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Bunuel
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Peddi
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nick1816
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In Triangle AEC

Cosecant(x)= EC/CA

In Triangle BDE

Cosecant(x)= DE/BE




Peddi
nick1816
AE= \(\sqrt{13^2-12^2}\)=5

BE=12.2 - 5=7.2


Triangle AEC is similar to Triangle BDE

Hence, \(\frac{EC}{CA}=\frac{DE}{BE}\)

\(\frac{13}{12}=\frac{DE}{7.2}\)

DE=7.8


Bunuel

What is the length of hypotenuse k?

(A) 9
(B) 7.8
(C) 7
(D) 6.9
(E) 6.8


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Attachment:
1b.jpg
EC/CA=ED/DB , what you have done with the similarity property is wrong

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Sorry nick, didnt pay much attention to the angle90degrees between two triangles so that makes one angle x and other angle 90-x right?so 90-x is between AC and CE, and DEand BE

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Bunuel

What is the length of hypotenuse k?

(A) 9
(B) 7.8
(C) 7
(D) 6.9
(E) 6.8


Attachment:
1b.jpg

(12.2-5)/12 = k/13

7.2/12 = k/13
k = 7.2*13/12

7.2*12/12 = 7.2 < 7.2*13/12, so k must be greater than 7.2. C, D, E is out

7.2*13/12 < 8*13/12 = 8 + 2/3, which is smaller than 9. So k <9. A is out.

IMO B.
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applying similarity does the job i guess 13/12 = x/7.2 = 7.8
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