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Spherical Trigonometry, for the Use of Colleges and Schools

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Problems

Exercise XV

  1. Exercise XV, problem 1, p. 155

    Find the locus of the vertices of all right-angled spherical triangles having the same hypotenuse; and from the equation obtained, prove that the locus is a circle when the radius of the sphere is infinite.

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  2. Exercise XV, problem 10, p. 155

    The arc of a great circle bisecting the sides $AB$, $AC$ of a spherical triangle cuts $BC$ produced at $Q$: shew that AQ a2 = c - b2 c + b2.

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  3. Exercise XV, problem 11, p. 155

    If $ABCD$ be a spherical quadrilateral, and the opposite sides $AB$, $CD$ when produced meet at $E$, and $AD$, $BC$ meet at $F$, the ratio of the sines of the arcs drawn from $E$ at right angles to the diagonals of the quadrilateral is the same as the ratio of those from $F$.

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  4. Exercise XV, problem 12, p. 155

    If $ABCD$ be a spherical quadrilateral whose sides $AB$, $DC$ are produced to meet at $P$, and $AD$, $BC$ at $Q$, and whose diagonals $AC$, $BD$ intersect at $R$, then AB CD P = AD BC Q = AC BD R.

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  5. Exercise XV, problem 13, p. 155

    If $A'$ be the angle of the chordal triangle which corresponds to the angle $A$ of a spherical triangle, shew that A’ = (S - A) a2.

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  6. Exercise XV, problem 14, p. 155

    If the tangent of the radius of the circle described about a spherical triangle is equal to twice the tangent of the radius of the circle inscribed in the triangle, the triangle is equilateral.

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  7. Exercise XV, problem 15, p. 155

    The arc $AP$ of a circle of the same radius as the sphere is equal to the greater of two sides of a spherical triangle, and the arc $AQ$ taken in the same direction is equal to the less; the sine $PM$ of $AP$ is divided at $E$, so that $\dfrac{EM}{PM} =$ the natural cosine %-----File: 158.png------------------------------------------------ of the angle included by the two sides, and $EZ$ is drawn parallel to the tangent to the circle at $Q$. Shew that the remaining side of the spherical triangle is equal to the arc $QPZ$.

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  8. Exercise XV, problem 16, p. 155

    If through any point $P$ within a spherical triangle $ABC$ great circles be drawn from the angular points $A$, $B$, $C$ to meet the opposite sides at $a$, $b$, $c$ respectively, prove that Pa PAAa + Pb PBBb + Pc PCCc = 1.

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  9. Exercise XV, problem 17, p. 155

    $A$ and $B$ are two places on the Earth’s surface on the same side of the equator, $A$ being further from the equator than $B$. If the bearing of $A$ from $B$ be more nearly due East than it is from any other place in the same latitude as $B$, find the bearing of $B$ from $A$.

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  10. Exercise XV, problem 18, p. 155

    From the result given in example 18 of Chapter V. infer the possibility of a regular dodecahedron.

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  11. Exercise XV, problem 19, p. 155

    $A$ and $B$ are fixed points on the surface of a sphere, and $P$ is any point on the surface. If $a$ and $b$ are given constants, shew that a fixed point $S$ can always be found, in $AB$ or $AB$ produced, such that a AP + b BP = s SP, where $s$ is a constant.

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  12. Exercise XV, problem 2, p. 155

    $AB$ is an arc of a great circle on the surface of a sphere, $C$ its middle point: shew that the locus of the point $P$, such that the angle $APC =$ the angle $BPC$, consists of two great circles at right angles to one another. Explain this when the triangle becomes plane.

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  13. Exercise XV, problem 20, p. 155

    $A$, $B$, $C$,…are fixed points on the surface of a sphere; $a$, $b$, $c$,…are given constants. If $P$ be a point on the surface of the sphere, such that a AP + b BP + c CP + …= constant, shew that the locus of $P$ is a circle.

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  14. Exercise XV, problem 3, p. 155

    On a given arc of a sphere, spherical triangles of equal area are described: shew that the locus of the angular point opposite to the given arc is defined by the equation multline* ^-1 (+ ) + ^-1 (- ) + ^-1 (+ ) + ^-1 (- ) = , multline* where $2\alpha$ is the length of the given arc, $\theta$ the arc of the great circle drawn from any point $P$ in the locus perpendicular to the given arc, $\phi$ the inclination of the great circle on which $\theta$ is %-----File: 156.png------------------------------------------------ measured to the great circle bisecting the given arc at right angles, and $\beta$ a constant.

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  15. Exercise XV, problem 4, p. 155

    In any spherical triangle c = A a + B b a b - A B.

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  16. Exercise XV, problem 5, p. 155

    If $\theta$, $\phi$, $\psi$ denote the distances from the angles $A$, $B$, $C$ respectively of the point of intersection of arcs bisecting the angles of the spherical triangle $ABC,$ shew that (b - c) + (c - a) + (a - b) = 0.

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  17. Exercise XV, problem 6, p. 155

    If $A'$, $B'$, $C'$ be the poles of the sides $BC$, $CA$, $AB$ of a spherical triangle $ABC$, shew that the great circles $AA'$, $BB'$, $CC'$ meet at a point $P$, such that PA BC = PB CA = PC AB.

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  18. Exercise XV, problem 7, p. 155

    If $O$ be the point of intersection of arcs $AD$, $BE$, $CF$ drawn from the angles of a triangle perpendicular to the opposite sides and meeting them at $D$, $E$, $F$ respectively, shew that ADOD, BEOE, CFOF are respectively equal to 1+AB C, 1+BA C, 1+CA B.

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  19. Exercise XV, problem 8, p. 155

    If $p$, $q$, $r$ be the arcs of great circles drawn from the angles of a triangle perpendicular to the opposite sides, $(\alpha, \alpha')$, $(\beta, \beta')$, $(\gamma, \gamma')$ the segments into which these arcs are divided, shew that flalign* &&& ’ = ’ = ’; && [2ex] &and&2$\displaystyle \frac{\cos p}{\cos \alpha \cos \alpha'} = \frac{\cos q}{\cos \beta \cos \beta' } = \frac{\cos r}{\cos \gamma \cos \gamma'}. $ &and & flalign*

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  20. Exercise XV, problem 9, p. 155

    In a spherical triangle if arcs be drawn from the angles to the middle points of the opposite sides, and if $\alpha$, $\alpha'$ be the two parts of the one which bisects the side $a$, shew that ’ = 2 a2.

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