In a triangle $ ABC $, the $ ∠A = 53 ° $ and the circumference measures $ 20 $, calculates the double of the distance to the $\overline{BC}\quad$side.

I do not understand the question why it asks for the distance of the circumcision to the $\overline{BC}\quad$side, and this distance varies according to the point on the $\overline{BC}\quad$side of the triangle. On the other hand, if it would be the distance from the cicuncentro to one of the points between $B$ and $C$ of the circumference, it would be $40$ ... But the answer is $24$.


  • $\begingroup$ Did you mean the radius is 20? $\endgroup$
    – David K
    Jul 30, 2018 at 17:51

1 Answer 1


Let $O$ be a center of the circle and $OD$ be a perpendicular from $O$ to $BC$.

I think it means to find $2OD$.

Since $$\measuredangle COD=\frac{1}{2}\measuredangle BOC=53^{\circ},$$ we obtain: $$2OD=2\cdot20\cos53^{\circ}=24.07...$$

  • $\begingroup$ Excuse me it was my mistake, they asked me twice the distance of the circumcenter to the $\overline{BC}\quad$ side. $\endgroup$
    – Payo
    Jul 30, 2018 at 16:25
  • $\begingroup$ @Payo It's exactly, which I got. $\endgroup$ Jul 30, 2018 at 16:26
  • $\begingroup$ @Michael- My mistake $\endgroup$ Jul 30, 2018 at 16:30
  • $\begingroup$ Yes, Thank you @Michael $\endgroup$
    – Payo
    Jul 30, 2018 at 16:40
  • $\begingroup$ You are welcome! $\endgroup$ Jul 30, 2018 at 17:26

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