Spherical Trigonometry, for the Use of Colleges and Schools

Solution of Oblique-Angled Triangles

Excerpts

Solution of Oblique-Angled Triangles
The solution of oblique-angled triangles may be made in some cases to depend immediately on the solution of right-angled triangles; we will indicate these cases before considering the subject generally.
Solution of Oblique-Angled Triangles
these determine $\tfrac{1}{2}(A + B)$ and $\tfrac{1}{2}(A - B)$, and thence $A$ and $B$.
Solution of Oblique-Angled Triangles
Thus, in the present case, there is no real ambiguity, and the triangle is always possible.
Solution of Oblique-Angled Triangles
In this case, since $B$ is found from its sine, there will sometimes be two solutions; and sometimes there will be no solution at all, namely, when the value found for $\sin B$ is greater than unity.
Solution of Oblique-Angled Triangles
Hence, when $a = b$, there will be no solution at all, unless $A$ and $a$ are of the same affection, and then there will be only one solution; except when $A$ and $a$ are both right angles, and then $\cot \tfrac{1}{2}C$ and $\tan\tfrac{1}{2}c$ are indeterminate, and there is an infinite number of solutions.
Solution of Oblique-Angled Triangles
If $\sin b \sin A$ be greater than $\sin a$, there is no triangle which satisfies the given conditions;

