Question

in triangle abc, point e lies on line segment bc and point d lies on line segment ab. what is the sine of angle c?

Explanation:

Step1: Identify triangle type

Triangle \( ABC \) is a right triangle with \( \angle A = 90^\circ \), \( AB \) and \( AC \) as legs, \( BC \) as hypotenuse? Wait, no, wait: Wait, \( AD \) and \( DE \): Wait, \( D \) is on \( AB \), \( E \) is on \( BC \), \( DE = 12 \), \( AC = 24 \), \( BD = 5 \), \( BE = 13 \)? Wait, no, \( BE \) is 13? Wait, \( BD = 5 \), \( DE = 12 \), so triangle \( BDE \) is right-angled at \( D \), so \( BE = \sqrt{5^2 + 12^2} = 13 \), which matches. So \( DE \parallel AC \) (since both are perpendicular to \( AB \)? Wait, \( \angle A = 90^\circ \), \( \angle D = 90^\circ \), so \( DE \parallel AC \). Therefore, triangle \( BDE \sim \) triangle \( BAC \) (by AA similarity, since \( \angle B \) is common, \( \angle BDE = \angle BAC = 90^\circ \)).

Step2: Find \( AB \) length

From triangle \( BDE \), \( BD = 5 \), \( DE = 12 \), \( BE = 13 \). Since \( DE \parallel AC \), the ratio of similarity: \( \frac{DE}{AC} = \frac{BD}{AB} \). \( DE = 12 \), \( AC = 24 \), so \( \frac{12}{24} = \frac{5}{AB} \)? Wait, no, that would be \( \frac{BD}{BA} = \frac{DE}{AC} \). So \( \frac{BD}{BA} = \frac{12}{24} = \frac{1}{2} \), so \( BD = \frac{1}{2} BA \), so \( BA = 2 \times BD = 2 \times 5 = 10 \). Wait, \( BD = 5 \), so \( BA = BD + DA = 5 + DA \), but if similarity ratio is \( \frac{1}{2} \), then \( BD = \frac{1}{2} BA \), so \( BA = 10 \), so \( DA = 5 \).

Step3: Find \( BC \) or use \( \sin C \)

In right triangle \( ABC \), \( \sin C = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{BC} \). Wait, \( AB = 10 \) (since \( BA = 10 \)), \( AC = 24 \), so \( BC = \sqrt{AB^2 + AC^2} = \sqrt{10^2 + 24^2} = \sqrt{100 + 576} = \sqrt{676} = 26 \). Therefore, \( \sin C = \frac{AB}{BC} = \frac{10}{26} = \frac{5}{13} \). Wait, or alternatively, in triangle \( BDE \), since \( DE \parallel AC \), \( \angle C = \angle BED \), so \( \sin C = \sin \angle BED = \frac{BD}{BE} = \frac{5}{13} \). That's simpler. Because \( \angle BED \) and \( \angle C \) are equal (corresponding angles, since \( DE \parallel AC \)), so \( \sin C = \sin \angle BED = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BD}{BE} = \frac{5}{13} \).

Answer:

\( \frac{5}{13} \)