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In ∆ABC, m∠ACB = 90°, m∠A = 40°, and D ∈ AB such that CD is perpendicular to side AB. Find m∠DBC and m∠BCD.

In ABC mACB 90 mA 40 and D AB such that CD is perpendicular to side AB Find mDBC and mBCD class=

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tramserran

Answer:  ∠B = 50°

               ∠BCD = 40°

Step-by-step explanation:

ACB is a right triangle where ∠A = 40° and ∠C = 90°.

Use the Triangle Sum Theorem for ΔABC to find ∠B:

∠A + ∠B + ∠C = 180°

40° + ∠B + 90° = 180°

         ∠B + 130° = 180°

                  ∠B = 50°

BCD is a right triangle where ∠B = 50° and ∠D = 90°.

Use the Triangle Sum Theorem for ΔBCD to find ∠C:

∠B + ∠C + ∠D = 180°

50° + ∠C + 90° = 180°

         ∠C + 140° = 180°

                  ∠C = 40°

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