# Thread: Very simple trig problem

1. ## Very simple trig problem

Hey, I need some help with this VERY basic trig/geometry problem. I just had my first trig class yesterday and we got this problem that is fairly basic, yet I can't seem to be able to do it without more advanced trig. We haven't learned anything with sin, cosin, or tangent yet and although I've already learned that in another class we're supposed to be able to solve this in a very basic way. Here's the problem:

An astronomical object has an actual diameter of 128 km and an apparent size of 24'16" as viewed from earth. How far is this object from earth?

So, I converted 24'16" into DD format and got .40444444 (with the last four repeating). I assume that the diagram should look something like this:

math

How should I find the lengths of the sides? Am I just going about this completely wrong? I wish we were allowed to use more advanced techniques but there's got to be a way to do this without them, or he wouldn't have assigned it. Thanks in advance.

2. Hello, redeemys24!

An astronomical object has an actual diameter of 128 km
and an apparent size of 24'16" as viewed from earth.
How far is this object from earth?

So, I converted $24'16''$ into DD format and got $0.40444...^o$

Bisect the vertex angle of that isosceles triangle.

You will have a right triangle.
One side runs from earth (E) to the center of the object (C).
. . Let $d = EC$
From the center directly to the right, CA is the radius of the object: 64 km.
. . And EA is the hypotenuse.

Let $\theta \:= \:\angle CEA \:=\:\frac{1}{2}(0.40444...^o) \:=\:0.20222...^o$

We have: . $\tan\theta \:=\:\frac{64}{d}\quad\Rightarrow\quad d \:=\:\frac{64}{\tan\theta}$

Hence: . $d \;=\;\frac{64}{\tan(0.20222...)} \:=\:18133.09449 \:=\:18133\text{ km}$

3. Originally Posted by redeemys24
...We haven't learned anything with sin, cosin, or tangent yet .... Here's the problem:

An astronomical object has an actual diameter of 128 km and an apparent size of 24'16" as viewed from earth. How far is this object from earth?

So, I converted 24'16" into DD format and got .40444444 (with the last four repeating). I assume that the diagram should look something like this:

math

...
Hello,

this is only a guess:

With very small angles the diameter of the object and the corresponding arc are nearly the same and the distance to the object and the length of the legs of the angle are nearly the same too.
Let a be the arc corresponding with the angle of 0°24'16'' = 91/225°. Then you get:

$\frac{a}{2 \pi} = \frac{\frac{91}{360}}{360}\ \Longrightarrow \ a \approx 0.007058887...$

Let d be the diameter of the object and R the length of one leg of the angle, then the arc a is calculated by:

$\frac{d}{R} = a\ \Longrightarrow \ R = \frac{d}{a}$

Plug in the values you know and you'll get $R \approx 18.100 \ km$

(Personal remark: If I haven't made some ridiculous mistake then this object flies between the surface of the earth and the cloud of geo-stationary satelites and it is remarkably big: It nearly can cover the sun completly (31') so there will happen some additional eclipses of the sun every day)