From a plane flying due east at 265 m above sea level, the angles of depression of two ships sailing due east measure 35 degrees and 25 degrees. How far apart are the two ships?

Answer:189.8 mStep-by-step explanation:See the diagram attached. Now, the plane is at B and the two ships are at C and at D. So, angle of depression of C from B is ∠ OBC = ∠ ACB = 35° Again, the angle of depression of D from B is ∠ OBD = ∠ ADB = 25° , Now, from the right triangle Δ ACB,  [tex]\tan 35 = \frac{AB}{AC} = \frac{265}{AC}[/tex] ⇒ [tex]AC = \frac{265}{\tan 35} = 378.5[/tex] m. Similarly, from the right triangle Δ ADB, [tex]\tan 25 = \frac{AB}{AD} = \frac{265}{AD}[/tex] ⇒ [tex]AD = \frac{265}{\tan 25} = 568.3[/tex] m. Hence, the distance between the ships = CD = AD - AC = 568.3 - 378.5 = 189.8 m (Approx.) (Answer)