Answer:
[tex]AD \approx 6.4\\BC= 4\\AB \approx 6.4\\\angle B = \angle D \approx 51.3\°[/tex]
Step-by-step explanation:
The image attached shows the situation described in the problem.
Notice that we can use Pythagorean's Theorem to find AD, which is hypothenuse of the right triangle ACD
[tex]AD^{2}=5^{2}+4^{2}\\ AD=\sqrt{25+16}\\AD= \sqrt{41}\approx 6.4[/tex]
Additionally, triangle ABD is isosceles, its height is perpendicular bisector of side BD, that means BC = 4, and AB = 6.4.
Now, we can use trigonometric reasons to find angle D
[tex]tan D=\frac{5}{4}\\ D=tan^{-1}(\frac{5}{4} ) \approx 51.3 \°[/tex]
And, [tex]B \approx 51.3 \°[/tex], because it's an isosceles triangle.
Therefore, the values we found were
[tex]AD \approx 6.4\\BC= 4\\AB \approx 6.4\\\angle B = \angle D \approx 51.3\°[/tex]
