Question
Question: The angle of depression of two ships from an aeroplane flying at the height of 7500m are \({30^ \cir...
The angle of depression of two ships from an aeroplane flying at the height of 7500m are 30∘ and 45∘ . If both the ships are in the same line and one ship is exactly behind the other, find the distance between the ships.
Solution
We will first sketch the situation and then using trigonometric formula of tanθ=baseperpendicular, we will calculate the distance between the ships.
_Complete step-by-step answer: _
We are given that an aeroplane is flying at a height of 7500m from the sea level and the angle of depression of two ships from the aeroplane are 30∘ and 45∘. Then, the figure of this condition will be:
Consider the figure given above, from it we can say that ∠ADB=30∘ ∠ACB=45∘ . It is so because of the property that alternate angles between two parallel lines are equal.
Let DC, the distance between the ships, be x and CB be y (which is the distance of the the first ship from the foot of the perpendicular drawn from the aeroplane)
∴ DB= x + y
Now in △ABC, tan45∘=BCAB=y7500
⇒1=y7500 ⇒y=7500
Similarly, in the △ABD, tan30∘=BDAB=x+y7500
⇒31=x+y7500 ⇒x+y=75003 ⇒x+7500=75003 ⇒x=7500(3−1) ⇒x=7500(1.73−1) ⇒x=7500×0.73 ⇒x=5475m (we know that 3=1.73)
Hence, the distance between the two ships is 5475m.
Note: In such problems where the angle of depression is given, students might get confused by how to solve it further. So, you need to draw a figure first hand and then you can have a clear idea about what is asked from the question. Converting the angle of depression into an angle of elevation will be helpful in calculating the tangent of the angles.