Subjects trigonometry

Boat Distance A62F25

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1. **State the problem:** We have a triangle formed by points A, B (the boat), and C. - Point C is 1.8 km east of B. - Point A is at a bearing of 12° from B (measured clockwise from north). - The distance between A and C is 2.5 km. We want to find the distance from the boat (B) to point A. 2. **Set up the problem:** - Let’s place point B at the origin $(0,0)$. - Since C is 1.8 km east of B, coordinates of C are $(1.8,0)$. - The bearing of 12° means the line from B to A makes a 12° angle clockwise from north (the positive y-axis). 3. **Coordinates of A:** - The direction from B to A forms an angle of $90^\circ - 12^\circ = 78^\circ$ from the positive x-axis (east). - Let the distance from B to A be $d$ km. - Then coordinates of A are: $$ (x_A,y_A) = (d \cos 78^\circ, d \sin 78^\circ) $$ 4. **Use the distance formula between A and C:** Given $AC = 2.5$ km, $$ AC^2 = (x_A - x_C)^2 + (y_A - y_C)^2 $$ Substitute values: $$ 2.5^2 = (d \cos 78^\circ - 1.8)^2 + (d \sin 78^\circ - 0)^2 $$ 5. **Simplify:** $$ 6.25 = (d \cos 78^\circ - 1.8)^2 + (d \sin 78^\circ)^2 $$ Expand the squares: $$ 6.25 = (d^2 \cos^2 78^\circ - 2 \times 1.8 \times d \cos 78^\circ + 1.8^2) + d^2 \sin^2 78^\circ $$ 6. **Combine terms:** Since $\cos^2 78^\circ + \sin^2 78^\circ = 1$, $$ 6.25 = d^2 - 3.6 d \cos 78^\circ + 3.24 $$ 7. **Rearrange to form a quadratic in $d$:** $$ d^2 - 3.6 d \cos 78^\circ + 3.24 - 6.25 = 0 $$ $$ d^2 - 3.6 d \cos 78^\circ - 3.01 = 0 $$ 8. **Calculate $\cos 78^\circ$:** $$\cos 78^\circ \approx 0.2079$$ So the equation becomes: $$ d^2 - 3.6 \times 0.2079 d - 3.01 = 0$$ $$ d^2 - 0.7484 d - 3.01 = 0$$ 9. **Solve quadratic equation:** Using quadratic formula: $$ d = \frac{0.7484 \pm \sqrt{0.7484^2 + 4 \times 3.01}}{2}$$ Calculate discriminant: $$ 0.7484^2 + 4 \times 3.01 = 0.5601 + 12.04 = 12.60 $$ $$ \sqrt{12.60} \approx 3.55 $$ So, $$ d = \frac{0.7484 \pm 3.55}{2}$$ Two solutions: - $$d = \frac{0.7484 + 3.55}{2} = \frac{4.2984}{2} = 2.1492$$ - $$d = \frac{0.7484 - 3.55}{2} = \frac{-2.8016}{2} = -1.4008$$ (discard negative distance) 10. **Final answer:** The distance from the boat to point A is approximately **2.15 km**.