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# If the angle of the sun is 34' and te sun is 92,919,800 miles away, what is the measure of the diameter of the sun?

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If the angle of the sun is 34' and te sun is 92,919,800 miles away, what is the measure of the diameter of the sun?

Mar 6, 2015

#5
+27342
+5

With this small angle, sines and tans are largely irrelevant!  The diameter is just the angle multiplied by the distance - as long as the angle is expressed in radians.

$${\mathtt{diameter}} = {\frac{{\frac{{\mathtt{34}}}{{\mathtt{60}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{180}}}}{\mathtt{\,\times\,}}{\mathtt{92\,919\,800}} \Rightarrow {\mathtt{diameter}} = {\mathtt{918\,995.321\: \!833\: \!621\: \!997\: \!390\: \!4}}$$

diameter ≈ 9.19*105 miles

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Mar 7, 2015

#1
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Since the actual angle that the sun subtends is 0.53 degrees, you are going to get a v. large and different number (using 34 degrees).

You can use the Law of Sines or the Pythagorean Theorem (with certain assumptions).

Mar 7, 2015
#2
+95179
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Thanks for answering anon.

34' means 34 minutes and

1 minute is 1/60 of a degree so

34' =  $${\frac{{\mathtt{34}}}{{\mathtt{60}}}} = {\frac{{\mathtt{17}}}{{\mathtt{30}}}} = {\mathtt{0.566\: \!666\: \!666\: \!666\: \!666\: \!7}}$$    degrees

This is bigger than the google value but not as much bigger as you thought.    :))

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tan(17/60)=r/92919800

r=tan(17/60)*92919800

d=2*tan(17/60)*92919800

$${\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{tan}}{\left({\frac{{\mathtt{17}}}{{\mathtt{60}}}}\right)}{\mathtt{\,\times\,}}{\mathtt{92\,919\,800}} = {\mathtt{919\,002.812\: \!968\: \!937\: \!2}}$$

diameter of the sun = $$9.19*10^5\;\; miles$$

I could have used sin if I wanted to the answer would have been close enough to the same.

Effectively that is what anon beneath here has done.  Only they did it using the sine rule.  Makes no difference. :)

According to google it is  $$8.64*10^5 \;\;miles$$     so my answer is ok.

Mar 7, 2015
#3
+5

I'm still holding out for the Law of Sines.

34' of arc means that the other angles of the triangle must be about 89.71 degrees.

$$\frac {Diameter}{sin 0.567}=\frac{92919800}{sin 89.71}$$

I get 919532.

Approximating the distance to the edge of the sun is the same as the other (frontal) distance.

Mar 7, 2015
#4
+95179
+5

I am a goose - I used atan instead of tan that is why my answer was so wrong,

I will go back and fix it.

Mar 7, 2015
#5
+27342
+5

With this small angle, sines and tans are largely irrelevant!  The diameter is just the angle multiplied by the distance - as long as the angle is expressed in radians.

$${\mathtt{diameter}} = {\frac{{\frac{{\mathtt{34}}}{{\mathtt{60}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{180}}}}{\mathtt{\,\times\,}}{\mathtt{92\,919\,800}} \Rightarrow {\mathtt{diameter}} = {\mathtt{918\,995.321\: \!833\: \!621\: \!997\: \!390\: \!4}}$$

diameter ≈ 9.19*105 miles

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Alan Mar 7, 2015