Answer:
59°
Step-by-step explanation:
In triangle DAB, by Pythagoras theorem:
[tex] DB= \sqrt {6^2 +5^2} =\sqrt{36+25}=\sqrt{61}[/tex]
Next, in triangle DCB, by Pythagoras theorem:
[tex] DB= \sqrt {(\sqrt {61} )^2 - 4^2} =\sqrt{61-16}=\sqrt{45}[/tex]
Now, by sin rule.
[tex] \frac{ \sin \: p}{ \sqrt{45} } = \frac{ \sin 90 \degree}{ \sqrt{61} } \\ \\ \frac{ \sin \: p}{ 6.70820393} = \frac{ 1}{ 7.81024968 } \\ \\ \sin \: p = \frac{ 6.70820393}{7.81024968} \\ \\ \sin \: p = 0.858897501 \\ \\p = 59.19301883 \degree \\ \\ p \approx 59 \degree[/tex]
