1. **Problem statement:** A balloon is observed from two stations, X and Y, which are 300 meters apart. At station X, the horizontal angle between the balloon and point C is $75^\circ 25'$ and the angle of elevation to the balloon is $18^\circ$. At station Y, the horizontal angle between the balloon and station X is $64^\circ 30'$. We need to find the height of the balloon. 2. **Understanding the problem:** We have a triangle formed by stations X, Y, and the balloon. The distance between X and Y is 300 m. We know some angles and want to find the vertical height of the balloon. 3. **Convert angles to decimal degrees:** - $75^\circ 25' = 75 + \frac{25}{60} = 75.4167^\circ$ - $64^\circ 30' = 64 + \frac{30}{60} = 64.5^\circ$ 4. **Use the Law of Sines to find distances:** Let $B$ be the balloon, $X$ and $Y$ the stations. - The horizontal angle at $X$ between balloon and $C$ is $75.4167^\circ$. - The horizontal angle at $Y$ between balloon and $X$ is $64.5^\circ$. Since $X$ and $Y$ are 300 m apart, the angle at $B$ is: $$180^\circ - 75.4167^\circ - 64.5^\circ = 40.0833^\circ$$ 5. **Apply Law of Sines:** $$\frac{XY}{\sin(\angle B)} = \frac{XB}{\sin(\angle Y)} = \frac{YB}{\sin(\angle X)}$$ Given $XY = 300$ m, $\angle B = 40.0833^\circ$, $\angle X = 75.4167^\circ$, $\angle Y = 64.5^\circ$. Calculate $XB$: $$XB = \frac{XY \times \sin(\angle Y)}{\sin(\angle B)} = \frac{300 \times \sin(64.5^\circ)}{\sin(40.0833^\circ)}$$ Calculate values: - $\sin(64.5^\circ) \approx 0.9004$ - $\sin(40.0833^\circ) \approx 0.6441$ So, $$XB = \frac{300 \times 0.9004}{0.6441} \approx 419.3 \text{ m}$$ 6. **Find height using angle of elevation at X:** Angle of elevation $= 18^\circ$. Height $h = XB \times \tan(18^\circ)$ Calculate $\tan(18^\circ) \approx 0.3249$ So, $$h = 419.3 \times 0.3249 \approx 136.2 \text{ m}$$ **Final answer:** The height of the balloon is approximately **136.2 meters**.