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Arcs and Triangles Help (page 3)

By — McGraw-Hill Professional
Updated on Oct 3, 2011

The Long Way Around

The foregoing example is not the only instance of “weird spherical triangles” that can be conjured up. Imagine a spherical triangle whose vertices are close to each other, but whose sides go the long way around (Fig. 11-8).

Global Trigonometry Arcs and Triangles The Long Way Around

Fig. 11–8. A spherical triangle in which each of the sides goes nearly all the way around the world.

As long as we are going to be extreme, why stop now? Suppose we free ourselves of the constraint that each side of a spherical n-gon must make less than one circumnavigation of the sphere. Any spherical polygon can then have sides that go more than once around, maybe hundreds of times, maybe millions of times. It’s not easy to envision what constitutes the interior of such a monstrosity; we might think of it as a globe wrapped up like a mummy in layer upon layer of infinitely thin tape. And what about the exterior? Perhaps we can think of the mummy-globe again, but this time, wrapped up in infinitely thin tape made of anti-matter.

Mind games like this can be fun, but they reduce to nonsense if taken too seriously. It’s a good idea to keep this sort of thing under control, if only for the sake of our sanity. Therefore, when we talk about a spherical polygon, we should insist that its size be limited as follows:

  • The perimeter cannot be greater than the circumference of the sphere
  • The interior area cannot be greater than half the surface area of the sphere

Any object that violates either of these two criteria should be regarded as “illegal” or “non-standard” unless we are dealing with some sort of exceptional case.

Spherical Law Of Sines

For any spherical triangle, there is a relationship among the angular lengths (in radians) of the sides and the measures of the interior spherical angles. Let ∠ sph QRS be a spherical triangle whose vertices are points Q, R, and S . Let the lengths of the sides opposite each vertex point, expressed in radians, be q, r, and s respectively, as shown in Fig. 11-9. Let the interior spherical angles ∠ sph RQS , ∠ sph SRQ, and ∠ sph QRS be denoted ψ q , ψ r , and ψ s respectively. (The symbol ψ is an italicized, lowercase Greek letter psi; we use this instead of θ to indicate spherical angles.) Then:

(sin q )/(sin ψ q ) = (sin r )/(sin ψ r ) = (sin s )/(sin ψ s )

That is to say, the sines of the angular lengths of the sides of any spherical triangle are in a constant ratio relative to the sines of the spherical angles opposite those sides. This rule is known as the spherical law of sines. It bears some resemblance to the law of sines for ordinary triangles in a flat plane.

Global Trigonometry Arcs and Triangles Spherical Law Of Sines

Fig. 11–9. The law of sines and the law of cosines for spherical triangles.

Spherical Law Of Cosines

The spherical law of cosines is another useful rule for dealing with spherical triangles. Suppose a spherical triangle is defined as above and in Fig. 11-9. Suppose you know the angular lengths of two of the sides, say q and r , and the measure of the spherical angle ψ s between them. Then the cosine of the angular length of the third side, s , can be found using the following formula:

cos s = cos q cos r + sin q sin r cos ψ s

This formula doesn’t look much like the law of cosines for ordinary triangles in a flat plane.

Global Trigonometry Arcs and Triangles Spherical Law Of Sines

Fig. 11–9. The law of sines and the law of cosines for spherical triangles.

Arcs and Triangles Practice Problems

Practice 1

A great-circle arc on the earth has a measure of 1.500 rad. What is the length of this arc in kilometers?

Solution 1

Multiply 6371, the radius of the earth in kilometers, by 1.500, obtaining 9556.5 kilometers. Round this off to 9557 kilometers, because the input data is accurate to four significant figures.

Practice 2

Describe and draw an example of a spherical triangle on the surface of the earth in which two interior spherical angles are right angles. Then describe and draw an example of a spherical triangle on the surface of the earth in which all three interior spherical angles are right angles.

Solution 2

To solve the first part of the problem, consider the spherical triangle Δ sph QRS such that points Q and R lie on the equator, and point S lies at the north geographic pole (Fig. 11-7A). The two interior spherical angles ∠ sph SQR and ∠ sph SRQ are right angles, because sides SQ and SR of Δ sph QRS lie along meridians, while side QR lies along the equator. (Remember that all of the meridian arcs intersect the equator at right angles.)

To solve the second part of the problem, we construct Δ sph QRS such that points Q and R lie along the equator and are separated by 90° of longitude (Fig. 11-7B). In this scenario, the measure of ∠ sph QSR, whose vertex is at the north pole, is 90°. We already know that the measures of angles ∠ sph SQR and ∠ sph SRQ are 90°. So all three of the interior spherical angles of ∠ sph QRS are right angles.

Global Trigonometry Arcs and Triangles The Case Of The Expanding Triangle

Fig. 11–7. Solution 2. At A, points Q and R are on the equator, and point S is at the north pole. At B, points Q and R are on the equator and are separated by 90° of longitude, while point S is at the north pole.

Practice 3

What are the measures of the interior spherical angles, in degrees, of an equilateral spherical triangle whose sides each have an angular span of 1.00000 rad? Express the answer to the nearest hundredth of a degree.

Solution 3

Let’s call the spherical triangle Δ sph QRS, with vertices Q, R, and S, and sides q = r = s = 1.00000 rad. Then:

cos q = 0.540302 cos r = 0.540302 cos s = 0.540302 sin q = 0.841471 sin r = 0.841471

Plug these values into the formula for the law of cosines to solve for cos ψ s , where ψ s is the measure of the angle opposite side s . It goes like this:

Global Trigonometry Arcs and Triangles Equilateral Spherical Triangle Principles

This means that ψ s = arccos 0.350777 = 69.4652°. Rounding to the nearest hundredth of a degree gives us 69.47°. Because the triangle is equilateral, we know that all three interior spherical angles ψ have the same measure: approximately 69.47°.

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