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avatar+893 

Answer: 18

 Mar 16, 2019

Best Answer 

 #2
avatar+5662 
+2

 

 

 

 

 

\(\text{Consider a line from the right bottom corner to the center of the circle}\\ \text{Convince yourself that it bisects the angle the triangle makes with the right side}\\ \text{and thus it is a }15^\circ \text{ angle}\)

 

\(\tan(15^\circ) = \dfrac{r}{s-r}\\ s = \dfrac{r(1+\tan(15^\circ))}{\tan(15^\circ)}\)

 

\(\text{I'll let you plug and chug to see that it's }s=18\)

.
 Mar 18, 2019
 #1
avatar+893 
+1

Is there an auxiliary line or something to solve this problem?

 Mar 16, 2019
 #2
avatar+5662 
+2
Best Answer

 

 

 

 

 

\(\text{Consider a line from the right bottom corner to the center of the circle}\\ \text{Convince yourself that it bisects the angle the triangle makes with the right side}\\ \text{and thus it is a }15^\circ \text{ angle}\)

 

\(\tan(15^\circ) = \dfrac{r}{s-r}\\ s = \dfrac{r(1+\tan(15^\circ))}{\tan(15^\circ)}\)

 

\(\text{I'll let you plug and chug to see that it's }s=18\)

Rom Mar 18, 2019
 #3
avatar+893 
0

Thanks!!!

dgfgrafgdfge111  Mar 18, 2019

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