1. **Problem statement:** A surveyor stands at point A. The true bearing of a tree from A is 070°. Another point B is 100 m due north of A, and from B the tree is observed at a bearing of 310°. Find the distance of the tree from points A and B. 2. **Understanding bearings and setup:** - Bearings are measured clockwise from the north direction. - Point B is 100 m north of A, so coordinates can be set as A at (0,0) and B at (0,100). - The tree lies somewhere such that from A the bearing is 070°, and from B the bearing is 310°. 3. **Convert bearings to angles from the positive x-axis (east):** - Bearing 070° means the line from A to the tree makes an angle of $90^\circ - 70^\circ = 20^\circ$ with the positive x-axis. - Bearing 310° means the line from B to the tree makes an angle of $90^\circ - 310^\circ = -220^\circ$, which is equivalent to $140^\circ$ (adding 360°). 4. **Set coordinates of the tree as $(x,y)$:** - From A (0,0), the tree lies along a line at 20°: $$ y = x \tan 20^\circ $$ - From B (0,100), the tree lies along a line at 140°: The slope is $\tan 140^\circ = \tan(180^\circ - 40^\circ) = -\tan 40^\circ$. Equation of line from B: $$ y - 100 = -\tan 40^\circ (x - 0) \implies y = 100 - x \tan 40^\circ $$ 5. **Find intersection of the two lines:** Set $$ x \tan 20^\circ = 100 - x \tan 40^\circ $$ Rearranged: $$ x (\tan 20^\circ + \tan 40^\circ) = 100 $$ Calculate values: $$ \tan 20^\circ \approx 0.3640, \quad \tan 40^\circ \approx 0.8391 $$ So: $$ x (0.3640 + 0.8391) = 100 \implies x (1.2031) = 100 \implies x = \frac{100}{1.2031} \approx 83.12 $$ Then: $$ y = x \tan 20^\circ = 83.12 \times 0.3640 \approx 30.25 $$ 6. **Calculate distances:** - Distance from A to tree: $$ d_A = \sqrt{x^2 + y^2} = \sqrt{83.12^2 + 30.25^2} \approx \sqrt{6909.9 + 915.1} = \sqrt{7825} \approx 88.46 \text{ m} $$ - Distance from B to tree: $$ d_B = \sqrt{(x - 0)^2 + (y - 100)^2} = \sqrt{83.12^2 + (30.25 - 100)^2} = \sqrt{6909.9 + 4765.6} = \sqrt{11675.5} \approx 108.06 \text{ m} $$ **Final answer:** - Distance from A to tree is approximately **88.46 m**. - Distance from B to tree is approximately **108.06 m**.