Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
Inactive
»
k12.ed.math
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic:
Geometry Problem  Extremely Hard
Replies:
3
Last Post:
May 21, 2002 12:27 PM




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.
 submissions: post to k12.ed.math or email to k12math@sd28.bc.ca private email to the k12.ed.math moderator: kemmoderator@thinkspot.net newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html



