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# Arcs and Triangles Help (page 4)

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

#### Practice 4

Suppose we have an equilateral spherical triangle Δ sph QRS on the surface of the earth, whose sides each measure 1.00000 rad in angular length, as in the previous problem. Let vertex Q be at the north pole (latitude +90.0000°) and vertex R be at the Greenwich meridian (longitude 0.0000°). Suppose vertex S is in the western hemisphere, so its longitude is negative. What are the latitude and longitude coordinates of each vertex to the nearest hundredth of a degree?

#### Solution 4

Figure 11-10 shows this situation. We are told that the latitude of point Q (Lat Q ) is +90.0000°. The longitude of Q (Lon Q ) is therefore undefined. We are told that Lon R = 0.0000°. Lat R must be equal to +90.0000° minus the angular length of side s. This is +90.0000° = 1.00000 rad. Note that 1.00000 rad is approximately equal to 57.2958°. Therefore:

Rounded off to the nearest hundredth of a degree, Lat R = +32.70°. This must also be the latitude of vertex S , because the angular length of side r is the same as the angular length of side s . The longitude of vertex S is equal to the negative of the measure of the interior spherical angle at the pole, or – ψ . We know from Solution 11-5 that ψ = 69.47°. Therefore, we have these coordinates for the vertices of Δ sph QRS rounded off to the nearest hundredth of a degree:

Fig. 11–10. Illustration for Solution 4.

Lat Q = +90.00° Lon Q = (undefined) Lat R = +32.70° Lon R = 0.00° Lat S = +32.70° Lon S = –69.47°

Practice problems for these concepts can be found at: Global Trigonometry Practice Test

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