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 $24$.
Best Answer
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: $$2OD=2\cdot20\cos53^{\circ}=24.07...$$