Solution:

∵AD⊥BC

∴∠ ADC = 90 (vertical direction)

∫∠C = 70。

∴∠DAC= 180 -(∠ADC+∠C)

= 180 -(90 +70 )

= 20 (triangle interior angle sum theorem)

And < EAD = 10.

∴∠EAC=∠EAD+∠DAC

= 10 +20

=30

∫AE segmentation∠ ∠BAC

∴∠ BAE =∠ CAE = 30 (the nature of angular bisector)

∴∠BAC=2∠EAC=2×30 =60

∴∠B= 180 -∠BAC-∠C

= 180 -60 -70

= 50 (triangle interior angle sum theorem)

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