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Question: The angles of depression of two ships from the top of a lighthouse and on the same side of it are fo...

The angles of depression of two ships from the top of a lighthouse and on the same side of it are found to be 45{45^ \circ } and 30{30^ \circ } respectively. If the ships are 200m200{\text{m}} apart, find the height of the light house.

Explanation

Solution

Hint : We are asked to find the height of the light house. Draw a diagram for the problem using the given parameters. Observe the diagram carefully. Recall the concepts of a right angled triangle and use them to find the height of the light house with the given angles.

Complete step-by-step answer :
Given, the angles of depression of two ships from the top of a lighthouse on the same side of it are θ1=45{\theta _1} = {45^ \circ } and θ2=30{\theta _2} = {30^ \circ } respectively
Distance between the two ships is d=200md = 200{\text{m}} .
Let us first draw a diagram for the problem

Here, ABAB is the height of the lighthouse and CDCD is the distance between the two ships which is given as 200m200{\text{m}} . Therefore, CD=200mCD = 200{\text{m}}
We observe from the figure that there are two right angled triangles which are ΔABC\Delta ABC and ΔABD\Delta ABD .
And for right angled triangle we have,
tanθ=perpendicularbase\tan \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}
where θ\theta is the angle opposite the perpendicular side
We will use this concept for angles 45{45^ \circ } and 30{30^ \circ } in triangles ΔABC\Delta ABC and ΔABD\Delta ABD respectively.
For ΔABC\Delta ABC , we have
tan45=ABBC\tan {45^ \circ } = \dfrac{{AB}}{{BC}}
1=ABBC\Rightarrow 1 = \dfrac{{AB}}{{BC}}
BC=AB\Rightarrow BC = AB (i)
For ΔABD\Delta ABD , we have
tan30=ABBD\tan {30^ \circ } = \dfrac{{AB}}{{BD}} (ii)
BDBD can be written as,
BD=BC+CDBD = BC + CD
Putting the value of CDCD we get,
BD=BC+200BD = BC + 200 (iii)
Using equation (iii) in equation (ii) we get,
tan30=ABBC+200\tan {30^ \circ } = \dfrac{{AB}}{{BC + 200}}
13=ABBC+200\Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{{AB}}{{BC + 200}}
BC+200=3AB\Rightarrow BC + 200 = \sqrt 3 AB (iv)
Using equation (i) in (iv) we get,
AB+200=3ABAB + 200 = \sqrt 3 AB
3ABAB=200\Rightarrow \sqrt 3 AB - AB = 200
AB(31)=200\Rightarrow AB(\sqrt 3 - 1) = 200
AB=200(31)m\Rightarrow AB = \dfrac{{200}}{{(\sqrt 3 - 1)}}{\text{m}}
Therefore, the height of the light house will be 200(31)m\dfrac{{200}}{{(\sqrt 3 - 1)}}{\text{m}} .
So, the correct answer is “ 200(31)m\dfrac{{200}}{{(\sqrt 3 - 1)}}{\text{m}} ”.

Note : By angle of depression it means the downward angle from the horizontal line to the point of interest and angle of elevation means the upward angle from the horizontal line to the point of interest. Students usually get confused between the two terms, so always remember their difference.