It is a well known fact that during a total solar eclipes the disc of the moon almost completely covers the disc of the sun. From this fact and from the information you can gather from Solved Examples 3 and 4 on page 1//44, determine the approximate diameter of the moon.
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From examples 2.3 and 2.4, we get `theta=1920"` and `S=3.8452xx10^(8) m`. During the total solar eclipse, the disc of the moon completely covers the disc of the sun, so the angular diameter of both the sun and the moon must be equal. Angular diameter of the moon, `theta =` angular diameter of the sun <br> `=1920''=1920xx4.85xx10^(-6) rad [1''=4.85xx10^(-6) rad]` <br> the earth-moon distance, `S=3.8452xx10^(8) m`. the diameter of the moon, `D = theta xx S` <br> `= 1920xx4.85xx10^(-6)xx3.8452xx10^(8) m=35806.5024xx10^(2) m=3581xx10^(3) m 3581 km`.