SOLUTION: 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. m&#

Algebra ->  Trigonometry-basics -> SOLUTION: 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. m&#      Log On


   



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.
m∠C (Angle C)
m∠B (Angle B)
Line BC

Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
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.
m∠C (Angle C)
m∠B (Angle B)
Line BC



Draw radii (in green) to the three points of tangency,
OD, OE, and OF, which are perpendicular to the three
sides. Draw OC and OB. Extend FO till it intersects BC at G.
Then draw GH⊥AB



OD=4, so AF=4, and since AC=12, FC=8

∠FCO = ∠ECG because the center of inscribed circle O
is the intersection of the angle bisectors.

ΔCFO = ΔCEO hypotenuse and a side equal

∠FOC ≅ ∠EOC

∠FOE = 2∠FOC, CF = CE = 8

tan(∠FOC) = FC%2FOF = 8%2F4 = 2

CE = CF = 8

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

∠GOE = 180°-∠FOE

tan(∠GOE) = 180°-∠FOE

tan(∠GOE) = tan(180°-∠FOE) = -tan(∠FOE) = -%28-4%2F3%29 = 4%2F3

tan(∠GOE) = EG%2FOE = EG%2F4

EG%2F4 = 4%2F3

EG = 16%2F3
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

ΔOEG ≅ ΔGHB because DE=GH=4 and ∠GBH = ∠OGE
parallel lines FG,AB cut by
transversal BC

GB = OG = 20%2F3

BC = CE + EG + GB = 8 + 16%2F3 + 20%2F3 = 60%2F3 = 20
∠C = ∠ACB = cos-1%28AC%2FBC%29 = cos-1%2812%2F20%29 = cos-1%283%2F5%29, approximately 53.13°

∠B = ∠ABC = sin-1%28AC%2FBC%29 = sin-1%2812%2F20%29 = sin-1%283%2F5%29, approximately 36.87°
Edwin