Spherical Trigonometry, for the Use of Colleges and Schools

Solution of Right-angled Triangles

Excerpts

Solution of Right-angled Triangles
The solution of spherical triangles is the process by which, when the values of a sufficient number of the six elements are given, we calculate the values of the remaining elements.
Solution of Right-angled Triangles
These six formul comprise ten equations; and thus we can solve every case of right-angled triangles.
Solution of Right-angled Triangles
the side opposite the right angle is called the *hypotenuse:*
Solution of Right-angled Triangles
Napier was also the inventor of Logarithms, and the Rules of Circular Parts were first published by him in a work entitled *Mirifici Logarithmorum Canonis Descriptio*……Edinburgh, 1614.
Solution of Right-angled Triangles
We do not give them, because we are convinced that they only create confusion instead of assisting the memory.
Solution of Right-angled Triangles
There are limitations of the data in order to insure a possible triangle.
Solution of Right-angled Triangles
Thus if one triangle exists with the given parts, there will be *in general* two, and only two, triangles with the given parts.
Solution of Right-angled Triangles
From (4) it follows that $\tan a$ has the same sign as $\tan A$.

Equations

Solution of Right-angled Triangles
\sin b = \sin B \sin c
In a right-angled spherical triangle the sine of a side equals the sine of its opposite angle times the sine of the hypotenuse.
Solution of Right-angled Triangles
\sin a = \sin A \sin c
In a right-angled spherical triangle the sine of a side equals the sine of its opposite angle times the sine of the hypotenuse.
Solution of Right-angled Triangles
\cos B = \sin A \cos b
The cosine of an angle equals the sine of the other angle times the cosine of the side opposite the first angle.
Solution of Right-angled Triangles
\cos A = \sin B \cos a
The cosine of an angle equals the sine of the other angle times the cosine of the side opposite the first angle.
Solution of Right-angled Triangles
\tan b = \tan c \cos A
Given hypotenuse c and angle A, the tangent of side b equals tan c times cos A.
Solution of Right-angled Triangles
\cot B = \cos c \tan A
Given hypotenuse c and angle A, the cotangent of angle B equals cos c times tan A.
Solution of Right-angled Triangles
\sin a = \sin c \sin A
Given hypotenuse c and angle A, the sine of side a equals sin c times sin A.
Solution of Right-angled Triangles
\tan c=\dfrac{\tan b}{\cos A}
Given side b and adjacent angle A, the tangent of the hypotenuse equals tan b divided by cos A.
Solution of Right-angled Triangles
\tan a = \tan A \sin b
Given side b and adjacent angle A, the tangent of side a equals tan A times sin b.
Solution of Right-angled Triangles
\cos B = \cos b \sin A
Given side b and adjacent angle A, the cosine of angle B equals cos b times sin A.
Solution of Right-angled Triangles
\cos c = \cos a \cos b
Given two sides a and b, the cosine of the hypotenuse equals cos a times cos b.
Solution of Right-angled Triangles
\cot A = \cot a \sin b
Given two sides a and b, the cotangent of angle A equals cot a times sin b.
Solution of Right-angled Triangles
\cot B = \cot b \sin a
Given two sides a and b, the cotangent of angle B equals cot b times sin a.
Solution of Right-angled Triangles
\cos b=\dfrac{\cos c}{\cos a}
Given hypotenuse c and side a, the cosine of side b equals cos c divided by cos a.
Solution of Right-angled Triangles
\cos B=\dfrac{\tan a}{\tan c}
Given hypotenuse c and side a, the cosine of angle B equals tan a divided by tan c.
Solution of Right-angled Triangles
\sin A=\dfrac{\sin a}{\sin c}
Given hypotenuse c and side a, the sine of angle A equals sin a divided by sin c.
Solution of Right-angled Triangles
\cos c = \cot A \cot B
Given two angles A and B, the cosine of the hypotenuse equals cot A times cot B.
Solution of Right-angled Triangles
\cos a = \frac{\cos A}{\sin B}
Given two angles A and B, the cosine of side a equals cos A divided by sin B.
Solution of Right-angled Triangles
\cos b = \frac{\cos B}{\sin A}
Given two angles A and B, the cosine of side b equals cos B divided by sin A.
Solution of Right-angled Triangles
\sin c = \dfrac{\sin a}{\sin A}
Given side a and its opposite angle A, the sine of the hypotenuse equals sin a divided by sin A.
Solution of Right-angled Triangles
\sin b = \tan a\, \cot A
Given side a and its opposite angle A, the sine of side b equals tan a times cot A.
Solution of Right-angled Triangles
\sin B = \dfrac{\cos A}{\cos a}
Given side a and its opposite angle A, the sine of angle B equals cos A divided by cos a.
Solution of Right-angled Triangles
a_1 + p_1 = a_2 + p_2 = a_5 + p_5 = \dfrac{\pi}{2}
Auxiliary quantities p_1, p_2, p_5 are defined so that each pairs with a_1, a_2, a_5 to make a right angle; this characterises the five allied triangles of Napier's Rules.
Solution of Right-angled Triangles
p_3 = a_3
The auxiliary quantity p_3 is set equal to the element a_3.
Solution of Right-angled Triangles
p_4 = a_4
The auxiliary quantity p_4 is set equal to the element a_4.

