Spherical Trigonometry, for the Use of Colleges and Schools

Numerical Solution of Spherical Triangles

Excerpts

Numerical Solution of Spherical Triangles
We shall give in this Chapter examples of the numerical solution of Spherical Triangles.
Numerical Solution of Spherical Triangles
We shall first take right-angled triangles, and then oblique-angled triangles.
Numerical Solution of Spherical Triangles
Here $\tan c$ is *negative*; and therefore $\tan b$ will be negative and $b$ greater than a quadrant.
Numerical Solution of Spherical Triangles
The numerical value of $\cos c$ is the same as that of $\cos 81^\circ\, 45'\, 36''$.
Numerical Solution of Spherical Triangles
Here $\cos c$ *is negative*; and therefore $\cot B$ will be negative, and $B$ greater than a right angle.
Numerical Solution of Spherical Triangles
since $\sin C$ is greater than $\sin A$ we shall obtain two values for $c$ both greater than $a$, and we shall not know which is the value to be taken.
Numerical Solution of Spherical Triangles
Thus by taking only the nearest number of seconds in the tables the two methods give values of $c$ which differ by $1''$; if, however, we estimate fractions of a second both methods will agree in giving about $43\tfrac12$ as the number of seconds.
Numerical Solution of Spherical Triangles
The student can obtain more examples, which can be easily verified, from those here worked out, by interchanging the given and required quantities, or by making use of the polar triangle.

Equations

This equation is in Solution of Right-angled Triangles (Solution of Right-angled Triangles)
This equation is in Solution of Right-angled Triangles (Solution of Right-angled Triangles)
This equation is in Solution of Right-angled Triangles (Solution of Right-angled Triangles)
This equation is in Solution of Right-angled Triangles (Solution of Right-angled Triangles)
This equation is in Solution of Right-angled Triangles (Solution of Right-angled Triangles)
This equation is in Solution of Right-angled Triangles (Solution of Right-angled Triangles)
Numerical Solution of Spherical Triangles
\sin c=\dfrac{\sin a}{\sin A}
In a right spherical triangle, the sine of the hypotenuse c equals the sine of side a divided by the sine of angle A; this gives two values of c, one acute and one obtuse.
Numerical Solution of Spherical Triangles
\sin b = \tan a \cot A
In a right spherical triangle, the sine of side b equals the tangent of side a times the cotangent of angle A.
This equation is in Solution of Right-angled Triangles (Solution of Right-angled Triangles)
Numerical Solution of Spherical Triangles
\tan\tfrac12 A = \Surd{\left\{\frac{\sin(s - b)\sin(s - c)}{\sin s \sin(s - a)}\right\}}
For an oblique spherical triangle, the tangent of half angle A equals the square root of sin(s−b) sin(s−c) divided by sin s sin(s−a), where s is the semi-perimeter.
Numerical Solution of Spherical Triangles
\tan\dfrac12 (A - B) = \dfrac{\sin\tfrac12 (a - b)}{\sin\tfrac12 (a + b)}\cot\tfrac12 C
For an oblique spherical triangle, the tangent of half the difference of angles A and B equals the ratio of sines of half the difference and half the sum of the sides a and b, times the cotangent of half of C.
Numerical Solution of Spherical Triangles
\tan\tfrac12 (A + B) = \dfrac{\cos\tfrac12 (a - b)}{\cos\tfrac12 (a + b)}\cot\tfrac12 C
For an oblique spherical triangle, the tangent of half the sum of angles A and B equals the ratio of cosines of half the difference and half the sum of sides a and b, times the cotangent of half of C.
Numerical Solution of Spherical Triangles
\sin c = \dfrac{\sin a \sin C}{\sin A}
In an oblique spherical triangle, the sine of side c equals the sine of a times the sine of C divided by the sine of A; when sin C exceeds sin A the formula gives two values, which is the ambiguous case.
Numerical Solution of Spherical Triangles
\cos \tfrac{1}{2} c = \dfrac{\cos \tfrac{1}{2} (a + b) \sin \tfrac{1}{2}C}{\cos\tfrac{1}{2}(A+B)}
In an oblique spherical triangle, the cosine of half of side c equals the cosine of half the sum of a and b times the sine of half of C, divided by the cosine of half the sum of A and B; this determines c without ambiguity.
Numerical Solution of Spherical Triangles
\tan \theta = \tan b \cos C
The auxiliary angle theta is defined by the tangent of theta equals the tangent of b times the cosine of C; a negative cos C makes theta greater than a right angle.
Numerical Solution of Spherical Triangles
\cos c = \dfrac{\cos b \cos (a - \theta)}{\cos \theta}
In an oblique spherical triangle, the cosine of side c equals the cosine of b times the cosine of a minus theta, divided by the cosine of theta.
Numerical Solution of Spherical Triangles
\sin B = \dfrac{\sin b}{\sin a}\sin A
In an oblique spherical triangle, the sine of angle B equals the sine of b over the sine of a, times the sine of A.
Numerical Solution of Spherical Triangles
\tan \tfrac12 C = \dfrac{\cos \tfrac12 (b - a)}{\cos \tfrac12 (b + a)}\cot \tfrac12 (B + A)
Napier's analogy: the tangent of half of C equals the ratio of cosines of half the difference and half the sum of sides b and a, times the cotangent of half the sum of B and A.
Numerical Solution of Spherical Triangles
\tan \tfrac12 c = \dfrac{\cos \tfrac12 (B + A)}{\cos \tfrac12 (B - A)}\tan \tfrac12 (b + a)
Napier's analogy: the tangent of half of side c equals the ratio of cosines of half the sum and half the difference of angles B and A, times the tangent of half the sum of b and a.

