Online Course Discussion Forum

Math Challenge II geometry 1.28 and 1.29

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Math Challenge II geometry 1.28 and 1.29
by Jialin Wang - Wednesday, September 8, 2021, 9:07 PM
1.28: Let ABC be a triangle with AB=AC and ∠BAC=20∘, and let P be a point on side AB such that AP=BC. Find ∠ACP.

I am stuck at the step that bca=cba=80 and I can't get further. The congruent sides given has no correlation and no aux line can be drawn.

1.29: Given that one of the angles of the triangle with sides (5,7,8) is 60∘, show that one of the angles of the triangle with sides (3,5,7) is 120∘.

I have no idea what this is

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Re: Math Challenge II geometry 1.28 and 1.29
by Dr. Kevin Wang - Thursday, September 9, 2021, 6:44 AM

Without checking the textbook, I cannot be sure (on the road now), but please check the example questions in the same chapter to see if there are related questions that usually can give you some hints on how to solve these.

Nevertheless, here are some hints.  Both of these questions involve angles, and it is beneficial to find or construct special triangles, such as equilateral triangle or 30-60-90 triangles.

For 1.28, considering the difference between 20 degrees and 80 degrees is 60 degrees, you can try to construct an equilateral triangle outside the side AC, and try to find congruent triangles from there.

For 1.29: you are told that  a triangle with the side lengths 5,7,8 has an angle of 60 degrees. This is indeed true, and it can be verified with law of cosines (but that proof is not needed for this question, since  we haven’t learned Law of cosines yet).  Given that fact, You are required to find a triangle with side lengths 3,5,7 and show that one of its angles is 120 degrees. To start, you can think about the 5,7,8 triangle, and figure out which angle is the 60 degree one.  Then see if you can construct an equilateral triangle from there.