \frac{1}{2} A
The inscribed-circle radius r equals (s-a) times the tangent of half A; the same holds with B and C in place of A.
SOLUTION OF TRIANGLES
r^2 = \frac{(s - a)(s - b)(s - c)}{s}
The square of the inscribed-circle radius equals (s-a)(s-b)(s-c) divided by s.
SOLUTION OF TRIANGLES
\log r = \frac{1}{2} \bigl\{ \log (s - a) + \log (s - b) + \log (s - c) - \log s \bigr\}
The common logarithm of r is half of log(s-a)+log(s-b)+log(s-c)-log s, the logarithmic form for solving a triangle from its three sides.
SOLUTION OF TRIANGLES
\tab\log\tan \frac{1}{2} A &= 10 + \log r - \log (s - a)
The common logarithm of tan of half A equals 10 plus log r minus log(s-a), giving angle A from the three sides.
SOLUTION OF TRIANGLES
\tab\log\tan \frac{1}{2} B &= 10 + \log r - \log (s - b)
The common logarithm of tan of half B equals 10 plus log r minus log(s-b), giving angle B from the three sides.
SOLUTION OF TRIANGLES
\tab\log\tan \frac{1}{2} C &= 10 + \log r - \log (s - c)
The common logarithm of tan of half C equals 10 plus log r minus log(s-c), giving angle C from the three sides.

Problems

No exercises in this chapter.