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here is the question

 Oct 14, 2018
 #2
avatar+98130 
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Call O the center of the circle.....draw radius OC perpendicular  to AB, and connect OB  and OA

 

In triangle OBC...angle OBC  = 30°  and angle BOC  = 60°

So....BC  = 6√3  cm

 

And in triangle OAC  = 15“  and angle COA  = 75°

Using the law of sines

 

CA / sin 75  = OC / sin 15

 

CA /  sin (30 + 45)  = 6 / sin (45 -30)

 

CA  =  6 sin (30 + 45)  / sin (45 - 30)

 

CA = 6 [ sin 30cos 45 + sin 45cos30 ] / [ sin 45cos 30 - sin 30 cos 45 ]

 

CA = 6  [ 1/2 * √2/2  + √2/2 * √3/2 ]  / [ √2/2 * √3 / 2 * 1/2 * √2/2 ]

 

CA = 6 [ √2/4 + √6/4 ] / [ √6 / 4  - √2/4]

 

CA = 6 [ √6 + √2 ] / [√6  - √2]

 

CA  = 6 [ 6 + 2√12 + 2 ] / [6 - 2]

 

CA = 6 [8 + 4√3 ] / [ 4]

 

CA = (3/2) [ 8 + 4√3]  =  3 [ 4 + 2√3]  = 12 + 6√3  cm

 

So....AB = BC + CA  =  6√3  + 12 + 6√3  =   12 + 12√3  cm 

 

 

cool cool cool

 Oct 14, 2018
edited by CPhill  Oct 14, 2018
 #3
avatar+464 
+1

Thank you Chris 

 Oct 14, 2018

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