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 the right triangle

Triangle \( ABC \) is a right triangle with \( \angle A = 90^\circ \), \( AB \) and \( AC \) as the legs, and \( BC \) as the hypotenuse? Wait, no, wait. Wait, \( AD \) and \( AC \) – wait, no, \( \angle A \) is right, so \( AB \) and \( AC \) are perpendicular. Wait, but in the diagram, \( BD = 5 \), \( DE = 12 \), \( AC = 24 \), \( BE = 13 \)? Wait, no, \( BE \) is 13? Wait, \( DE = 12 \), \( BD = 5 \), so triangle \( BDE \) is a right triangle with legs 5 and 12, hypotenuse 13 (since \( 5^2 + 12^2 = 25 + 144 = 169 = 13^2 \)). So \( \triangle BDE \sim \triangle BAC \) by AA similarity (both right triangles, and \( \angle B \) is common). So corresponding sides are proportional. But to find \( \sin C \), we can use the right triangle \( ABC \). In right triangle \( ABC \), \( \sin C = \frac{\text{opposite side to } C}{\text{hypotenuse}} \). The opposite side to \( C \) is \( AB \), and the hypotenuse is \( BC \). Wait, but we can also use \( \triangle BDE \) and \( \triangle BAC \) similarity. Since \( DE \parallel AC \) (because \( \angle A = \angle D = 90^\circ \), so \( DE \parallel AC \)), so \( \triangle BDE \sim \triangle BAC \). So \( \frac{BD}{BA} = \frac{DE}{AC} \). We know \( BD = 5 \), \( DE = 12 \), \( AC = 24 \). So \( \frac{5}{BA} = \frac{12}{24} \), so \( \frac{5}{BA} = \frac{1}{2} \), so \( BA = 10 \). Then \( AB = AD + BD \), but \( AD = AB - BD = 10 - 5 = 5 \)? Wait, no, wait, \( BD = 5 \), so \( BA = BD + DA \), but if \( \frac{BD}{BA} = \frac{12}{24} = \frac{1}{2} \), then \( BD = \frac{1}{2} BA \), so \( BA = 2 \times BD = 10 \). So \( AB = 10 \), \( AC = 24 \). Then \( BC \) can be found by Pythagoras: \( BC = \sqrt{AB^2 + AC^2} = \sqrt{10^2 + 24^2} = \sqrt{100 + 576} = \sqrt{676} = 26 \). Then \( \sin C = \frac{AB}{BC} = \frac{10}{26} = \frac{5}{13} \)? Wait, no, wait. Wait, in right triangle \( ABC \), angle \( C \) has opposite side \( AB \) and adjacent side \( AC \). Wait, no: in right triangle \( ABC \), \( \angle A = 90^\circ \), so the sides: \( AB \) is vertical leg, \( AC \) is horizontal leg, \( BC \) is hypotenuse. So angle \( C \) is at \( C \), so the sides: opposite to \( C \) is \( AB \), adjacent is \( AC \), hypotenuse is \( BC \). So \( \sin C = \frac{AB}{BC} \). We found \( AB = 10 \), \( BC = 26 \), so \( \sin C = \frac{10}{26} = \frac{5}{13} \). Alternatively, since \( \triangle BDE \) is similar to \( \triangle BAC \), \( \angle C = \angle BED \), so \( \sin C = \sin \angle BED \). In \( \triangle BDE \), \( \sin \angle BED = \frac{BD}{BE} = \frac{5}{13} \), which matches. So that's the sine of angle \( C \).

Step1: Confirm the right triangle and sides

In right triangle \( ABC \), \( \angle A = 90^\circ \). We can find \( AB \) using the similarity of \( \triangle BDE \) and \( \triangle BAC \). Since \( DE \parallel AC \), \( \triangle BDE \sim \triangle BAC \) (AA similarity: \( \angle B \) is common, \( \angle BDE = \angle BAC = 90^\circ \)). Thus, \( \frac{BD}{BA} = \frac{DE}{AC} \). Given \( BD = 5 \), \( DE = 12 \), \( AC = 24 \), substitute:
\( \frac{5}{BA} = \frac{12}{24} \)
Simplify \( \frac{12}{24} = \frac{1}{2} \), so:
\( \frac{5}{BA} = \frac{1}{2} \implies BA = 10 \).

Step2: Calculate hypotenuse \( BC \)

In right triangle \( ABC \), by Pythagoras:
\( BC = \sqrt{AB^2 + AC^2} = \sqrt{10^2 + 24^2} = \sqrt{100 + 576} = \sqrt{676} = 26 \).

Step3: Find \( \sin C \)

In right triangle \( ABC \), \( \sin C = \frac{\text{opposite side to } C}{\text{hypotenuse}} = \frac{AB}{BC} \). Substitute \( AB = 10 \) and \( BC = 26 \):
\( \sin C = \…

Answer:

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