the angle of elevation of an aeroplane from a point on the ground is 45 degree After 15 seconds - Maths - Some Applications of Trigonometry

Question

the angle of elevation of an aeroplane from a point on the
ground is 45 degree After 15 seconds of flight the angle changes to
30 degree f the plane is flying at a constant height of 2500m,find
the speed of the plane

Ayush Jha · Accepted Answer

Let the initial and final position of the plane be A and B Let O be the point of observation

AAÂ´ and BBÂ´ denote the height of the plane from the ground âˆ´ AAÂ´ = BBÂ´ = 2500 m From âˆ†OAAÂ´ $\tan 45 ° = \frac{A A^{'}}{{O A}^{'}} \Rightarrow {O A}^{'} = \frac{A A^{'}}{\tan 45 °} \Rightarrow {O A}^{'} = \frac{2500}{1} \Rightarrow {O A}^{'} = 2500 m$ From âˆ†OBBÂ´ $\tan 30 ° = \frac{B B^{'}}{{O B}^{'}} \Rightarrow {O B}^{'} = \frac{B B^{'}}{\tan 30 °} \Rightarrow {O B}^{'} = \frac{2500}{(\frac{1}{\sqrt{3}})} \Rightarrow {O B}^{'} = 2500 \sqrt{3} m$ Distance covered by the plane from A to B is AB or AÂ´BÂ´ â‡’ AÂ´BÂ´ = OBÂ´ â€“ OAÂ´ $= (2500 \sqrt{3} - 2500) m = 2500 (\sqrt{3} - 1) m = 2500 (1 73 - 1) m = (2500 \times 0 73) m = 1825 m$ Time taken by the plane in moving from A to B = 15 seconds Speed of the plane $= \frac{1825}{15} m / \sec = 121 66 m / s$

the angle of elevation of an aeroplane from a point on the ground is 45 degree. After 15 seconds of flight the angle changes to 30 degree.f the plane is flying at a constant height of 2500m,find the speed of the plane.