Proposed Problem
Click the figure below to see the complete problem 420 about Triangle, Angles, Altitude, Sides, Measurement.
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Complete Problem 420
Level: High School, SAT Prep, College geometry
Tuesday, January 19, 2010
Problem 420. Triangle, Angles, Altitude, Sides, Measurement
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Another way would be to extend AB to E such that BE=3c. Now it's easy to see that Tr. EAC is isosceles with BD // to altitude from E to AC which gives us: c/4c= 1/((x+1)/2) or x=7
ReplyDeleteAjit
Let AD = AE, E being on DC.
ReplyDeleteBE bisects < ABC, hence c/3c = 2/(x-1) from which we have x = 7
Sumith Peiris
Moratuwa
Sri Lanka
Problem 420
ReplyDeleteSuppose that the point E is symmetric of A with respect to the D. Then BE is the bisector of the angle ABC.Is c/3c=2/(x-1) or x=7.
[For easy typing, I use a instead of alpha]
ReplyDeletecosa=BD/c
cos3a=BD/3c
cos3a/cosa=1/3
4(cosa)^2-3=1/3
(cosa)^2=5/6
(sina)^2=1/6
(1/c)^2=1/6
c=sqrt6
In triangle BDC
sin3a=x/3c
3sina-4(sina)^3=x/3c
3/c-4(1/c)^3=x/3c
3c^2-4=(x*c^2)/3
18-4=2x
x=7