Question

11.
14
22

1. 25

78
13.
36
31
14.
10
19

1. find de.

c
24
f
19
d
e

Explanation:

To solve for \( DE \), we analyze the given diagram. We assume triangle \( CFE \) and triangle \( FDE \) are right triangles (since \( \angle CFE \) and \( \angle FDE \) appear to be right angles). First, we find the length of \( FD \) using the right triangle \( CFE \) with \( CF = 24 \) and \( \angle C = 41^\circ \). Then we use the right triangle \( FDE \) with \( FD \) and \( \angle F = 19^\circ \) to find \( DE \).

Step 1: Find \( FD \)

In right triangle \( CFE \), \( \sin(41^\circ) = \frac{FD}{CF} \) (opposite over hypotenuse).
Given \( CF = 24 \), we have:
\[
FD = CF \cdot \sin(41^\circ)
\]
\[
FD = 24 \cdot \sin(41^\circ)
\]
Using a calculator, \( \sin(41^\circ) \approx 0.6561 \), so:
\[
FD \approx 24 \cdot 0.6561 \approx 15.7464
\]

Step 2: Find \( DE \)

In right triangle \( FDE \), \( \sin(19^\circ) = \frac{DE}{FD} \) (opposite over hypotenuse).
We know \( FD \approx 15.7464 \), so:
\[
DE = FD \cdot \sin(19^\circ)
\]
\[
DE \approx 15.7464 \cdot \sin(19^\circ)
\]
Using a calculator, \( \sin(19^\circ) \approx 0.3256 \), so:
\[
DE \approx 15.7464 \cdot 0.3256 \approx 5.12
\]

Answer:

\( DE \approx \boldsymbol{5.12} \) (rounded to two decimal places)