Problems

Exercise XVI

Exercise XVI, problem 1, p. 168
Given $b$ & = 137^ 3’ 48”, A = 147^ 2’ 54”, C = 90^.
Printed answer:
- *Results.* c & = 47^ 57’ 15”, a = 156^ 10’ 34”, B = 113^ 28’.
unverified: no computed check settled this one (yet)
How it was checked
- other: the record may be misread
Exercise XVI, problem 2, p. 168
Given $c$ & = 61^ 4’ 56”, a = 40^ 31’ 20”, C = 90^.
Printed answer:
- *Results.* b & = 50^ 30’ 29”, B = 61^ 50’ 28”, A = 47^ 54’ 21”.
unverified: no computed check settled this one (yet)
How it was checked
- other: the record may be misread
Exercise XVI, problem 3, p. 168
Given $A$ & = 36^, B = 60^, C = 90^.
Printed answer:
- *Results.* a & = 20^ 54’ 18”.5, b = 31^ 43’ 3”, c = 37^ 21’ 38”.5.
unverified: no computed check settled this one (yet)
How it was checked
- other: the record may be misread
Exercise XVI, problem 4, p. 168
Given $a$ & = 59^ 28’ 27”, A = 66^ 7’ 20”, C = 90^.
Printed answer:
- *Results.* c & = 70^ 23’ 42”, b = 48^ 39’ 16”, B = 52^ 50’ 20”, && or,*Results.* c & = 109^ 36’ 18”, b = 131^ 20’ 44”, B = 127^ 9’ 40”.
unverified: no computed check settled this one (yet)
How it was checked
- other: the record may be misread
Exercise XVI, problem 5, p. 168
Given $c$ & = 90^, a = 138^ 4’, b = 109^ 41’.
Printed answer:
- *Results.* C & = 113^ 28’ 2”, A = 142^ 11’ 38”, B = 120^ 15’ 57”.
unverified: no computed check settled this one (yet)
How it was checked
- other: the record may be misread
Exercise XVI, problem 6, p. 168
Given $c$ & = 90^, A = 131^ 30’, B = 120^ 32’.
Printed answer:
- *Results.* C & = 109^ 40’ 20”, a = 127^ 17’ 51”, b = 113^ 49’ 31”.
unverified: no computed check settled this one (yet)
How it was checked
- other: the record may be misread
Exercise XVI, problem 7, p. 168
Given $a$ & = 76^ 35’ 36”, b = 50^ 10’ 30”, c = 40^ 0’ 10”.
Printed answer:
- *Results.* A & = 121^ 36’ 20”, B = 42^ 15’ 13”, C = 34^ 15’ 3”.
unverified: no computed check settled this one (yet)
How it was checked
- other: the record may be misread
Exercise XVI, problem 8, p. 168
Given $A$ &= 129^ 5’ 28”, B= 142^ 12’ 42”, C= 105^ 8’ 10”.
Printed answer:
- *Results.* a & = 135^ 49’ 20”, b = 144^ 37’ 15”, c = 60^ 4’ 54”.
unverified: no computed check settled this one (yet)
How it was checked
- other: the record may be misread