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-2
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avatar+61 

[asy] size(5cm);pair A,B,C,D;A=(0,0);B=(5,0);C=(3,3);D=(0,1);draw(A--C,0.5*red + 0.5*white);draw(A--B--C--D--A);draw(rightanglemark(B,A,D)^^rightanglemark(D,C,B));add(pathticks(B--C, 1, .5, 6, 10));add(pathticks(C--D, 1, .5, 6, 10));label(

Find AC

 

I figured out that CB and DC are 5, but I can't figure out AC.

 Apr 11, 2022
 #1
avatar+124598 
+3

You are correct in that CB = DC  = 5  ......so....let's carry this a  little further....

 

Note that the four interior angles must sum to 360°

 

This means that  angles  D  and B   are supplementary   (since A and C  = 90° )  

 

This also means  that the  cos  B  =  -cos D     (since D)  is  > 90°

 

Put  another way..... -cosB  = cos D 

 

Since AC forms two triangles we can use the Law of Cosines twice and we have  that

 

AC^2  =  1^2  + 5^2  -  2 ( 1)(5) (-cos B)      (1)

AC^2  =  5^2  +  7^2 - 2 ( 5) (7) (cos B)       (2)

 

Which implies  that

 

1^2 + 5^2  + 10cos B   =  5^2 + 7^2 - 70 cos B

 

26  + 10cos B  =  74 - 70 cos B

 

80cos B  = 48

 

cos B  =   48 / 80  =    3/5 

 

So  using (2)

 

AC^2  =    74 - 70 (3/5)

 

AC^2  =  32

 

AC  = sqrt (32)  =  4sqrt (2)  ≈   5.66 

 

 

cool cool cool

 Apr 11, 2022

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