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Point D lies on side AC of equilateral triangle ABC such that the measure of angle DBC is 45 degrees. What is the ratio of the area of triangle ADB to the area of triangle CDB? Express your answer as a common fraction in simplest radical form.

Unavailable  Jun 30, 2018
 #1
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Call the side of the equilateral triangle , S

 

Angle ABC  = 60°

Angle DBC  = 45°

And angle  ABD  =  ABC  - DBC  =  60 - 45  = 15°

 

The area of triangle  CDB  =

(1/2) * S * BD * sin 45

 

And the area of triangle ADB =

(1/2) * S * BD * sin 15

 

So...the area  of  triangle ADB  to the area of triangle CDB  =

 

(1/2) * S * BD * sin 15           sin  15

_________________   =     ______

(1/2 * S * BD  * sin 45          sin  45

 

 

sin 15   = sin (45 - 30)  = 

sin45 cos30  - sin 30 cos 45  =

√2/2 * √3/2    -  (1/2)(√2/2) =

[ √6  - √2 ] / 4

 

And sin 45  = √2/2

 

So    sin 15  / sin 45   =

 

[ √6  - √2 ] / 4   *  2 /  √2   =

 

[ √6  - √2 ]  / [ 2√2 ] =

 

√6 / [ 2√2 ]  - √2 / [ 2√2] = 

 

√3 / 2  -  1/ 2  =

 

(√3  - 1 )

______

     2

 

 

 

cool cool cool

CPhill  Jun 30, 2018

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