1. **Problem Statement:** Given $AB=11$, $\angle ABD=10^\circ$, $\angle ADB=20^\circ$, and $\angle ACD=60^\circ$, find the value of $BC^2 + AD^2 - AC^2$. 2. **Understanding the problem:** We have points $A$, $B$, $C$, and $D$ with $C$ as the midpoint of $BD$. We want to find an expression involving lengths $BC$, $AD$, and $AC$. 3. **Key facts:** Since $C$ is midpoint of $BD$, $BC = CD = \frac{BD}{2}$. 4. **Step 1: Find $BD$ using triangle $ABD$** - In triangle $ABD$, angles are $\angle ABD=10^\circ$, $\angle ADB=20^\circ$, so $\angle BAD = 180^\circ - 10^\circ - 20^\circ = 150^\circ$. - Using Law of Sines: $$\frac{AB}{\sin \angle ADB} = \frac{BD}{\sin \angle BAD}$$ $$\Rightarrow \frac{11}{\sin 20^\circ} = \frac{BD}{\sin 150^\circ}$$ - Since $\sin 150^\circ = \sin 30^\circ = 0.5$, $$BD = \frac{11 \times 0.5}{\sin 20^\circ} = \frac{5.5}{\sin 20^\circ}$$ 5. **Step 2: Calculate $BC$** - Since $C$ is midpoint of $BD$, $$BC = \frac{BD}{2} = \frac{5.5}{2 \sin 20^\circ} = \frac{2.75}{\sin 20^\circ}$$ 6. **Step 3: Use Law of Cosines in triangle $ACD$ to find $AC^2$** - Given $\angle ACD = 60^\circ$ and $CD = BC = \frac{BD}{2}$. - We want $AC^2 = AD^2 + CD^2 - 2 \times AD \times CD \times \cos 60^\circ$. 7. **Step 4: Express $BC^2 + AD^2 - AC^2$** - Substitute $AC^2$ from Law of Cosines: $$BC^2 + AD^2 - AC^2 = BC^2 + AD^2 - (AD^2 + CD^2 - 2 AD \times CD \times \cos 60^\circ)$$ - Simplify: $$= BC^2 + AD^2 - AD^2 - CD^2 + 2 AD \times CD \times \cos 60^\circ = BC^2 - CD^2 + 2 AD \times CD \times \cos 60^\circ$$ - Since $BC = CD$, $BC^2 - CD^2 = 0$, so $$BC^2 + AD^2 - AC^2 = 2 AD \times CD \times \cos 60^\circ$$ 8. **Step 5: Calculate $AD$ using Law of Sines in triangle $ABD$** - Using Law of Sines: $$\frac{AB}{\sin \angle ADB} = \frac{AD}{\sin \angle ABD}$$ $$\Rightarrow \frac{11}{\sin 20^\circ} = \frac{AD}{\sin 10^\circ}$$ $$AD = \frac{11 \sin 10^\circ}{\sin 20^\circ}$$ 9. **Step 6: Substitute values into expression** - Recall $CD = BC = \frac{5.5}{\sin 20^\circ}$ - $\cos 60^\circ = 0.5$ - So, $$BC^2 + AD^2 - AC^2 = 2 \times AD \times CD \times 0.5 = AD \times CD$$ - Substitute $AD$ and $CD$: $$= \left(\frac{11 \sin 10^\circ}{\sin 20^\circ}\right) \times \left(\frac{5.5}{\sin 20^\circ}\right) = \frac{11 \times 5.5 \times \sin 10^\circ}{\sin^2 20^\circ} = \frac{60.5 \sin 10^\circ}{\sin^2 20^\circ}$$ 10. **Step 7: Calculate numerical value** - $\sin 10^\circ \approx 0.1736$ - $\sin 20^\circ \approx 0.3420$ - So, $$= \frac{60.5 \times 0.1736}{0.3420^2} = \frac{10.5}{0.117} \approx 89.7$$ **Final answer:** $$BC^2 + AD^2 - AC^2 \approx 89.7$$