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Chapter 5: Problem 218

Let \(\triangle A B C\) be a triangle on the unit sphere with a right angle at \(A\). Let \(A^{\prime}\) lie on the arc \(B A\) produced, and \(C^{\prime}\) lie on the arc \(B C\) produced so that \(\triangle A^{\prime} B C^{\prime}\) is right angled at \(A^{\prime} .\) With the usual labelling (so that \(x\) denotes the length of the side of a triangle opposite vertex \(X\), with arc \(A C=b,\) arc \(B C=a,\) arc \(B C^{\prime}=a^{\prime},\) and \(\operatorname{arc} A^{\prime} C^{\prime}=b^{\prime},\) prove that: $$\frac{\sin b}{\sin a}=\frac{\sin b^{\prime}}{\sin a^{\prime}}$$

Short Answer

Expert verified
Using the spherical law of sines, both triangles yield the same ratio, hence \(\frac{\sin b}{\sin a} = \frac{\sin b'}{\sin a'}\). This is because they reduce to the same angle \(\sin B\).

Step by step solution

01

Understand the Given Problem

We are given a spherical triangle \(\triangle ABC\) on a unit sphere with a right angle at \(A\). Points \(A'\) and \(C'\) lie on the extended arcs \(BA\) and \(BC\) respectively, forming another triangle \(\triangle A'BC'\) which is also right-angled at \(A'\). We need to prove that \(\frac{\sin b}{\sin a}=\frac{\sin b'}{\sin a'}\).
02

Utilize Spherical Law of Sines

The Spherical Law of Sines states that for any spherical triangle, \(\frac{\sin a}{\sin A} = \frac{\sin b}{\sin B} = \frac{\sin c}{\sin C}\), where \(a, b, c\) are sides, and \(A, B, C\) are the corresponding opposite angles. We will apply this to both triangles \(\triangle ABC\) and \(\triangle A'BC'\).
03

Apply to Triangle \(\triangle ABC\)

For \(\triangle ABC\), the right angle at \(A\) implies \(\sin A = 1\). Thus, \(\frac{\sin b}{\sin a} = \sin B\) since \(C = 90^\circ\). Hence, \(\frac{\sin b}{\sin a} = \sin B\).
04

Apply to Triangle \(\triangle A'BC'\)

Similarly, for \(\triangle A'BC'\), the right angle at \(A'\) implies \(\sin A' = 1\). Thus, \(\frac{\sin b'}{\sin a'} = \sin B'\) where \(B'\) represents the same angle as \(B\) since the triangles share this angle at vertex \(B\).
05

Establish Relation Between Angles

Since both triangles share the same angle \(B\), we have \(\sin B = \sin B'\). Therefore, using Steps 3 and 4, we equate \(\frac{\sin b}{\sin a}\) and \(\frac{\sin b'}{\sin a'}\) as they both equal \(\sin B\).
06

Conclusion

Thus, we conclude that \(\frac{\sin b}{\sin a} = \frac{\sin b'}{\sin a'}\), as both are equal to \(\sin B\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spherical Triangle
A spherical triangle is formed by three arcs of great circles on a sphere. Unlike planar triangles, spherical triangles are represented on a curved surface which makes their properties quite unique. In a spherical triangle, the sum of the angles is greater than 180 degrees, differing from the sum of angles in a flat triangle.
Since we are working with a unit sphere (a sphere with radius 1), the sides of these triangles are usually expressed in radians, which correspond to the arc length on the sphere's surface.
Understanding the characteristics of spherical triangles is fundamental in spherical trigonometry, as they appear in navigation and astronomy applications.
Spherical Law of Sines
The Spherical Law of Sines is a critical formula that is used to solve spherical triangles. It ties together the sides and angles of a spherical triangle in a consistent way. This law is expressed as \[ \frac{\sin a}{\sin A} = \frac{\sin b}{\sin B} = \frac{\sin c}{\sin C} \]where \(a, b, c\) are the sides of the spherical triangle, and \(A, B, C\) are the respective opposite angles.
This relationship is similar to the planar law of sines, but it accounts for the curvature of the sphere. For spherical triangles, this law helps in calculating unknown sides or angles when some values are known. It can be particularly useful because it allows for the solution of triangles when we are given data that would otherwise be insufficient in planar geometry.
Unit Sphere Geometry
The unit sphere, with a radius of 1, is a common model used when dealing with spherical geometry. In this context, distances are expressed in radians since on a unit sphere the arc length directly corresponds to the angle in radians.
The unit sphere is crucial for simplifying complex calculations in spherical trigonometry because it removes the need for radius multiplication when applying trigonometric identities or formulas. All computations can be directly related to angles and arc lengths, making it more intuitive and straightforward for solving problems involving spherical triangles.
  • Angles are often measured in radians
  • Arcs and sides of triangles are seen as angles at the sphere's center
By using a unit sphere, we gain a consistent and simpler framework to explore the complexities of spherical trigonometry.
Right Angle Triangle on Sphere
A right angle spherical triangle has one angle equal to 90 degrees. This type of triangle simplifies the use of trigonometric identities and formulas due to its special properties. On the unit sphere, a right-angled triangle means one of the angles subtends a great circle arc that is a quarter of the whole circle's circumference on the sphere's surface.
In the given exercise, triangles \triangle ABC\ and \triangle A'BC'\ both have a right angle. This makes applying the Spherical Law of Sines straightforward because \(\sin A\) and \(\sin A'\) become 1 due to \(A\) and \(A'\) being right angles.
This specific property is particularly powerful when solving spherical triangles with known right angles, just as planar geometry benefits from rectangular solutions. The principle allows for significant simplifications in many real-world spherical calculations like in astronomical navigation or geolocation.

