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diameter from B to D. Tangent at A. BC=CD, BH=10, AH=5, TE=8, HD=6. Diagram attached
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|dw:1370478703893:dw| FInd TA to the nearest tenth
I know that TA^2= TE(TE*EB) But I dont know how to find EH (I forgot to label H but it is the intersect of AD and TB)
Is \(\overline{TA}\|\overline{DB}\)?
No it doesn't say that it is. I was thinking that at first tho
Just wondering if there would be a similar triangle relationship hiddein in there somewhere.
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I could use the law of sines to find BD right? But what then
DO you have any idea?
Not any good ones at the moment.
ok thanks :)
@jim_thompson5910 Might have a better idea.
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|dw:1370479482276:dw|
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