1. **Problem Statement:** We have two figures with an inclined pole. In Figure 1, the pole makes an angle $x$ with the ground and the height of the top of the pole from the ground is 14 meters. In Figure 2, the pole is raised to an angle $y$ with the ground, and the height increases by 4 meters, making it 18 meters. Given $\tan(x) = \frac{7}{6}$, find $\tan(y)$. 2. **Understanding the problem:** The height of the pole corresponds to the vertical side of a right triangle formed by the pole and the ground. If the length of the pole is $L$, then height $= L \sin(\theta)$ and the base $= L \cos(\theta)$. 3. **Using the given information:** From Figure 1: $$\text{height}_1 = L \sin(x) = 14$$ From Figure 2: $$\text{height}_2 = L \sin(y) = 18$$ 4. **Express $L$ in terms of $x$:** $$L = \frac{14}{\sin(x)}$$ 5. **Find $\sin(x)$ and $\cos(x)$ from $\tan(x)$:** Given: $$\tan(x) = \frac{7}{6}$$ Recall: $$\tan(x) = \frac{\sin(x)}{\cos(x)}$$ Let $\sin(x) = 7k$ and $\cos(x) = 6k$ for some $k$. Using Pythagoras: $$\sin^2(x) + \cos^2(x) = 1$$ $$ (7k)^2 + (6k)^2 = 1$$ $$49k^2 + 36k^2 = 1$$ $$85k^2 = 1$$ $$k = \frac{1}{\sqrt{85}}$$ So: $$\sin(x) = \frac{7}{\sqrt{85}}, \quad \cos(x) = \frac{6}{\sqrt{85}}$$ 6. **Calculate $L$:** $$L = \frac{14}{\sin(x)} = \frac{14}{7/\sqrt{85}} = 14 \times \frac{\sqrt{85}}{7} = 2 \sqrt{85}$$ 7. **Use $L$ to find $\sin(y)$:** $$L \sin(y) = 18 \implies \sin(y) = \frac{18}{L} = \frac{18}{2 \sqrt{85}} = \frac{9}{\sqrt{85}}$$ 8. **Find $\cos(y)$ using Pythagoras:** $$\cos(y) = \sqrt{1 - \sin^2(y)} = \sqrt{1 - \left(\frac{9}{\sqrt{85}}\right)^2} = \sqrt{1 - \frac{81}{85}} = \sqrt{\frac{4}{85}} = \frac{2}{\sqrt{85}}$$ 9. **Calculate $\tan(y)$:** $$\tan(y) = \frac{\sin(y)}{\cos(y)} = \frac{9/\sqrt{85}}{2/\sqrt{85}} = \frac{9}{2}$$ **Final answer:** $$\boxed{\frac{9}{2}}$$ This corresponds to option A.