find lengths of all the unknown members of the roof truss in this diagram. The figure is symmetric. The diagram is #50?
http://books.google.com/books?id=g1Bul7oPMF4C&pg=PA362&lpg=PA362&dq=Find+all+the+unknown+lengths+of+the

HELPP PLEASEE

Question

find lengths of all the unknown members of the roof truss in this diagram. The figure is symmetric. The diagram is #50?
 http://books.google.com/books?id=g1Bul7oPMF4C&pg=PA362&lpg=PA362&dq=Find+all+the+unknown+lengths+of+the

HELPP PLEASEE

anonymous · Answer

As suggested by moderators I am helping you with hints.....calculate cos 35 for the length of the hypotenuse.You could repeat this to claculate whole length

amriju · Answer

is the figure symettric ..?

anonymous · Answer

yes

amriju · Answer

ok...on it

anonymous · Answer

@campbell_st do you think you could elaborate more?

anonymous · Answer

@campbell_st is x hypotenuse? what is perpendicular bisector?

amriju · Answer

if u dont want trigonometry....u can do by similarity....

anonymous · Answer

which angle?

anonymous · Answer

@amriju i dont really know what yur talking about

amriju · Answer

use the properties of similar triangles and u wont need trigonometry

anonymous · Answer

@campbell_st sorry to bother you but could i have like a step by step help maybe?

amriju · Answer

|dw:1350936306962:dw|...

anonymous · Answer

@campbell_st well i got x =sqrt{117}

anonymous · Answer

@amriju i dont mean to sound rude but idk what your doing

anonymous · Answer

@campbell_st where is angle x? i thought you wanted me to find the hypotenuse for that is where you put the x in your drawing above

anonymous · Answer

just kidding x=1.5

anonymous · Answer

cosx=9/6=1.5?

amriju · Answer

|dw:1350936537891:dw| BD=6...CD=9..BC=$$\sqrt{117}$$.....then u have triangles BDC and AFC similar...by property...since AF is twice of BD...so AC is twice BC...by symmetry..AH=AC=2$$\sqrt{117}$$ thats a..now u have tan angle BCD= 6/9...cot BED=6/9..=BD/ED...find BE=b from here....rest is pythagoras for c

anonymous · Answer

@amriju what?

amriju · Answer

what..?....look at the diag...and proceed

anonymous · Answer

the labels arent veru clear

amriju · Answer

ok...i am redrawing..

amriju · Answer

|dw:1350937525280:dw|

amriju · Answer

BD=6...CD=9..BC=$$\sqrt{117}$$.....then u have triangles BDC and AFC similar...by property...since AF is twice of BD...so AC is twice BC...by symmetry..AH=AC=2$$\sqrt{117}$$ thats a..now u have tan angle BCD= 6/9...cot BED=6/9..=BD/ED...find BE=b from here....rest is pythagoras for c

anonymous · Answer

@amriju that redraw doesnt make a difference with the first?

amriju · Answer

WELL...u cant have a betr diagram...whats bothering u in the diag..?

anonymous · Answer

whats c?whats a, b , d , e, f ,g ,h

amriju · Answer

capital A,B,C,...i used to denote the vertices.....lowercase a,b,c...are the lenghts as asked in the ques of ur link...a=AH...like thats...

amriju · Answer

did u get it..?