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I have been having trouble on this question, help is appreciated

 Apr 15, 2020
 #1
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Call BE = x         then AC = x

Call DE = y

Call EC = z

 

Triangle(ACB) is similar to triangle(DEC).

--->   y / x  =  x / (x + z)

 

Right triangle   --->   y  =  sqrt(1 - x2)

 

x + y + z  =  2   --->   z  = 2 - x - y   --->   z  =  2 - x - sqrt(1 - x2)

 

y / x  =  x / (x + z)   --->   sqrt(1 - x2) / x  =  x / [ x + (2 - x - sqrt(1 - x2) ) ]

--->   sqrt(1 - x2) / x  =  x / [ 2 - sqrt(1 - x2) ]

Cross-multiplying:

--->   x2  =  2sqrt(1 - x2) - (1 - x2)

--->   x2  =  2sqrt(1 - x2) - 1 + x2

--->     0  =  2sqrt(1 - x2) - 1

--->    1  =  2sqrt(1 - x2)

--->   ½  =  sqrt(1 - x2)

--->   ¼  =  1 - x2

--->   x2  =  3/4

--->     x  =  sqrt(3)/2       [the negative answer can be ignored]

 

cos(ABC)  =  [ sqrt(3)/2  ] / 1

angle(ABC)  =  cos-1[ sqrt(3)/2 ]

angle(ABC)  =  30o 

 Apr 16, 2020

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