1. **State the problem:** Given a right triangle ABC with right angle at C, and segment CE perpendicular to AB, creating two right triangles BEC (with angle B = 60°) and CEA (with angle A = 30°). Given $x = BE = 7\sqrt{3}$, find $a = BC$, $CE$, $y = EA$, and $b = CA$. 2. **Recall special right triangle ratios:** - In a 30°-60°-90° triangle, the sides are in ratio $1 : \sqrt{3} : 2$ opposite to angles $30^\circ : 60^\circ : 90^\circ$ respectively. 3. **Analyze triangle BEC (30°-60°-90°):** - Angle at B is $60^\circ$, angle at C is $90^\circ$, so angle at E is $30^\circ$. - Given $BE = x = 7\sqrt{3}$ is opposite the $30^\circ$ angle at E. - In 30°-60°-90°, side opposite 30° is half the hypotenuse. - So hypotenuse $BC = a = 2 \times BE = 2 \times 7\sqrt{3} = 14\sqrt{3}$. 4. **Find $CE$ in triangle BEC:** - Side opposite $60^\circ$ is $CE$. - Using ratio, $CE = BE \times \sqrt{3} = 7\sqrt{3} \times \sqrt{3} = 7 \times 3 = 21$. 5. **Analyze triangle CEA (30°-60°-90°):** - Angle at A is $30^\circ$, angle at C is $90^\circ$, so angle at E is $60^\circ$. - $CE$ is hypotenuse of triangle CEA, so $CE = 21$. - Side opposite $30^\circ$ is $y = EA$. - Side opposite $60^\circ$ is $b = CA$. 6. **Find $y = EA$:** - Side opposite $30^\circ$ is half the hypotenuse. - So $y = \frac{CE}{2} = \frac{21}{2} = 10.5$. 7. **Find $b = CA$:** - Side opposite $60^\circ$ is $b = y \times \sqrt{3} = 10.5 \times \sqrt{3} = 10.5\sqrt{3}$. **Final answers:** $$a = 14\sqrt{3}, \quad CE = 21, \quad y = 10.5, \quad b = 10.5\sqrt{3}$$