Answer:
Here we have right triangles, which we can solve using trigonometric reasons, bceause we know the acute angles and the hypothenuse of the bigger right triangle.
Let's the sin function
[tex]sin30=\frac{x}{15}\\ x=15(sin30)\\x=15(\frac{1}{2} )\\x=\frac{15}{2}[/tex]
Then, we use the cosine function
[tex]cos30=\frac{z}{15}\\ z=15(\frac{\sqrt{3} }{2} )[/tex]
Now, we use the sin to find [tex]y[/tex] in the middle size right triangle
[tex]sin30=\frac{y}{15\frac{\sqrt{3} }{2} }\\ y=\frac{1}{2} \times 15\frac{\sqrt{3} }{2}\\ y=15\frac{\sqrt{3} }{4}[/tex]
Then, we use cosine in the smaller triangle
[tex]cos60=\frac{a}{\frac{15}{2} } \\a=\frac{15}{2} \times \frac{1}{2}\\ a=\frac{15}{4}[/tex]
Also, we know that
[tex]a+b=15[/tex], and [tex]a=\frac{15}{4}[/tex], so
[tex]b=15-a\\b=15-\frac{15}{4}\\ b=\frac{60-15}{4}\\ b=\frac{45}{4}[/tex]
Therefore, all the answers are
[tex]x= \frac{15}{2}\\ y=15\frac{\sqrt{3} }{4}\\ z=15\frac{\sqrt{3} }{2}\\a=\frac{15}{4}\\ b=\frac{45}{4}[/tex]
