Question

solve for x using properties of special right triangles
(images of right triangles with various angles and side lengths, each labeled to solve for x)
solve for x using properties of special right triangles
(images of right triangles with various angles and side lengths, each labeled to solve for x)
solve for x using properties of special right triangles
(images of right triangles with various angles and side lengths, each labeled to solve for x)
solve for x using properties of special right triangles
(images of right triangles with various angles and side lengths, each labeled to solve for x)
solve for x using properties of special right triangles
(images of right triangles with various angles and side lengths, each labeled to solve for x)
solve for x using properties of special right triangles
(images of right triangles with various angles and side lengths, each labeled to solve for x)
solve for x using properties of special right triangles
(images of right triangles with various angles and side lengths, each labeled to solve for x)

Explanation:

Step1: Identify triangle type (top-left)

30-60-90 triangle, hypotenuse=10. Opposite 60° is x: $x = 10 \cdot \sin60^\circ = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3}$

Step2: Identify triangle type (top-right)

Isosceles right triangle (legs equal), hypotenuse=7: $x = \frac{7}{\sqrt{2}} = \frac{7\sqrt{2}}{2}$

Step3: Identify triangle type (middle-left)

45-45-90 triangle, leg=6: $x = 6\sqrt{2}$

Step4: Identify triangle type (middle-right)

30-60-90 triangle, hypotenuse=18. Opposite 30° is x: $x = 18 \cdot \sin30^\circ = 18 \cdot \frac{1}{2} = 9$

Step5: Identify triangle type (bottom-left, first)

45-45-90 triangle, hypotenuse=$20\sqrt{41}$: $x = \frac{20\sqrt{41}}{\sqrt{2}} = 10\sqrt{82}$

Step6: Identify triangle type (bottom-left, second)

30-60-90 triangle, adjacent to 60° is $30\sqrt{3}$: $x = 30\sqrt{3} \cdot \tan60^\circ = 30\sqrt{3} \cdot \sqrt{3} = 90$

Step7: Identify triangle type (bottom-right)

30-60-90 triangle, adjacent to 60° is 12: $x = 12 \cdot \tan60^\circ = 12\sqrt{3}$

Answer:

Top-left: $5\sqrt{3}$
Top-right: $\frac{7\sqrt{2}}{2}$
Middle-left: $6\sqrt{2}$
Middle-right: 9
Bottom-left (first): $10\sqrt{82}$
Bottom-left (second): 90
Bottom-right: $12\sqrt{3}$