Question

Solve for xx. Round to the nearest tenth, if necessary.
D
E
F
18°
x
52

dF 62
e-d = x
e has a 90 degree angle
Answer
Attempt 1 out of 2

Answer 1

To solve for \( x \) in the given triangle problem, we can apply the sine rule since we have angles and sides given.

Let's denote the triangle as \( \triangle DEF \):
- Angle \( D \) is \( 18^\circ \)
- Angle \( E \) is \( 90^\circ \)
- Angle \( F \) can be found since the sum of angles in a triangle is \( 180^\circ \).

Calculating angle \( F \):
\[
F = 180^\circ - D - E = 180^\circ - 18^\circ - 90^\circ = 72^\circ
\]

Now, we have angles:
- \( \angle D = 18^\circ \)
- \( \angle E = 90^\circ \)
- \( \angle F = 72^\circ \)

Given the sides adjacent to these angles:
- If we denote:
- \( d = DF = 62 \)
- \( e = DE = x \)
- \( f = EF = 52 \)

Now, we will use the sine rule:
\[
\frac{d}{\sin D} = \frac{e}{\sin E} = \frac{f}{\sin F}
\]

Using the sine rule for sides \( d \) and \( e \):
\[
\frac{62}{\sin(18^\circ)} = \frac{x}{\sin(90^\circ)}
\]

Since \( \sin(90^\circ) = 1 \), we can simplify this to:
\[
\frac{62}{\sin(18^\circ)} = x
\]

Now calculating \( \sin(18^\circ) \):
\[
\sin(18^\circ) \approx 0.309
\]

Now plugging this value back into the equation:
\[
x = \frac{62}{0.309} \approx 200.65
\]

Rounding to the nearest tenth gives:
\[
x \approx 200.7
\]

Thus, the final answer for \( x \) is:
\[
\boxed{200.7}
\]