Mathematics: Find the missing sides of the figure picture below

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Find the missing sides of the figure picture below

broang23

20.01.2022

The missing sides are w = 6 cm, $x=3\sqrt{2}$ cm, $y=6\sqrt{2}$ cm, $z=3\sqrt{6}$ cm.

Solution:

In a large right triangle, θ = 45° and hypotenuse = y

Using basic trigonometric ratios,

$$\sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}$

$$\sin45^\circ=\frac{6}{y}$

$$\frac{1}{\sqrt{2} } =\frac{6}{y}$

Do cross multiplication, we get

$y=6\sqrt{2}$ cm

$$\cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}$

$$\cos45^\circ=\frac{w}{6\sqrt{2} }$

$$\frac{1}{\sqrt{2} } =\frac{w}{6\sqrt{2} }$

Do cross multiplication, we get

w = 6 cm

In a small right triangle, θ = 30° and hypotenuse = w

Using basic trigonometric ratios,

$$\sin \theta=\frac{\text { Opposite side of } \theta}{\text { Hypotenuse }}$

$$\sin30^\circ=\frac{x}{6\sqrt{2} }$

$$\frac{1}{2 } =\frac{x}{6\sqrt{2} }$

Do cross multiplication, we get

$$\frac{6\sqrt{2} }{2}=x$

$x=3\sqrt{2}$ cm

$$\cos \theta=\frac{\text { Adjacent side of } \theta}{\text { Hypotenuse }}$

$$\cos30^\circ=\frac{z}{6\sqrt{2} }$

$$\frac{\sqrt{3}}{2 } =\frac{z}{6\sqrt{2} }$

Do cross multiplication, we get

$$\frac{6\sqrt{6}}{2}=z$

$z=3\sqrt{6}$ cm

The missing sides are w = 6 cm, $x=3\sqrt{2}$ cm, $y=6\sqrt{2}$ cm, $z=3\sqrt{6}$ cm.

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