Right triangle \( A B C \) has side lengths \( A B=6, B C=6 \sqrt{3} \), and \( A C=12 \). A second right triangle, \( A^{\prime} B^{\prime} C^{\prime} \), has side lengths of \( 8 \sqrt{3}, 8 \), and 16 . Find the ratio of the side opposite \( \angle C \) ) the hypotenuse of triangle \( A B C \). Then use this ratio to identify the location of point \( C^{\prime} \) in the second right triangle. (1 point)
Solución de tutoría real
Responder
Solución
Revisado y aprobado por el equipo de tutoría de UpStudy
Extra Insights
To find the ratio of the side opposite \( \angle C \) to the hypotenuse of triangle \( ABC \), we identify the components. In triangle \( ABC \), the side opposite \( \angle C \) is \( BC = 6\sqrt{3} \) and the hypotenuse is \( AC = 12 \). Thus, the ratio is: \[ \text{Ratio} = \frac{BC}{AC} = \frac{6\sqrt{3}}{12} = \frac{\sqrt{3}}{2} \] In triangle \( A' B' C' \), we need to find where \( C' \) would be located based on this ratio. Here, the hypotenuse \( A'C' = 16 \) and we can apply the same ratio to identify the side opposite \( \angle C' \): \[ \text{Side opposite } \angle C' = \text{Ratio} \times A'C' = \frac{\sqrt{3}}{2} \times 16 = 8\sqrt{3} \] So, point \( C' \) is located such that it is opposite \( 8\sqrt{3} \) in triangle \( A' B' C' \).