I got stuck on a problem on this sample math worksheet. The question reads "Triangle ABC has AB=2*AC. Let D and E be on line segments AB and BC, respectively, such that angle BAE= angle ACD. Let F be the intersection of segments AE and CD, and suppose that triangle CFE is equilateral. What is angle ACB?" The choices are 60 degrees, 75 degrees, 90 degrees, 105 degrees, or 120 degrees. I can't find any work that I think is useful. Thanks for helping.
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AMC 10 Math Question
- Thread starter Gamest
- Start date
Hello, Gamest!
I found a solution . . . but there must be a better way!
\(\displaystyle \text{Let }\theta \,=\,\angle BAE \,=\,\angle ACD.\)
\(\displaystyle \text{Let }AB \,=\,2,\;AC\,=\,1.\)
\(\displaystyle \Delta CFE\text{ is equilateral; note the 60}^o\text{ angles.}\)
\(\displaystyle \text{In }\Delta ACE:\;\angle ACE = \theta + 60,\;\angle CEA = 60 \quad\Rightarrow\quad \angle CAE \,=\,60-\theta\)
. . \(\displaystyle \text{Hence: }\,\angle CAB \,=\,60^o\)
\(\displaystyle \text{In }\Delta ABC\text{, we have: }\:AB = 2,\;AC = 1,\;\angle A = 60^o\)
\(\displaystyle \text{Law of Cosines: }\:BC^2 \;=\;2^1+1^2 - 2(2)(1)\cos60^o \;=\;3\)
. . \(\displaystyle \text{Hence: }\:BC \,=\,\sqrt{3}\)
\(\displaystyle Hey\,!\;\hdots\;\Delta ABC\text{ is a 30-60 right triangle!}\)
\(\displaystyle \text{Therefore: }\:\angle ACB \:=\:90^o\)
I found a solution . . . but there must be a better way!
\(\displaystyle \text{Triangle }ABC\text{ has: } AB = 2\!\cdot\! AC.\)
\(\displaystyle \text{Let }D\text{ and }E\text{ be on line segments }AB\text{ and }BC\text{, resp., such that: }\,\angle BAE\,=\, \angle ACD.\)
\(\displaystyle \text{Let }F\text{ be the intersection of segments }AE\text{ and }CD\text{, and suppose that }\Delta CFE\text{ is equilateral.}\)
\(\displaystyle \text{What is }\angle ACB\,?\)
\(\displaystyle \text{The choices are: }\; 60^o,\;75^o,\;90^o,\;105^o,\;120^o\)
Code:
o C
** 60 *
* * *
*@ * * E
* * 60 o
* * * *
* * * *
* * 60 * *
* o *
* * @ * F *
A o * o * * * * * * * * * o B
D\(\displaystyle \text{Let }\theta \,=\,\angle BAE \,=\,\angle ACD.\)
\(\displaystyle \text{Let }AB \,=\,2,\;AC\,=\,1.\)
\(\displaystyle \Delta CFE\text{ is equilateral; note the 60}^o\text{ angles.}\)
\(\displaystyle \text{In }\Delta ACE:\;\angle ACE = \theta + 60,\;\angle CEA = 60 \quad\Rightarrow\quad \angle CAE \,=\,60-\theta\)
. . \(\displaystyle \text{Hence: }\,\angle CAB \,=\,60^o\)
\(\displaystyle \text{In }\Delta ABC\text{, we have: }\:AB = 2,\;AC = 1,\;\angle A = 60^o\)
\(\displaystyle \text{Law of Cosines: }\:BC^2 \;=\;2^1+1^2 - 2(2)(1)\cos60^o \;=\;3\)
. . \(\displaystyle \text{Hence: }\:BC \,=\,\sqrt{3}\)
\(\displaystyle Hey\,!\;\hdots\;\Delta ABC\text{ is a 30-60 right triangle!}\)
\(\displaystyle \text{Therefore: }\:\angle ACB \:=\:90^o\)
"Apparently"?Gamest said:Apparently there is an easier way than this.
Soroban's can be simply/easily worded this way:
let u = angleBAE = angleACD (given)
since angleAFC=120, then angleCAF = 60 - u
since angleBAE = u, then angleBAC = 60 - u + u = 60
since AB = 2AC (given) and angle BAC = 60, then triangle ABC = 30-60-90 right triangle
so angle ACB = 90