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Topic: Geometry Problem - Extremely Hard
Replies: 3   Last Post: May 21, 2002 12:27 PM

 Messages: [ Previous | Next ]
 Sheldon Ackerman Posts: 28 Registered: 12/6/04
Re: Geometry Problem - Extremely Hard
Posted: May 20, 2002 11:27 PM

>If the measures of the sides of a triangle are consecutive even
>integers and the measure of the greatest angle is twice that of the
>smallest angle, find the measures of the sides of the triangle.
>

How about setting up the diagram. The sides of the triangle will be y, y+2,
and y+4. You specify that the greatest angle is twice the smallest.
Therefore, let the greatest angle be 2x and the smallest x. The greatest
angle, 2x, will be opposite the greatest side of y+4. The smallest angle x
will be opposite the smallest side y.

Use the law of sines and you will be able to arrive at cos x = (y+4)/2y
(remember that sin 2x= 2 sin x cos x)

Next use the law of cosines and you will arrive at an equivalent expression
for cos x. Set them equal and solve for y. I came up with an answer of 8,
so the other 2 sides are 10 and 12. I
did not bother doing any checks so I certainly may have made some
computational error or errors :-)

You can use the law of cosines now to figure out the angles and then see
whether or not the smaller and larger angles are in a ration of 1:2.

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Date Subject Author
5/19/02 Tough Geometry Problem
5/20/02 Paul Tanner
5/20/02 Sheldon Ackerman
5/21/02 Rob Morewood