Question

fencing x - braces are also used to provide support in rectangular fencing. if $ab = 6$ feet, $ad = 2$ feet, and $mangle dae = 65^{circ}$, find $mangle ceb$. round to the nearest tenth, if necessary.

Explanation:

Step1: Find length of AE

First, in right triangle \(DAE\), use cosine to find \(AE\):
\(\cos(\angle DAE) = \frac{AD}{AE}\)
\(AE = \frac{AD}{\cos(\angle DAE)} = \frac{2}{\cos(65^\circ)}\)
Calculate \(\cos(65^\circ) \approx 0.4226\), so \(AE \approx \frac{2}{0.4226} \approx 4.733\) feet.

Step2: Identify congruent segments

Since \(ABCD\) is a rectangle, the diagonals bisect each other, so \(AE = CE\) and \(BE = DE\). Also, \(AB = DC = 6\) feet, and \(BE = AE \approx 4.733\) feet, \(CE \approx 4.733\) feet.

Step3: Use Law of Cosines on \(\triangle CEB\)

We know \(CE = BE \approx 4.733\) ft, \(BC = AD = 2\) ft. Let \(\angle CEB = \theta\):
\(BC^2 = CE^2 + BE^2 - 2 \cdot CE \cdot BE \cdot \cos\theta\)
Substitute values:
\(2^2 = (4.733)^2 + (4.733)^2 - 2 \cdot 4.733 \cdot 4.733 \cdot \cos\theta\)
\(4 = 22.401 + 22.401 - 44.802 \cos\theta\)
\(4 = 44.802 - 44.802 \cos\theta\)

Step4: Solve for \(\cos\theta\)

Rearrange to isolate \(\cos\theta\):
\(44.802 \cos\theta = 44.802 - 4\)
\(44.802 \cos\theta = 40.802\)
\(\cos\theta = \frac{40.802}{44.802} \approx 0.9107\)

Step5: Calculate \(\theta\)

Find the arccosine to get \(\theta\):
\(\theta = \arccos(0.9107) \approx 24.0^\circ\)

Answer:

\(24.0^\circ\)