The first step is to find angleC∠C.
We know that angleA = 30∠A=30 and angleB = 90∠B=90. Since all triangles have angles that add up to 180,
angleC = 180 - 30 - 90∠C=180−30−90
angleC = 60∠C=60
We can use this to complete the next step: finding angleCEB∠CEB (this is complementary with angleCED∠CED, so we'll be able to find angleCED∠CED by subtracting angleCEB∠CEB from 90).
We can do this by taking triangleCEB△CEB as its own triangle. angleB∠B is still 90. angleECB∠ECB is angleC-:2∠C÷2, since bar(CE)¯¯¯¯¯¯CE bisects angleC∠C. So angleECB = 30∠ECB=30.
angleCEB = 180 - 90- 30∠CEB=180−90−30
angleCEB = 60∠CEB=60
angleCED = 90 - 60∠CED=90−60
color(red)(angleCED = 30)∠CED=30
