triangle length vs angles: from phytagoras, AC = 25/3 and AD = 7/3 considring AB = AC

ketanco

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In the attached I can figure from phytagoras, AC = 25/3 and AD = 7/3 considring AB = AC
and and agin from phtagoras BC = 10.
But then I am stuck.
Answer should be 1720181108_212012 - 3.jpg
 
In the attached I can figure from phytagoras, AC = 25/3 and AD = 7/3 considring AB = AC
and and agin from phtagoras BC = 10.
But then I am stuck.
Answer should be 17View attachment 10467
attachment.php


Now you need to use similar triangle theorems.

Let DA= p and AC = h;

Then:

p/8 = (6+p)/x ... and

(h+9)/x = b/8 .... and

See where does those take you along with Pythagoras.
 
Don't you have a straight edge (or something!)
to draw straight lines with :confused:
 
Now you need to use similar triangle theorems.

Let DA= p and AC = h;

Then:

p/8 = (6+p)/x ... and

(h+9)/x = b/8 .... and

See where does those take you along with Pythagoras.

Are you assuming that ABE is a right angle? It isn't.
 
In the attached I can figure from phytagoras, AC = 25/3 and AD = 7/3 considring AB = AC
and and agin from phtagoras BC = 10.
But then I am stuck.
Answer should be 17

Yes, the answer is 17. And I got the same values for AC and AD.

I found it by using the law of cosines applied to triangle BCE, obtaining the cosine of BCE from triangle ABC.

What theorems have you learned, that might be used if you can't use the law of cosines?

For confirmation, here is a picture:

FMH113276.jpg

Here's an alternative way: drop a perpendicular from B to AE.
 
Last edited:
But you DO ENJOY funny distortions if angleABE is forced to equal 90;
everything works out a-la-Pythagoras, but 2 1/3 > 6 and 6 1/3 > 9 :rolleyes:
 
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