Questions: Solve the following problems. Given: AB ≅ BC and AE=10 in m ∠ FEC=90° m ∠ ABC=130° 30′ Find: m ∠ EBC, AC m ∠ EBC= . AC= in.

Transcript text: Solve the following problems. Given: $\overline{\mathrm{AB}} \cong \overline{\mathrm{BC}}$ and $\mathrm{AE}=10 \mathrm{in}$ \[ \begin{array}{l} \mathrm{m} \angle \mathrm{FEC}=90^{\circ} \\ \mathrm{m} \angle \mathrm{ABC}=130^{\circ} 30^{\prime} \end{array} \] Find: $m \angle E B C, A C$ $\mathrm{m} \angle E B C=$ $\square$ . $A C=$ $\square$ in.

Solution

Solution Steps

Step 1: Identify the Given Information

$ \overline{AB} \cong \overline{BC} $
$ AE = 10 $ inches
$ m\angle FEC = 90^\circ $
$ m\angle ABC = 130^\circ 30' $

Step 2: Determine the Measure of $ \angle EBC $

Since $ \overline{AB} \cong \overline{BC} $, triangle $ \triangle ABC $ is isosceles with $ AB = BC $. Therefore, $ \angle BAC = \angle BCA $.

Given $ m\angle ABC = 130^\circ 30' $, the sum of the angles in triangle $ \triangle ABC $ is $ 180^\circ $. Let $ x $ be the measure of $ \angle BAC $ and $ \angle BCA $.

\[ x + x + 130^\circ 30' = 180^\circ \]

\[ 2x + 130^\circ 30' = 180^\circ \]

\[ 2x = 180^\circ - 130^\circ 30' \]

\[ 2x = 49^\circ 30' \]

\[ x = 24^\circ 45' \]

Thus, $ m\angle BAC = m\angle BCA = 24^\circ 45' $.

Step 3: Calculate $ AC $

Since $ \overline{AB} \cong \overline{BC} $ and $ AE = 10 $ inches, $ E $ is the midpoint of $ \overline{AC} $. Therefore, $ E $ divides $ \overline{AC} $ into two equal segments.

Given $ m\angle FEC = 90^\circ $, triangle $ \triangle AEC $ is a right triangle with $ AE = 10 $ inches and $ \angle FEC = 90^\circ $.

Using the Pythagorean theorem in $ \triangle AEC $:

\[ AC = 2 \times AE = 2 \times 10 = 20 \text{ inches} \]