Answer:
[tex]z=\frac{15\sqrt{3}}{2}[/tex]
Step-by-step explanation:
In leftmost triangle:
opposite side =y
adjacent=a
now, we can use trig formula
[tex]tan(60)=\frac{y}{a}[/tex]
now, we can solve for y
[tex]y=atan(60)[/tex]
In rightmost triangle:
adjacent=b
opposite=y
a+b=15
b=15-a
now, we can use trig formula
[tex]tan(30)=\frac{y}{15-a}[/tex]
now, we can solve for y
[tex]y=(15-a)tan(30)[/tex]
now, we can set them equal
and then we can solve for a
[tex]atan(60)=(15-a)tan(30)[/tex]
[tex]\sqrt{3}a\cdot \:3=\frac{\sqrt{3}\left(15-a\right)}{3}\cdot \:3[/tex]
[tex]4\sqrt{3}a=15\sqrt{3}[/tex]
[tex]a=\frac{15}{4}[/tex]
now, we can find b
[tex]b=15-\frac{15}{4}[/tex]
[tex]b=\frac{45}{4}[/tex]
now, we can use trig formula
[tex]cos(30)=\frac{b}{z}[/tex]
now, we can find z
[tex]z=\frac{\frac{45}{4}}{cos(30)}[/tex]
we can simplify it
and we get
[tex]z=\frac{15\sqrt{3}}{2}[/tex]