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]
