This worksheet involves solving for missing variables in triangles based on angles and side lengths. The triangles appear to involve special right triangles, such as:

1. 45°-45°-90° triangles (isosceles right triangles, where legs are equal, and the hypotenuse is $\sqrt{2}$ times the leg).
2. 30°-60°-90° triangles (where the relationships between the sides are 1 : $\sqrt{3}$ : 2).

Let me solve these step-by-step:

Problem 1:

45°-45°-90° triangle:

• Hypotenuse = 13.
• Relationship: Leg = $\frac{\text{hypotenuse}}{\sqrt{2}} = \frac{13}{\sqrt{2}} = \frac{13\sqrt{2}}{2}$.

Thus:

• $x = y = \frac{13\sqrt{2}}{2}$.

Problem 3:

30°-60°-90° triangle:

• Shorter leg ($y'$) = 3 (opposite 30°).
• Relationship:
  • Longer leg = $y' \cdot \sqrt{3} = 3\sqrt{3}$.
  • Hypotenuse = $2y' = 6$.

Thus:

• $x = 6$, $y = 3\sqrt{3}$.

Problem 5:

45°-45°-90° triangle:

• Hypotenuse = $10\sqrt{2}$.
• Relationship: Leg = $\frac{\text{hypotenuse}}{\sqrt{2}} = \frac{10\sqrt{2}}{\sqrt{2}} = 10$.

Thus:

• $x = y = 10$.

Problem 7:

45°-45°-90° triangle:

• Leg = $2\sqrt{14}$.
• Relationship: Hypotenuse = $\text{leg} \cdot \sqrt{2} = 2\sqrt{14} \cdot \sqrt{2} = 2\sqrt{28} = 4\sqrt{7}$.

Thus:

• $x = y = 2\sqrt{14}$, $y = 4\sqrt{7}$.

Problem 9:

30°-60°-90° triangle:

• Hypotenuse = $22\sqrt{3}$.
• Relationship:
  • Shorter leg = $\frac{\text{hypotenuse}}{2} = \frac{22\sqrt{3}}{2} = 11\sqrt{3}$.
  • Longer leg = Shorter leg $\cdot \sqrt{3} = 11\sqrt{3} \cdot \sqrt{3} = 33$.

Thus:

• $x = 33$, $y = 11\sqrt{3}$.

Problem 11:

45°-45°-90° triangle:

• Hypotenuse = $\sqrt{10}$.
• Relationship: Leg = $\frac{\text{hypotenuse}}{\sqrt{2}} = \frac{\sqrt{10}}{\sqrt{2}} = \sqrt{5}$.

Thus:

• $x = y = \sqrt{5}$.

Problem 13:

Combination of 30°-60°-90° and 45°-45°-90° triangles:

• Longer leg of 30°-60°-90° is shared with the hypotenuse of the 45°-45°-90° triangle.

1. 30°-60°-90° triangle:

   • Hypotenuse = 17.
   • Shorter leg = $\frac{\text{hypotenuse}}{2} = \frac{17}{2}$.
   • Longer leg = $\text{shorter leg} \cdot \sqrt{3} = \frac{17}{2} \cdot \sqrt{3} = \frac{17\sqrt{3}}{2}$.

2. 45°-45°-90° triangle:

   • Hypotenuse = Longer leg of the 30°-60°-90° triangle = $\frac{17\sqrt{3}}{2}$.
   • Legs = $\frac{\text{hypotenuse}}{\sqrt{2}} = \frac{\frac{17\sqrt{3}}{2}}{\sqrt{2}} = \frac{17\sqrt{6}}{4}$.

Thus:

• $x = \frac{17}{2}$,
• $y = \frac{17\sqrt{3}}{2}$,
• $z = \frac{17\sqrt{6}}{4}$.

Let me know if you'd like detailed explanations for specific steps!

Here are 5 related questions for further practice:

1. What are the side ratios of a 30°-60°-90° triangle?
2. How do you rationalize a denominator like $\frac{5}{\sqrt{2}}$?
3. Can the Pythagorean Theorem verify side relationships in special triangles?
4. How do you derive the side lengths for a 45°-45°-90° triangle?
5. If one leg of a 30°-60°-90° triangle is 7, what are the other two sides?

Tip: Always memorize the ratios for special right triangles—they save time and reduce errors!