Length Ad E4E7A1

1. **Problem Statement:** We want to find the approximate length of the wire $AD$ that crosses the canyon from point $A$ to point $D$. 2. **Given Data:** - Height of tower $AE = 1.3$ km - Height of tower $BC = 1.7$ km - Length $ED = 0.9$ km - Angle at $B$ between tower $BC$ and wire $BD$ is $27^\circ$ - Other distances: $EB = 0.9$ km, $EC = 1.6$ km, $BD$ unknown, $AD$ unknown 3. **Approach:** We will use trigonometry and the Law of Cosines to find $AD$. 4. **Step 1: Find length $BD$ using the right triangle at $B$** Since $BC = 1.7$ km and angle between $BC$ and $BD$ is $27^\circ$, we can find $BD$ using: $$BD = BC \times \tan(27^\circ) = 1.7 \times \tan(27^\circ)$$ Calculate $\tan(27^\circ)$: $$\tan(27^\circ) \approx 0.5095$$ So, $$BD = 1.7 \times 0.5095 = 0.8662 \text{ km}$$ 5. **Step 2: Find length $AD$ using the triangle $ABD$** We know: - $AB = AE + EB = 1.3 + 0.9 = 2.2$ km (vertical plus horizontal) - $BD = 0.8662$ km (from step 1) - Angle at $B$ between $AB$ and $BD$ is $90^\circ - 27^\circ = 63^\circ$ (since $BC$ is vertical) Use Law of Cosines: $$AD^2 = AB^2 + BD^2 - 2 \times AB \times BD \times \cos(63^\circ)$$ Calculate: $$AB^2 = 2.2^2 = 4.84$$ $$BD^2 = 0.8662^2 = 0.7503$$ $$\cos(63^\circ) \approx 0.4540$$ So, $$AD^2 = 4.84 + 0.7503 - 2 \times 2.2 \times 0.8662 \times 0.4540$$ Calculate the product: $$2 \times 2.2 \times 0.8662 \times 0.4540 = 1.732$$ Therefore, $$AD^2 = 4.84 + 0.7503 - 1.732 = 3.8583$$ 6. **Step 3: Calculate $AD$** $$AD = \sqrt{3.8583} \approx 1.964 \text{ km}$$ **Final answer:** The approximate length of $AD$ is **1.96 km**.