1. **Problem Statement:** Find the cosecant of angle $H$ in right triangle $GHF$ where $\angle F$ is the right angle, $GF = 7\sqrt{3}$, and $FH = 7$. 2. **Recall the definition:** Cosecant of an angle is the reciprocal of the sine of that angle. $$\csc(H) = \frac{1}{\sin(H)}$$ 3. **Find the hypotenuse:** Since $\triangle GHF$ is right angled at $F$, the hypotenuse is $GH$. Use the Pythagorean theorem: $$GH = \sqrt{GF^2 + FH^2} = \sqrt{(7\sqrt{3})^2 + 7^2} = \sqrt{49 \times 3 + 49} = \sqrt{147 + 49} = \sqrt{196} = 14$$ 4. **Find $\sin(H)$:** Sine of angle $H$ is opposite side over hypotenuse. The side opposite $H$ is $GF = 7\sqrt{3}$. $$\sin(H) = \frac{GF}{GH} = \frac{7\sqrt{3}}{14}$$ 5. **Simplify $\sin(H)$:** $$\sin(H) = \frac{7\sqrt{3}}{14} = \frac{\cancel{7}\sqrt{3}}{2\cancel{7}} = \frac{\sqrt{3}}{2}$$ 6. **Find $\csc(H)$:** $$\csc(H) = \frac{1}{\sin(H)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$$ 7. **Rationalize the denominator:** $$\csc(H) = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$ **Final answer:** $$\boxed{\csc(H) = \frac{2\sqrt{3}}{3}}$$