1. The problem involves using the SOH CAH TOA mnemonic to solve for $x$ in a right triangle. 2. The given equation is $a \times \tan 62^\circ = \frac{x}{7} \times 9$ and the value of $x$ is given as 16.9 cm. 3. SOH CAH TOA helps us remember the trigonometric ratios: - Sine = Opposite / Hypotenuse - Cosine = Adjacent / Hypotenuse - Tangent = Opposite / Adjacent 4. Here, $\tan 62^\circ = \frac{\text{Opposite}}{\text{Adjacent}}$. 5. The equation can be rewritten as: $$a \times \tan 62^\circ = \frac{x}{7} \times 9$$ 6. Simplify the right side: $$a \times \tan 62^\circ = \frac{9x}{7}$$ 7. To solve for $x$, multiply both sides by $\frac{7}{9}$: $$x = a \times \tan 62^\circ \times \frac{7}{9}$$ 8. If $a$ is the adjacent side length, and $x$ is the opposite side, this formula correctly relates them using tangent. 9. Given $x = 16.9$ cm, you can check if the equation holds by substituting values. 10. Your teacher's method is correct if you follow the tangent ratio properly and solve for the unknown side accordingly. Therefore, the approach using SOH CAH TOA and the tangent function as shown is correct for this problem.