Answer:
Step-by-step explanation:
From the given information, consider DB=x be the height of the ship, BC= 50 m, then DE=50 m.
Now, From ΔABC,
[tex]\frac{AC}{BC}=tan45^{\circ}[/tex]
⇒[tex]AC=50 m[/tex]
Also, from ΔADE, DE=50 m, we have
[tex]\frac{AE}{DE}=tan30^{\circ}[/tex]
⇒[tex]\frac{AE}{50}=\frac{1}{\sqrt{3}}[/tex]
[tex]AE=\frac{50}{\sqrt{3}}[/tex]
Now, AC=AE+EC
⇒[tex]50=\frac{50}{\sqrt{3}}+EC[/tex]
⇒[tex]EC=50-\frac{50}{\sqrt{3}}[/tex]
⇒[tex]EC=50-\frac{50}{1.732}[/tex]
⇒[tex]EC=50-28.86=21.14 m[/tex]
Also, we know that EC=BD, therefore BD=21.14 meters.
Thus,the height of the ship= 21.14 meters.
