1. The problem involves finding the height of a cliff using the sine rule in triangle ABC with angles $B=72^\circ$, $C=58^\circ$, and side $AB=18.4$ m. 2. The sine rule states: $$\frac{BC}{\sin A} = \frac{BA}{\sin C}$$ where $A$, $B$, and $C$ are angles opposite sides $BC$, $AB$, and $AC$ respectively. 3. Given $\sin A = \sin(180^\circ - (B+C)) = \sin(180^\circ - 130^\circ) = \sin 50^\circ$, so $\sin A = \sin 50^\circ$. 4. Using the sine rule: $$BC = \frac{BA \times \sin A}{\sin C} = \frac{18.4 \times \sin 50^\circ}{\sin 58^\circ}$$ 5. Calculate the sines: $\sin 50^\circ \approx 0.7660$, $\sin 58^\circ \approx 0.8480$. 6. Substitute values: $$BC = \frac{18.4 \times 0.7660}{0.8480} = \frac{14.0944}{0.8480} \approx 16.62$$ m. 7. To find the height $CD$, use $CD = BC \times \sin B = 16.62 \times \sin 72^\circ$. 8. Calculate $\sin 72^\circ \approx 0.9511$. 9. Compute height: $$CD = 16.62 \times 0.9511 \approx 15.8$$ m. 10. Your calculation of height as $-17.7$ m is incorrect because height cannot be negative and the sine rule application had errors. 11. The correct height of the cliff is approximately $15.8$ m. Your approach was close but needed careful application of the sine rule and angle values.