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You are an astronomer on planet tirth, which orbits a distant star. it has recently been accepted that tirth is spherical in shape, though no one knows its size. one day, you learn that on the equinox your sun is directly overhead in the city of tyene, located 750 kilometers due north of you. on the equinox, you go outside in alectown and observe that the altitude of your sun is 87.0 ∘.

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Answer:

The circumference of planet tirth is [tex]C = 90000 \ km[/tex]

Explanation:

From the question we are told that

   The distance of tyene from the observer is  [tex]d = 750 \ km[/tex]

   The altitude of the sun from the equinox is   [tex]\theta_s = 87.0^o[/tex]

Generally given that on the equinox your sun is directly overhead in the city of tyene , then the altitude of the tyene from the equinox is  [tex]\theta_e = 90 ^o[/tex]

Generally the angular distance between tyene and the sun as observed from the equinox is mathematically represented as

         [tex]\theta _d = \theta_e - \theta_s[/tex]

=>       [tex]\theta _d = 90 -87[/tex]

=>       [tex]\theta _d =3^o[/tex]

Generally the ratio of [tex]\theta _d =3^o[/tex] to  [tex]360^o[/tex] which is the total angle in a circle is equal to the ratio of  [tex]d = 750 \ km[/tex] to the circumference of the planet

So

         [tex]\frac{\theta_d}{360} = \frac{750}{C}[/tex]

=>      [tex]\frac{3}{360} = \frac{750}{C}[/tex]

=>     [tex]C = 90000 \ km[/tex]

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