Problems

Exercise V

Exercise V, problem 1, p. 053
$\operatorname{Sin}^2\dfrac{c}{2} = \sin^2 \dfrac{a}{2}\, \cos^2 \dfrac{b}{2} + \cos^2 \dfrac{a}{2}\, \sin^2 \dfrac{b}{2}$.
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Exercise V, problem 10, p. 053
$OAA_1$ is a spherical triangle right-angled at $A_1$ and acute-angled at $A$; the arc $A_1A_2$ of a great circle is drawn perpendicular to $OA$, then $A_2A_3$ is drawn perpendicular to $OA_1$, and so on: shew that $A_nA_{n+1}$ vanishes when $n$ becomes infinite; and find the value of $\cos AA_1 \cos A_1A_2 \cos A_2A_3\ldots\ldots$ to infinity.
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Exercise V, problem 11, p. 053
$ABC$ is a right-angled spherical triangle, $A$ not being the right angle: shew that if $A = a$, then $c$ and $b$ are quadrants.
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Exercise V, problem 12, p. 053
If $\delta$ be the length of the arc drawn from $C$ perpendicular to $AB$ in *any* triangle, shew that = cosec c (^2 a + ^2 b - 2 a b c)^12.
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Exercise V, problem 13, p. 053
$ABC$ is a great circle of a sphere; $AA'$, $BB'$, $CC'$, are arcs of great circles drawn at right angles to $ABC$ and reckoned positive %-----File: 055.png------------------------------------------------ when they lie on the same side of it: shew that the condition of $A'$, $B'$, $C'$ lying in a great circle is AA’ BC + BB’ CA + CC’ AB = 0.
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Exercise V, problem 14, p. 053
Perpendiculars are drawn from the angles $A$, $B$, $C$ of any triangle meeting the opposite sides at $D$, $E$, $F$ respectively: shew that BD CE AF = DC EA FB.
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Exercise V, problem 15, p. 053
$Ox$, $Oy$ are two great circles of a sphere at right angles to each other, $P$ is any point in $AB$ another great circle. $OC = p$ is the arc perpendicular to $AB$ from $O$, making the angle $COx = a$ with $Ox$. $PM$, $PN$ are arcs perpendicular to $Ox$, $Oy$ respectively: shew that if $OM = x$ and $ON = y$, a x + a y = p.
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Exercise V, problem 16, p. 053
The position of a point on a sphere, with reference to two great circles at right angles to each other as axes, is determined by the portions $\theta$, $\phi$ of these circles cut off by great circles through the point, and through two points on the axes, each $\dfrac{\pi}{2}$ from their point of intersection: shew that if the three points ($\theta$, $\phi$), ($\theta'$, $\phi'$), ($\theta''$, $\phi''$) lie on the same great circle gather* (’ - ”) + ’ (” - ) + ” (- ’) = 0. gather*
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Exercise V, problem 17, p. 053
If a point on a sphere be referred to two great circles at right angles to each other as axes, by means of the portions of these axes cut off by great circles drawn through the point and two points on the axes each $90^\circ$ from their intersection, shew that the equation to a great circle is + = 1.
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Exercise V, problem 18, p. 053
In a spherical triangle, if $A = \dfrac{\pi}{5}$, $B = \dfrac{\pi}{3}$, and, $C = \dfrac{\pi}{2}$, shew that $a + b + c = \dfrac{\pi}{2}$.
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Exercise V, problem 2, p. 053
$\operatorname{Tan}\tfrac{1}{2}(c + a)\, \tan \tfrac{1}{2}(c - a) = \tan^2 \dfrac{b}{2}$.
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Exercise V, problem 3, p. 053
$\operatorname{Sin}(c - b) = \tan^2 \dfrac{A}{2}\,\sin(c + b)$.
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Exercise V, problem 4, p. 053
$\operatorname{Sin} a\, \tan \tfrac{1}{2}A - \sin b\, \tan \tfrac{1}{2}B = \sin (a - b)$.
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Exercise V, problem 5a, p. 053
&& Sin (c - a) &= b a 12B, && && Sin (c - a) &= b c 12B. &&
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Exercise V, problem 5b, p. 053
&& Sin (c - a) &= b a 12B, && && Sin (c - a) &= b c 12B. &&
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Exercise V, problem 6, p. 053
If $ABC$ be a spherical triangle, right-angled at $C$, and $\cos A = \cos^2 a$, shew that if $A$ be not a right angle $b + c = \tfrac{1}{2}\pi$ or $\dfrac{3}{2}\pi$, according as $b$ and $c$ are both less or both greater than $\dfrac{\pi}{2}$.
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Exercise V, problem 7, p. 053
If $\alpha$, $\beta$ be the arcs drawn from the right angle respectively perpendicular to and bisecting the hypotenuse $c$, shew that ^2 c2 (1 + ^2) = ^2.
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Exercise V, problem 8, p. 053
In a triangle, if $C$ be a right angle and $D$ the middle point of $AB$, shew that 4^2c2 ^2 CD = ^2 a + ^2 b.
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Exercise V, problem 9, p. 053
In a right-angled triangle, if $\delta$ be the length of the arc drawn from $C$ perpendicular to the hypotenuse $AB$, shew that = (^2a + ^2b).
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