1. **Problem Statement:** Two signposts A and B are 11 km apart on a straight highway. An airplane is flying above the highway. The angles of depression from the airplane to A and B are given as $\angle X = 27^\circ$ and $\angle Y = 53^\circ$. We need to find the height of the airplane above the highway and its position relative to the signposts. 2. **Define variables:** Let the height of the airplane be $h$ km. Let the horizontal distance from the airplane's vertical position to signpost A be $x$ km. Then, the distance to signpost B is $11 - x$ km, since A and B are 11 km apart. 3. **Use tangent relationships (angles of depression correspond to angles between horizontal flight level and lines to the points on ground):** From the angle of depression $\angle X = 27^\circ$ to A: $$\tan(27^\circ) = \frac{h}{x}$$ From the angle of depression $\angle Y = 53^\circ$ to B: $$\tan(53^\circ) = \frac{h}{11 - x}$$ 4. **Express height $h$ from both equations:** $$h = x \tan(27^\circ)$$ $$h = (11 - x) \tan(53^\circ)$$ 5. **Set equal and solve for $x$:** $$x \tan(27^\circ) = (11 - x) \tan(53^\circ)$$ $$x \tan(27^\circ) + x \tan(53^\circ) = 11 \tan(53^\circ)$$ $$x (\tan(27^\circ) + \tan(53^\circ)) = 11 \tan(53^\circ)$$ $$x = \frac{11 \tan(53^\circ)}{\tan(27^\circ) + \tan(53^\circ)}$$ 6. **Calculate numerical values:** $$\tan(27^\circ) \approx 0.5095$$ $$\tan(53^\circ) \approx 1.3270$$ $$x = \frac{11 \times 1.3270}{0.5095 + 1.3270} = \frac{14.597}{1.8365} \approx 7.95\text{ km}$$ 7. **Find height $h$:** $$h = x \tan(27^\circ) = 7.95 \times 0.5095 \approx 4.05\text{ km}$$ **Final answer:** The airplane is approximately **4.05 km** above the highway and located **7.95 km** from signpost A, towards B.