Geometry Problem
Click the figure below to see the complete problem 504 about Equilateral Triangle, Angles, 45, 60 Degrees.
See also:
Complete Problem 504
Level: High School, SAT Prep, College geometry
Sunday, August 15, 2010
Problem 504: Equilateral Triangle, Angles, 45, 60 Degrees
See also:
Complete Problem 504
Level: High School, SAT Prep, College geometry
Subscribe to:
Post Comments (Atom)
in triangle ABD:<A=45,<B=105,<D=30
ReplyDeletewith the sine law
BDsin30=ABsin45;BD=ABsqr(2)
in triangle BDC
with the cosine law
CD²=BC²+BD²-2BC.BD.cos45=AB²
CD=AC
triangle ACD is isoscele
x=15
.-.
let t=AB, it is known that BD=t\sqrt{2} (because tr ABD is 45-30-105).
ReplyDeleteConsecuently CD=t (by tracing CX, altitude of tr BCD and using Pithagorean Theorem).
So tr ACD is isosceles and finally <CDA= < CAD=15.
by mathreyes
draw DG perpendicular to AD, G on extensin of AC
ReplyDeleteABDG cyclic AC = CB = R => C midpoint => AC = CD
arc BG = 2∙A = 120° => arc AB = 60° => C midpoint
ReplyDeleteC origin, A,B and D circular
ReplyDeleteIBCI=ICDI=r
so trianle BCD isosceles and x=15
kadir Altintaş/emirdağ