Question

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 1

Answer:

189.8 m

Step-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,  

$\tan 35 = \frac{AB}{AC} = \frac{265}{AC}$

⇒ $AC = \frac{265}{\tan 35} = 378.5$ m.

Similarly, from the right triangle Δ ADB,

$\tan 25 = \frac{AB}{AD} = \frac{265}{AD}$

⇒ $AD = \frac{265}{\tan 25} = 568.3$ m.

Hence, the distance between the ships = CD = AD - AC = 568.3 - 378.5 = 189.8 m (Approx.) (Answer)