Answer to Question #97725 in Geometry for akash kumar

Question #97725
Consider an equal-sided triangle connected to a square with side a. A straight line from
Q intersects the square at x and y. You have given x, find an equation for the intersection at y(x).
1
Expert's answer
2019-10-31T12:20:58-0400


(This image is for reference and better understanding)

Draw a hypothetical line QCG such that it passes through middle and perpendicular to AD and EH.

"AC=\\frac{a}{2};AB=x;BC=\\frac{a}{2}-x" ;

"EG=\\frac{a}{2};EF=y;FG=\\frac{a}{2}-y;"

In "\\triangle QCA,QC=QA\\ sin 60\\degree=a\\frac{\\sqrt{3}}{2}"

"GC=HD=a" ;"QG=a+\\frac{\\sqrt{3}}{2}a"

As "\\triangle QBC" and "\\triangle QFG" are similar,

So,"\\frac{QC}{QG}=\\frac{BC}{FG}" ;

"\\frac{\\frac{\\sqrt{3}}{2}a}{(\\frac{\\sqrt{3}}{2}+1)a}=\\frac{\\frac{a}{2}-x}{\\frac{a}{2}-y}" ;

"\\frac{\\frac{\\sqrt{3}}{2}}{(\\frac{\\sqrt{3}}{2}+1)}=\\frac{a-2x}{a-2y}" ;

"\\frac{{\\sqrt{3}}}{(\\sqrt{3}+2)}=\\frac{a-2x}{a-2y}" ;

"\\sqrt{3}a-2\\sqrt{3}y=(2+\\sqrt{3})a-2(2+\\sqrt{3})x;"

"y=\\frac{2(2+\\sqrt{3})x-2a}{2\\sqrt{3}}"



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Assignment Expert
01.11.19, 15:45

According to conditions of the question, a triangle is equal-sided, then all internal angles of this triangle are 60 degrees. A derivation of the value of the sine of 60 degrees can be found at https://www.mathsisfun.com/geometry/unit-circle.html.

Ano
01.11.19, 15:29

How did you get sin60 ?

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS