step 1
In the smaller right triangle of the left we have that
tan(60)=y/a ------> opposite side divided by the adjacent side
Remember that
[tex]\tan (60^o)=\sqrt[]{3}[/tex]so
[tex]\begin{gathered} \frac{y}{a}=\sqrt[]{3} \\ \\ y=a\sqrt[]{3} \end{gathered}[/tex]step 2
In the right triangle of the right
we have
tan(30)=y/b
Remember that
[tex]\tan (30^o)=\frac{1}{\sqrt[]{3}}[/tex]so
[tex]\begin{gathered} \frac{1}{\sqrt[]{3}}=\frac{y}{b} \\ \\ y=\frac{b}{\sqrt[]{3}} \end{gathered}[/tex]step 3
Equate equation step 1 and equation step 2
[tex]\begin{gathered} a\sqrt[]{3}=\frac{b}{\sqrt[]{3}} \\ 3a=b \end{gathered}[/tex]Remember that
a+b=15
substitute the value of b =3a
a+3a=15
4a=15
a=15/4
therefore
b=3(15/4)
