Given:
A triangle
Required:
To calculate the value of v, w
Explanation:
Solving steps
[tex]\begin{gathered} sin(\angle BAC)=\frac{BC}{AB}\text{ substitute BC=v AB=15} \\ \\ sin(65\degree)=\frac{v}{15}\text{ v=13.595} \end{gathered}[/tex][tex]\begin{gathered} \frac{BD}{sin(\angle BAD)}=\frac{AB}{sin(\angle ADB)} \\ \\ \frac{21}{sin(65\degree)}=\frac{15}{sin(w)} \\ \\ w=40.343\degree \end{gathered}[/tex]
Required answer:
v= 13.595
w=40.343
