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Math Challenge II-A Geometry Problem 1.29

 
 
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Math Challenge II-A Geometry Problem 1.29
by Zeyin Wu - Tuesday, 12 October 2021, 5:10 AM
 

Math Challenge II-A Geometry Problem 1.29

Given that one of the angles of the triangle with sides (5,7,8)(5,7,8) is 6060∘, show that one of the angles of the triangle with sides (3,5,7)(3,5,7) is 120120∘.

Why is must the 60 degrees angle has rays of 5 and 8?

 
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Re: Math Challenge II-A Geometry Problem 1.29
by Areteem Professor - Tuesday, 12 October 2021, 1:37 PM
 

Good question! It is a pretty standard geometry result that if $\angle A$ is bigger than $\angle B$ in a triangle, then the side opposite $\angle A$ is longer than the side opposite $\angle B$. (This follows from the law of sines, but can also be proven using creating an isosceles triangle inside the triangle.)

The side of length 7 is in the middle in terms of length, so opposite from it must be the middle angle. Since the angles in a triangle add up to $180^\circ$, if $60^\circ$ is the smallest angle, then the angles add up to greater than $180^\circ$. Similarly, $60^\circ$ cannot be the largest angle. Hence, the $60^\circ$ angle must be the middle angle, so opposite from the middle side.