Proposed Problem

Click the figure below to see the complete problem 395 about Square, 15 Degree, Equilateral triangle, Congruence.

See also:

Complete Problem 395

Level: High School, SAT Prep, College geometry

## Saturday, November 28, 2009

### Problem 395: Square, 15 Degree, Equilateral triangle

Labels:
15 degree,
congruence,
equilateral,
square,
triangle

Subscribe to:
Post Comments (Atom)

http://www.wolframalpha.com/input/?i=%281-1%2F2*tg%2815%29%29%5E2%2B1%2F4

ReplyDeletehttp://geometri-problemleri.blogspot.com/2009/11/problem-52-ve-cozumu.html

ReplyDeletesolution plz some1

ReplyDeletein triangle ABE ang(ABE)=75;BE=AB/(2cos15)

ReplyDeletewith the law of cosine

AE²=AB²+BE²-2AB.BE cos75=

AB²(1+(1/(2cos15)²-cos75/cos15)= AB² then

cos(75)=cos(45+30)=(sqr6-sqr2)/4

cos(15)=cos(45-30)=(sqr6+sqr2)/4

same result for CE

CE=AE=AD

ADE is equilateral

build on each side of the square (from the inside) isosceles triangle with 15 degrees..from there its easy..

ReplyDeleteConsider the equilateral triangle BCF at the exterior of the square ABCD say of side a. The quandrilateral FEDC and FEAB are rhombuses, therefore

ReplyDeleteED = FC = a and EA = FB = a

(This holds because ang(FEC) = 75 = ang(ECF), so FE = FC = a and FE parallel to CD = a)

See

ReplyDeletehttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=472125