Equations

This equation is in Relations between the Trigonometrical Functions of the Sides and the Angles of a Spherical Triangle (Relations between the Trigonometrical Functions of the Sides and the Angles of a Spherical Triangle)
Solution of Oblique-Angled Triangles
\cos a = \dfrac{\cos A + \cos B \cos C}{\sin B \sin C}
Gives the cosine of a side of a spherical triangle from the three angles.
Solution of Oblique-Angled Triangles
\tan \tfrac{1}{2}(A + B) = \dfrac{\cos\tfrac{1}{2}(a - b)}{\cos\tfrac{1}{2}(a + b)}\cot\tfrac{1}{2}C
First of Napier's analogies: half the sum of two angles from two sides and the included angle.
Solution of Oblique-Angled Triangles
\tan \tfrac{1}{2}(A - B) = \dfrac{\sin\tfrac{1}{2}(a - b)}{\sin\tfrac{1}{2}(a + b)}\cot\tfrac{1}{2}C
Second of Napier's analogies: half the difference of two angles from two sides and the included angle.
Solution of Oblique-Angled Triangles
\sin c = \dfrac{\sin a\, \sin C}{\sin A}
Sine rule: the sine of side c is proportional to the sine of its opposite angle C.
Solution of Oblique-Angled Triangles
\cos c = \cos a\, \cos b + \sin a\, \sin b\, \cos C
Cosine rule for a side of a spherical triangle, from two sides and the included angle; free from ambiguity.
Solution of Oblique-Angled Triangles
\cos c = \cos b\, (\cos a + \sin a \tan b \cos C)
The cosine rule for c rearranged into a form suited to logarithms.
Solution of Oblique-Angled Triangles
\tan \theta = \tan b\, \cos C
Auxiliary angle theta defined by tan theta equal to tan b times cos C.
Solution of Oblique-Angled Triangles
\cos c = \cos b\, (\cos a + \sin a\, \tan \theta) = \dfrac{\cos b\, \cos (a - \theta)}{\cos \theta}
With the auxiliary angle theta, the cosine of side c is written as a single quotient adapted to logarithms.
Solution of Oblique-Angled Triangles
\tan CD = \tan b \cos C
In the right-angled decomposition, the tangent of the segment CD equals tan b times cos C.
Solution of Oblique-Angled Triangles
\tan AD = \tan C \sin CD
Right-angled relation giving the tangent of the perpendicular AD from the angle C and the segment CD.
Solution of Oblique-Angled Triangles
\tan ABD \sin DB = \tan C \sin \theta
Relation that finds angle B independently of A, from the right-angled decomposition.
Solution of Oblique-Angled Triangles
\tan \tfrac{1}{2} (a + b) = \dfrac{\cos \tfrac{1}{2} (A - B)}{\cos \tfrac{1}{2} (A + B)} \tan \tfrac{1}{2} c
Napier's analogy giving half the sum of two sides from two angles and the included side.
Solution of Oblique-Angled Triangles
\tan \tfrac{1}{2} (a - b) = \dfrac{\sin \tfrac{1}{2} (A - B)}{\sin \tfrac{1}{2} (A + B)} \tan \tfrac{1}{2} c
Napier's analogy giving half the difference of two sides from two angles and the included side.
Solution of Oblique-Angled Triangles
\sin C = \dfrac{\sin A \sin c}{\sin a}
Sine rule giving the angle C from its sine, with the usual ambiguity.
Solution of Oblique-Angled Triangles
\cos C = -\cos A \cos B + \sin A \sin B \cos c
Cosine rule for an angle of a spherical triangle from two angles and the included side; free from ambiguity.
Solution of Oblique-Angled Triangles
\cot \phi = \tan B\, \cos c
Auxiliary angle phi defined by cot phi equal to tan B times cos c.
Solution of Oblique-Angled Triangles
\cos C = \cos B (-\cos A + \cot \phi \sin A) = \dfrac{\cos B \sin (A-\phi)}{\sin \phi}
The cosine rule for C written with the auxiliary angle phi, adapted to logarithms.
Solution of Oblique-Angled Triangles
\cos c = \cot B \cot DAB
Right-angled relation giving cos c from the angle B and the angle DAB.
Solution of Oblique-Angled Triangles
\cos AD \sin CAD = \cos C
Right-angled relation between the perpendicular AD, the angle CAD and the angle C.
Solution of Oblique-Angled Triangles
\cos AD \sin BAD = \cos B
Right-angled relation between the perpendicular AD, the angle BAD and the angle B.
Solution of Oblique-Angled Triangles
\dfrac{\cos C}{\sin CAD} = \dfrac{\cos B}{\sin BAD}
Ratio relation from which the angle C is found in the perpendicular decomposition.
Solution of Oblique-Angled Triangles
\tan b \cos CAD = \tan c \cos \phi
Relation that finds side b independently of a, in the perpendicular decomposition.
Solution of Oblique-Angled Triangles
\sin B = \frac{\sin b}{\sin a} \sin A
Sine rule giving angle B from its sine; it may take two values, so the solution is ambiguous.
Solution of Oblique-Angled Triangles
\tan \tfrac{1}{2} C = \dfrac{\cos \tfrac{1}{2} (a - b)}{\cos \tfrac{1}{2} (a + b)} \cot \tfrac{1}{2} (A + B)
Napier's analogy giving half of angle C from two sides, one opposite angle, and the sum of the two angles.
Solution of Oblique-Angled Triangles
\tan \tfrac{1}{2} c = \dfrac{\cos \tfrac{1}{2} (A + B)}{\cos \tfrac{1}{2} (A - B)} \tan \tfrac{1}{2} (a + b)
Napier's analogy giving half of side c from two sides, one opposite angle, and the sum of the two angles.
Solution of Oblique-Angled Triangles
\cot a\, \sin b = \cos b\, \cos C + \sin C\, \cot A
Four-part (cotangent) relation linking two sides, the included angle C and the angle A.
Solution of Oblique-Angled Triangles
\cos (C - \phi) = \cos \phi \cot a \tan b
Equation determining C minus phi, from which C is found; it may have two admissible values.
This equation is in Relations between the Trigonometrical Functions of the Sides and the Angles of a Spherical Triangle (Relations between the Trigonometrical Functions of the Sides and the Angles of a Spherical Triangle)
Solution of Oblique-Angled Triangles
\tan\theta = \tan b \cos A
Auxiliary angle theta defined by tan theta equal to tan b times cos A.
Solution of Oblique-Angled Triangles
\cos(c - \theta) = \dfrac{\cos a \cos\theta}{\cos b}
Equation determining c minus theta, from which side c is found; may be ambiguous.
Solution of Oblique-Angled Triangles
\cos b = \cot A \cot ACD
Right-angled relation giving cos b from the angle A and the angle ACD.
Solution of Oblique-Angled Triangles
\sin b = \dfrac{\sin B \sin a}{\sin A}
Sine rule giving side b from its sine; the solution is ambiguous.
Solution of Oblique-Angled Triangles
\cos A= -\cos B\cos C + \sin B\sin C\cos a
Cosine rule for angle A from two angles and the included side a.
Solution of Oblique-Angled Triangles
\sin(C - \phi)=\dfrac{\cos A \sin \phi}{\cos B}
Equation determining C minus phi from its sine; may be ambiguous.
Solution of Oblique-Angled Triangles
\cot A \sin B = \cot a \sin c - \cos c \cos B
Relation between the cotangent of A, the sine of B and sides a and c, used to find side c.
Solution of Oblique-Angled Triangles
\cot \theta = \dfrac{\cot a}{\cos B}
Auxiliary angle theta defined by cot theta equal to cot a divided by cos B.
Solution of Oblique-Angled Triangles
\sin (c - \theta) = \cot A \tan B \sin \theta
Equation determining c minus theta from its sine; may be ambiguous.
Solution of Oblique-Angled Triangles
\cot \tfrac{1}{2}C = \tan A \cos a
In the special case a = b (so A = B), half of cot C is tan A times cos a.
Solution of Oblique-Angled Triangles
\tan \tfrac{1}{2}c = \tan a \cos A
In the special case a = b (so A = B), half of tan c is tan a times cos A.
Solution of Oblique-Angled Triangles
\sin B = \dfrac{\sin b \sin A}{\sin a}
Sine rule giving the two values of B, beta and beta', when two sides and the angle opposite one are given.
Solution of Oblique-Angled Triangles
\beta' = \pi - \beta
The two values of B from the sine equation are supplementary: beta' is pi minus beta.

Problems

Exercise VI

Exercise VI, problem 1, p. 068
The sides of a triangle are $105^\circ$, $90^\circ$, and $75^\circ$ respectively: find the sines of all the angles.
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Exercise VI, problem 10, p. 068
If $c_1$, $c_2$ be the two values of the third side when $A$, $a$, $b$ are given and the triangle is ambiguous, shew that c_12 c_22 = 12 (b - a) 12 (b + a).
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Exercise VI, problem 2a, p. 068
Shew that $\tan \tfrac{1}{2} A \tan \tfrac{1}{2} B= \dfrac{\sin(s-c)}{\sin s}$. Solve a triangle when a side, an adjacent angle, and the sum of the other two sides are given.
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Exercise VI, problem 2b, p. 068
Shew that $\tan \tfrac{1}{2} A \tan \tfrac{1}{2} B= \dfrac{\sin(s-c)}{\sin s}$. Solve a triangle when a side, an adjacent angle, and the sum of the other two sides are given.
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Exercise VI, problem 3, p. 068
Solve a triangle having given a side, an adjacent angle, and the sum of the other two angles.
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Exercise VI, problem 4, p. 068
A triangle has the sum of two sides equal to a semicircumference: find the arc joining the vertex with the middle of the base.
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Exercise VI, problem 5a, p. 068
If $a$, $b$, $c$ are known, $c$ being a *quadrant*, determine the angles: shew also that if $\delta$ be the perpendicular on $c$ from the opposite angle, $\cos^2 \delta = \cos^2 a + \cos^2 b$.
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Exercise VI, problem 5b, p. 068
If $a$, $b$, $c$ are known, $c$ being a *quadrant*, determine the angles: shew also that if $\delta$ be the perpendicular on $c$ from the opposite angle, $\cos^2 \delta = \cos^2 a + \cos^2 b$.
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Exercise VI, problem 6, p. 068
If one side of a spherical triangle be divided into four equal parts, and $\theta_1$, $\theta_2$, $\theta_3$, $\theta_4$, be the angles subtended at the opposite angle by the parts taken in order, shew that (_1 + _2) _2 _4 = (_3 + _4) _1 _3.
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Exercise VI, problem 7, p. 068
In a spherical triangle if $A = B = 2C$, shew that 8 (a + c2) ^2 c2 c2 = ^3 a.
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Exercise VI, problem 8, p. 068
In a spherical triangle if $A = B = 2C$, shew that 8 ^2 C2 (s + C2) c2a = 1.
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Exercise VI, problem 9, p. 068
If the equal sides of an isosceles triangle $ABC$ be bisected by an arc $DE$, and $BC$ be the base, shew that DE2 = 12 BC2 AC2.
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