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Most popular questions from this chapter

The point \(P\) lies inside a circle. Two secants from \(P\) meet the circle at \(A, B\) and at \(C, D\) respectively. Prove that $$\underline{P A} \times \underline{P B}=\underline{P C} \times \underline{P D}$$ We end our summary of the foundations of Euclidean geometry by deriving the familiar formula for the area of a trapezium and its 3-dimensional analogue, and a formulation of the similarity criteria which is often attributed to Thales (Greek \(6^{\text {th }}\) century BC).

You are given a pyramid \(A B C D\) with all three faces meeting at \(A\) being right angled triangles with right angles at \(A .\) Suppose \(\underline{A B}=b,\) \(\underline{A C}=c, \underline{A D}=d\) (a) Calculate the areas of \(\triangle A B C, \triangle A C D, \triangle A D B\) in terms of \(b, c, d\). (b) Calculate the area of \(\triangle B C D\) in terms of \(b, c, d\). (c) Compare your answer in part (b) with the sum of the squares of the three areas you found in part (a).

(a) What is the angle between the two hands of a clock at 1:35? Can you find another time when the angle between the two hands is the same as this? (b) How many times each day do the two hands of a clock 'coincide'? And at what times do they coincide? (c) If we add a second hand, how many times each day do the three hands coincide?

Given any triangle \(\triangle A B C,\) draw the line through \(A\) which is parallel to \(B C,\) the line through \(B\) which is parallel to \(A C,\) and the line through \(C\) which is parallel to \(A B .\) Let the first two constructed lines meet at \(C^{\prime},\) the second and third lines meet at \(A^{\prime},\) and the first and third lines meet at \(B^{\prime}\). (a) Prove that \(A\) is the midpoint of \(\underline{B^{\prime}} C^{\prime}\), that \(B\) is the midpoint of \(\underline{C^{\prime}} A^{\prime}\), and that \(C\) is the midpoint of \(\underline{A^{\prime}} B^{\prime}\) (b) Conclude that the perpendicular from \(A\) to \(B C\), the perpendicular from \(B\) to \(C A\), and the perpendicular from \(C\) to \(A B\) all meet in a single point \(H .(H\) is called the orthocentre of \(\triangle A B C .)\)

(a)(i) Sketch a unit 2 D-cube as follows. Starting with two unit \(1 \mathrm{D}\) -cubes one directly above the other. Then join up each vertex in the upper 1D-cube to the vertex it corresponds to in the lower 1 D-cube (directly beneath it). (ii) Label each vertex of your sketch with coordinates \((x, y)(x, y=0\) or 1) so that the lower \(2 \mathrm{D}\) -cube has the equation " \(y=0\) " and the upper 2D-cube has the equation " \(y=1 "\). (b)(i) Sketch a unit 3D-cube, starting with two unit \(2 \mathrm{D}\) -cubes \(-\) one directly above the other. Then join up each vertex in the upper \(2 \mathrm{D}\) -cube to the vertex it corresponds to in the lower \(2 \mathrm{D}\) -cube (directly beneath it). (ii) Label each vertex of your sketch with coordinates \((x, y, z)\) (where each \(x, y, z=0\) or 1 ) so that the lower \(2 \mathrm{D}\) -cube has the equation \(" z=0 "\) and the upper \(2 \mathrm{D}\) -cube has the equation \(" z=1 "\). (c)(i) Now sketch a unit \(4 \mathrm{D}\) -cube in the same way \(-\) starting with two unit 3D-cubes, one "directly above" the other. (ii) Label each vertex of your sketch with coordinates \((w, x, y, z)\) (where each \(w, x, y, z=0\) or 1\()\) so that the lower 3 D-cube has the equation \(" z=0\) " and the upper 3 D-cube has the equation " \(z=1\) ".
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