Bearing Distance 3Eb0B7

1. **State the problem:** We have points $A$, $B$, and $C$ on level ground with given bearings and distances: - Bearing of $C$ from $A$ is $160^\circ$. - Bearing of $A$ from $B$ is $225^\circ$. - $AB = 140$ m, $AC = 195$ m. We need to find: (a)(i) $\angle BAC$, (a)(ii) length $BC$, (a)(iii) bearing of $C$ from $B$. (b) Distance Linda walked from $B$ to $S$, where $S$ is the closest point on $BC$ to $A$. --- 2. **Calculate $\angle BAC$:** - Bearing of $C$ from $A$ is $160^\circ$. - Bearing of $B$ from $A$ is the opposite of bearing of $A$ from $B$ plus $180^\circ$: $$\text{Bearing of } B \text{ from } A = 225^\circ - 180^\circ = 45^\circ$$ - The angle between $AB$ and $AC$ at $A$ is the difference between these bearings: $$\angle BAC = 160^\circ - 45^\circ = 115^\circ$$ --- 3. **Calculate length $BC$ using the Law of Cosines:** Formula: $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(\angle BAC)$$ Substitute values: $$BC^2 = 140^2 + 195^2 - 2 \times 140 \times 195 \times \cos(115^\circ)$$ Calculate: $$BC^2 = 19600 + 38025 - 2 \times 140 \times 195 \times \cos(115^\circ)$$ Calculate $\cos(115^\circ)$: $$\cos(115^\circ) \approx -0.4226$$ So: $$BC^2 = 19600 + 38025 - 2 \times 140 \times 195 \times (-0.4226)$$ Calculate the product: $$2 \times 140 \times 195 = 54600$$ Then: $$BC^2 = 19600 + 38025 + 54600 \times 0.4226$$ Calculate: $$54600 \times 0.4226 \approx 23074$$ So: $$BC^2 = 19600 + 38025 + 23074 = 80699$$ Therefore: $$BC = \sqrt{80699} \approx 284.0 \text{ m}$$ --- 4. **Calculate the bearing of $C$ from $B$:** - We know bearing of $A$ from $B$ is $225^\circ$. - Find $\angle ABC$ using Law of Cosines: $$\cos(\angle ABC) = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC}$$ Substitute values: $$\cos(\angle ABC) = \frac{140^2 + 284^2 - 195^2}{2 \times 140 \times 284}$$ Calculate numerator: $$19600 + 80699 - 38025 = 62274$$ Calculate denominator: $$2 \times 140 \times 284 = 79520$$ So: $$\cos(\angle ABC) = \frac{62274}{79520} \approx 0.783$$ Calculate angle: $$\angle ABC = \cos^{-1}(0.783) \approx 38.5^\circ$$ - Bearing of $C$ from $B$ is bearing of $A$ from $B$ plus $\angle ABC$ (since $C$ is clockwise from $A$ at $B$): $$\text{Bearing of } C \text{ from } B = 225^\circ + 38.5^\circ = 263.5^\circ$$ --- 5. **Find distance Linda walked from $B$ to $S$ where $S$ is closest point on $BC$ to $A$:** - $S$ is the foot of perpendicular from $A$ to line $BC$. - Use area of triangle $ABC$ to find height from $A$ to $BC$: $$\text{Area} = \frac{1}{2} AB \times AC \times \sin(\angle BAC)$$ Calculate $\sin(115^\circ)$: $$\sin(115^\circ) \approx 0.9063$$ Calculate area: $$\text{Area} = \frac{1}{2} \times 140 \times 195 \times 0.9063 \approx 12350 \text{ m}^2$$ - Height $h$ from $A$ to $BC$: $$h = \frac{2 \times \text{Area}}{BC} = \frac{2 \times 12350}{284} \approx 87.0 \text{ m}$$ - To find distance $BS$, use Pythagoras in triangle $BSC$: $$BS = \sqrt{BC^2 - h^2} = \sqrt{284^2 - 87^2}$$ Calculate: $$BS = \sqrt{80699 - 7569} = \sqrt{73130} \approx 270.5 \text{ m}$$ - Linda walked from $B$ to $S$, so distance walked is approximately $270.5$ m. --- **Final answers:** (a)(i) $\angle BAC = 115^\circ$ (a)(ii) $BC \approx 284.0$ m (a)(iii) Bearing of $C$ from $B$ is approximately $263.5^\circ$ (b) Distance Linda walked from $B$ to $S$ is approximately $270.5$ m.