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reviewing homework for lesson 3 math challenge I-C questions

 
 
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reviewing homework for lesson 3 math challenge I-C questions
by Bober Yang - Thursday, 16 January 2020, 8:38 PM
 

I'm reviewing the past lessons then I still don't really get 3.30, after I reviewed the posted solutions, also Mr. John didn't explain it in the homework discussion.

can I get some help? Thank you.

 
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Re: reviewing homework for lesson 3 math challenge I-C questions
by John Lensmire - Friday, 17 January 2020, 10:01 AM
 

Hi Bober,

Can you maybe say a little more about what you're confused about? To start, attached below is the diagram from class yesterday (with points A,B,C,D added in)

Diagram

B is the point the two cables are attached to the ground, with BD one cable and BC the other. $\angle ABD = 55^\circ$ while $\angle ABC = 75^\circ$. The problem states $CD = 3$ and we let $AD = x$ and $AB = y$. (Thus $AC = x+3$ is the height of the pole and $AB=y$ is the distance from the pole the cables are attached.

 Looking at the tangent equation for $\triangle ABD$ and $\triangle ABC$ gives the two equations above for $x$ and $y$. Since $\tan(55^\circ)$ and $\tan(75^\circ)$ are numbers, we can then solve these equations for $x$ and $y$.

Let me know if you still have questions from here!

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Re: reviewing homework for lesson 3 math challenge I-C questions
by Bober Yang - Tuesday, 21 January 2020, 3:43 PM
 

Mr. John, I'm confused about 3.30, not 4.30 sorry I didn't say it clearly.

Thank You

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Re: reviewing homework for lesson 3 math challenge I-C questions
by John Lensmire - Thursday, 23 January 2020, 9:45 AM
 

Sorry! It does clearly say 3.30, but it was right after we finished the chapter 4 discussion, so I was thinking of those problems.

For 3.30, we want the largest square in an equilateral triangle with side length $6$. This square (with side length $x$) will be as shown below:

Diagram

Since the square has $90^\circ$ angles, it divides the equilateral triangle into a smaller equilateral triangle, the square, and two 30-60-90 triangles. Since these 30-60-90 triangles have a ratio of sides $1 : \sqrt{3} : 2$, if the middle length is $x$, the sides are $x/\sqrt{3} : x : 2x/\sqrt{3}$.  This allows us to set up a linear equation and solve for $x$, as shown above.  Writing $x$ in simplest radical form gives $12\sqrt{3} - 18$.

Hope this helps a bit, otherwise let me know if there's a specific part that's confusing.

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Re: reviewing homework for lesson 3 math challenge I-C questions
by Bober Yang - Thursday, 23 January 2020, 3:46 PM
 

thank you