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
![\tan (60^o)=\sqrt[]{3}](https://img.qammunity.org/2023/formulas/mathematics/college/r15dx5xf73fmzmvu5otn28q6abglcxar9k.png)
so
![\begin{gathered} (y)/(a)=\sqrt[]{3} \\ \\ y=a\sqrt[]{3} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vak0xp8ww2bo5f4oavaksz513qby3b9tw8.png)
step 2
In the right triangle of the right
we have
tan(30)=y/b
Remember that
![\tan (30^o)=\frac{1}{\sqrt[]{3}}](https://img.qammunity.org/2023/formulas/mathematics/college/5nqowwqkgm42jiv9d3ysg89z2xcawk2frz.png)
so
![\begin{gathered} \frac{1}{\sqrt[]{3}}=(y)/(b) \\ \\ y=\frac{b}{\sqrt[]{3}} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4id0og8i7wmr5p7hr2fm80kico6aw0h2fk.png)
step 3
Equate equation step 1 and equation step 2
![\begin{gathered} a\sqrt[]{3}=\frac{b}{\sqrt[]{3}} \\ 3a=b \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bg3l2c7gn6fq2yu9c2wgevuwqp8xhv2z65.png)
Remember that
a+b=15
substitute the value of b =3a
a+3a=15
4a=15
a=15/4
therefore
b=3(15/4)
b=45/4
b=4 5/4