Question
Question: An aeroplane is flying at a height of 3000 m. An observer finds his angle of elevation to be \({60^ ...
An aeroplane is flying at a height of 3000 m. An observer finds his angle of elevation to be 60∘ and after 6 seconds it changes to 45∘. What is the speed of aeroplane?
Solution
We will begin by drawing the corresponding figure of the given statement. Then, we will use the value of tan60∘ to find the length of BC. Next, we will use the value of tan45∘ to determine the value of BE. Next, we will divide the distance travelled in 6 seconds by 6 to calculate the speed of the aeroplane.
Complete step-by-step answer:
We are given that aeroplane is flying at a height of 3000 m. Also, the angle of elevation is 60∘ but after 6 seconds, the angle of elevation is 45∘.
We will first draw the corresponding diagram.
We can see, there are two triangles in the figure,
Here, the plane from point A is moved to point D in 6 seconds. The angle of elevation ∠ACB changed to ∠DCB after 6 seconds.
The height or perpendicular distance is the same for the triangles, DE and AB is 3000 metres.
We will now write the value of tan60∘ from the given figure, where tanθ=BasePerpendicular,
tan60∘=BCAB
We know that tan60∘=3 and the value of AB is 3000 m
3=BC3000 ⇒BC=33000 ⇒BC=1.7323000≈1732m
We will now find the value of tan45∘
tan45∘=BEDE
We know that tan45∘=1 and the value of DE is 3000 m
1=BE3000 ⇒BE=13000 ⇒BE=3000m
Also, BE=BC+CE, where CE is the distance covered in 6 seconds by the plane.
3000=1732+CE ⇒CE=3000−1732 ⇒CE=1268m
Now, we know that speed is given as timedistance
Here, distance travelled 1268m in 6 seconds.
Therefore, speed is 61268=211.34m/s
Hence, the speed of the plane is 211.34 metres per second.
Note: Angle of elevation is formed when an object is above the sight of an observer. The upward angle from the point of sight is known as the angle of elevation, whereas the angle of depression is when an object is kept below the line of sight of an observer. Also, students must know how to write trigonometric ratios.