document.write( "Question 859829: Each side of the right triangle ABC is tangent to the circle with center o. The radius of the circle is 4 inches and the length AC is 12 inches. Find each of the following. \r
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Algebra.Com's Answer #518142 by Edwin McCravy(20060)\"\" \"About 
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Each side of the right triangle ABC is tangent to the circle with center O. The radius of the circle is 4 inches and the length AC is 12 inches. Find each of the following.
\n" ); document.write( "m∠C (Angle C)
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\n" ); document.write( "Draw radii (in green) to the three points of tangency,
\n" ); document.write( "OD, OE, and OF, which are perpendicular to the three
\n" ); document.write( "sides. Draw OC and OB. Extend FO till it intersects BC at G.
\n" ); document.write( "Then draw GH⊥AB
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\n" ); document.write( "OD=4, so AF=4, and since AC=12, FC=8
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\n" ); document.write( "∠FCO = ∠ECG because the center of inscribed circle O
\n" ); document.write( "is the intersection of the angle bisectors.
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\n" ); document.write( "ΔCFO = ΔCEO hypotenuse and a side equal
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\n" ); document.write( "∠FOC ≅ ∠EOC
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\n" ); document.write( "∠FOE = 2∠FOC, CF = CE = 8
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\n" ); document.write( "tan(∠FOC) = \"FC%2FOF\" = \"8%2F4\" = 2
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\n" ); document.write( "CE = CF = 8
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\n" ); document.write( "tan(∠FOE) = tan(∠FOC) = \"2tan%28FOC%29%2F%281-tan%5E2%28FOC%29%29\" = \"2%282%29%2F%281-2%5E2%29\" = \"4%2F%28-3%29\" = \"-4%2F3\"
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\n" ); document.write( "∠GOE = 180°-∠FOE
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\n" ); document.write( "tan(∠GOE) = 180°-∠FOE
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\n" ); document.write( "tan(∠GOE) = tan(180°-∠FOE) = -tan(∠FOE) = \"-%28-4%2F3%29\" = \"4%2F3\"
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\n" ); document.write( "tan(∠GOE) = \"EG%2FOE\" = \"EG%2F4\"
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\n" ); document.write( "\"EG%2F4\" = \"4%2F3\"
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\n" ); document.write( "EG = \"16%2F3\"\r
\n" ); document.write( "\n" ); document.write( "OG = \"sqrt%28OE%5E2%2BEG%5E2%29\" = \"sqrt%284%5E2%2B%2816%2F3%29%5E2%29\" = \"sqrt%2816%2B256%2F9%29\" = \"sqrt%28400%2F9%29\" = \"20%2F3\"
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\n" ); document.write( "ΔOEG ≅ ΔGHB because DE=GH=4 and ∠GBH = ∠OGE
\n" ); document.write( " parallel lines FG,AB cut by
\n" ); document.write( " transversal BC
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\n" ); document.write( "GB = OG = \"20%2F3\"
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\n" ); document.write( "BC = CE + EG + GB = 8 + \"16%2F3\" + \"20%2F3\" = \"60%2F3\" = 20\r
\n" ); document.write( "\n" ); document.write( "∠C = ∠ACB = cos-1\"%28AC%2FBC%29\" = cos-1\"%2812%2F20%29\" = cos-1\"%283%2F5%29\", approximately 53.13°
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\n" ); document.write( "∠B = ∠ABC = sin-1\"%28AC%2FBC%29\" = sin-1\"%2812%2F20%29\" = sin-1\"%283%2F5%29\", approximately 36.87°\r
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