The missing sides are w = 6 cm, [tex]x=3\sqrt{2}[/tex] cm, [tex]y=6\sqrt{2}[/tex] cm, [tex]z=3\sqrt{6}[/tex] cm.
Solution:
In a large right triangle, θ = 45° and hypotenuse = y
Using basic trigonometric ratios,
[tex]$\sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}[/tex]
[tex]$\sin45^\circ=\frac{6}{y}[/tex]
[tex]$\frac{1}{\sqrt{2} } =\frac{6}{y}[/tex]
Do cross multiplication, we get
[tex]y=6\sqrt{2}[/tex] cm
[tex]$\cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}[/tex]
[tex]$\cos45^\circ=\frac{w}{6\sqrt{2} }[/tex]
[tex]$\frac{1}{\sqrt{2} } =\frac{w}{6\sqrt{2} }[/tex]
Do cross multiplication, we get
w = 6 cm
In a small right triangle, θ = 30° and hypotenuse = w
Using basic trigonometric ratios,
[tex]$\sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}[/tex]
[tex]$\sin30^\circ=\frac{x}{6\sqrt{2} }[/tex]
[tex]$\frac{1}{2 } =\frac{x}{6\sqrt{2} }[/tex]
Do cross multiplication, we get
[tex]$\frac{6\sqrt{2} }{2}=x[/tex]
[tex]x=3\sqrt{2}[/tex] cm
[tex]$\cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}[/tex]
[tex]$\cos30^\circ=\frac{z}{6\sqrt{2} }[/tex]
[tex]$\frac{\sqrt{3}}{2 } =\frac{z}{6\sqrt{2} }[/tex]
Do cross multiplication, we get
[tex]$\frac{6\sqrt{6}}{2}=z[/tex]
[tex]z=3\sqrt{6}[/tex] cm
The missing sides are w = 6 cm, [tex]x=3\sqrt{2}[/tex] cm, [tex]y=6\sqrt{2}[/tex] cm, [tex]z=3\sqrt{6}[/tex] cm.
