Answer:
Step-by-step explanation:
∠2 is a right angle, so ∠3 is complementary to the 39° angle shown. It is ...
∠3 = 90° -39° = 51°
ΔDAB is isosceles, so ∠DBA ≅ ∠3 and ∠1 ≅ 39°.
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If you're starting completely from scratch, you might go to the trouble to show ΔABC ≅ ΔADC, hence AC is a bisector of angles A and C. If X is where the diagonals cross, then you can show ΔABX ≅ ΔADX and ∠2 is 90°.
I prefer to start with the fact that the diagonals of a (symmetrical) kite cross at right angles. That makes all the triangles right triangles, so the angles are easily found, given any one of them. (Acute angles of a right triangle are complementary.) Of course, the triangles on one side of the line of symmetry (AC) are congruent to those on the other side.
