Answer:
[tex]y=\frac{15\sqrt{3}}{4}[/tex]
Step-by-step explanation:
We are given that
[tex]\theta_1=60^{\circ}[/tex]
[tex]\theta_2=30^{\circ}[/tex]
We have to find the missing part.
[tex]\frac{x}{15}=cos\theta_1=cos60^{\circ}[/tex]
Using the formula
[tex]\frac{base}{hypotenuse}=cos\theta[/tex]
[tex]x=15cos60^{\circ}=\frac{15}{2}[/tex]
[tex]\frac{z}{15}=sin60^{\circ}[/tex]
Using the formula
[tex]\frac{Perpendicular\;arm}{hypotenuse}=sin\theta[/tex]
[tex]z=15\times \frac{\sqrt{3}}{2}=\frac{15\sqrt{3}}{2}[/tex]
Now,
[tex]\frac{a}{x}=cos60^{\circ}[/tex]
[tex]\frac{a}{\frac{15}{2}}=\frac{1}{2}[/tex]
[tex]a=\frac{1}{2}\times 15/2=\frac{15}{4}[/tex]
[tex]y=x sin60^{\circ}[/tex]
[tex]y=\frac{15}{2}\times \frac{\sqrt{3}}{2}[/tex]
[tex]y=\frac{15\sqrt{3}}{4}[/tex]
