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$tan \ \theta = \frac{5}{3} \ = 1.667$
$\theta \ = 59.0367$
$tan \ \alpha = \frac{3}{5} \ = 0.6$
$\alpha \ = 30.9638$
$\angle BCD = 59.0367 \times 2 \ = 118.073^o$
$\angle BAD = \angle BCD = 118.073^o$
$\angle ABC \ = 30.9638 \times 2 \ = 61.928^o$
$\angle ADC \ = \angle ABC \ = 61.928^o$
raphael answered the question on January 20, 2017 at 05:52
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