Answer: [tex]32.98^{\circ}[/tex]
Step-by-step explanation:
Since, By the given diagram,
In quadrilateral ABCD, all angles are right angles ⇒ DC = AB = 15 cm and CB = DA
Also, In triangle DFA,
[tex]sin 56^{\circ} = \frac{DA}{DF}[/tex]
⇒ [tex]sin 56^{\circ} = \frac{DA}{16}[/tex]
⇒ [tex]16\times sin 56^{\circ}=DA[/tex]
⇒ [tex]13.2646011609\text{ cm} = DA[/tex]
Now, In triangle, DEC,
[tex]tan x = \frac{EC}{DC}[/tex]
⇒ [tex]tan x = \frac{EB-CB}{DC}[/tex]
⇒ [tex]tan x = \frac{EB-DA}{DC}[/tex]
⇒ [tex]tan x = \frac{23-13.2646011609}{15}[/tex]
⇒ [tex]tanx = \frac{9.73539883912}{15}[/tex]
⇒ [tex]tanx = 0.64902658927[/tex]
⇒ [tex]x = 32.9846428556^{\circ} \approx 32.98^{\circ}[/tex]
