1. **Problem Statement:** We have two right triangles with sides labeled as follows: the first triangle has sides $x$, $y$, and $6$ with a height of $2$ corresponding to side $6$. The second triangle has sides $z$, $y$, and an unknown side, with angles labeled $\frac{1}{3}$ and $\frac{1}{6}$ opposite sides $z$ and $y$ respectively. We want to find relationships between these sides and angles. 2. **Step 1: Analyze the first triangle.** It is a right triangle with hypotenuse $6$ and height $2$ drawn to the hypotenuse. The height divides the hypotenuse into two segments. Using the property of right triangles, the height $h$ satisfies: $$h = \frac{xy}{6}$$ Given $h=2$, we have: $$2 = \frac{xy}{6} \implies xy = 12$$ 3. **Step 2: Use Pythagoras theorem in the first triangle.** Since the triangle is right angled, if $x$ and $y$ are legs, then: $$x^2 + y^2 = 6^2 = 36$$ 4. **Step 3: Analyze the second triangle.** It is also right angled with sides $z$, $y$, and an unknown side. The angles opposite $z$ and $y$ are $\frac{1}{3}$ and $\frac{1}{6}$ respectively (likely in radians or fractions of $\pi$). Assuming these are angles in radians, the sum of angles in a triangle is $\pi$, so the right angle is $\frac{\pi}{2}$. 5. **Step 4: Use trigonometric ratios in the second triangle.** Let the right angle be between sides $z$ and $y$. Then: - Opposite $z$ is angle $\frac{1}{3}$ - Opposite $y$ is angle $\frac{1}{6}$ Using tangent or sine ratios, for angle $\frac{1}{3}$ opposite $z$: $$\tan\left(\frac{1}{3}\right) = \frac{z}{\text{adjacent side}}$$ Similarly for angle $\frac{1}{6}$ opposite $y$: $$\tan\left(\frac{1}{6}\right) = \frac{y}{\text{adjacent side}}$$ Since both share the same adjacent side, the ratio of $z$ to $y$ is: $$\frac{z}{y} = \frac{\tan\left(\frac{1}{3}\right)}{\tan\left(\frac{1}{6}\right)}$$ 6. **Step 5: Express $z$ in terms of $y$ using the ratio:** $$z = y \cdot \frac{\tan\left(\frac{1}{3}\right)}{\tan\left(\frac{1}{6}\right)}$$ 7. **Step 6: Summary of key equations:** - From first triangle: $$xy = 12$$ and $$x^2 + y^2 = 36$$ - From second triangle: $$z = y \cdot \frac{\tan\left(\frac{1}{3}\right)}{\tan\left(\frac{1}{6}\right)}$$ These equations relate the sides $x$, $y$, and $z$ based on the given information. **Final answer:** $$xy = 12$$ $$x^2 + y^2 = 36$$ $$z = y \cdot \frac{\tan\left(\frac{1}{3}\right)}{\tan\left(\frac{1}{6}\right)}$$