Bearing Distance E332B3

1. **State the problem:** We have points A and B such that B is 10 cm due south of A. The bearing of point C from A is 150\degree, and the bearing of C from B is 070\degree. We need to draw the diagram and find the length of AC to 1 decimal place. 2. **Understanding bearings and setup:** - Bearing is measured clockwise from north. - B is 10 cm directly south of A, so if A is at (0,0), B is at (0,-10). - From A, C lies on a line at 150\degree bearing. - From B, C lies on a line at 070\degree bearing. 3. **Convert bearings to angles from the positive x-axis (east) for coordinate calculation:** - Bearing 150\degree means from north clockwise 150\degree, so angle from east is $90 - 150 = -60\degree$ or equivalently $300\degree$. - Bearing 070\degree means from north clockwise 70\degree, so angle from east is $90 - 70 = 20\degree$. 4. **Set coordinates:** - Let A be at $(0,0)$. - B is at $(0,-10)$. - Let C be at $(x,y)$. 5. **Write equations for lines from A and B to C:** - Line from A at 150\degree bearing: $$ y = x \tan(150\degree) = x \tan(150\degree) $$ Since $\tan(150\degree) = \tan(180\degree - 30\degree) = -\tan(30\degree) = -\frac{1}{\sqrt{3}} \approx -0.5774$, $$ y = -0.5774 x $$ - Line from B at 070\degree bearing: The slope is $\tan(70\degree) \approx 2.747$. Equation of line through B $(0,-10)$: $$ y + 10 = 2.747 (x - 0) \Rightarrow y = 2.747 x - 10 $$ 6. **Find intersection of the two lines:** Set $$ -0.5774 x = 2.747 x - 10 $$ $$ -0.5774 x - 2.747 x = -10 $$ $$ -3.3244 x = -10 $$ $$ x = \frac{10}{3.3244} \approx 3.008 $$ Find $y$: $$ y = -0.5774 \times 3.008 \approx -1.737 $$ So, $C$ is at approximately $(3.008, -1.737)$. 7. **Calculate length $AC$:** $$ AC = \sqrt{(3.008 - 0)^2 + (-1.737 - 0)^2} = \sqrt{3.008^2 + (-1.737)^2} $$ $$ = \sqrt{9.048 + 3.017} = \sqrt{12.065} \approx 3.5 \text{ cm} $$ **Final answer:** Length of $AC$ is approximately $3.5$ cm to 1 decimal place.