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Posts Made By: Jim Gleason

September 19, 2004 08:22 AM Forum: Beginning Astronomy?

What the heck is "tan"?!

Posted By Jim Gleason

Hello Sarah,

1st, relax and let`s backtrack a bit. Keeping it simple, (I hope), let`s take another look at the ex. given in the book. In that ex., the book means that the moon`s angular diameter is 1/2 of a degree or(1/2)*.

Also for the purpose of simplicity, let`s change the fraction 1/2 into an equivalent decimal which can be used more easily on your calculator,

Then we get 1/2 =.5

Now punch in .5 on your calculator and take the tangent of that. Then multiply this answer by 384,400 and you should get roughly 3350 just like the book states!

Other problems like this work out the same way. Take the tangent of the angular diameter converted to an equivalent decimal just like we did when we changed 1/2 to .5 and took the tangent of .5 on the calculator. Then multiply that answer by the distance to that object just like we multiplied that answer by the distance to the moon which is 384,400km. This will give you the physical diameter of the object just like it gave us an answer of roughly 3,350km. for the physical diameter of the moon.

In brief and perhaps more simply then, the book is saying to use this formula,

tang of angular diameter in degrees x distance to object in km. = object`s physical diameter in km. or in the moon ex.,

Tang. .5 x 384,400 = 3,350, (roughly),

Hope This Helped, 8) grin and let us know if you still don`t understand it. If you have any other questions, feel free to ask!!! 8) grin

Good Luck!!!