# The angle of elevation of an aeroplane from a point on the ground is 45°. After a flight of 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane.

In the fig let C be the initial position of the aeroplane. After 15 seconds the position of the aeroplane becomes E.

In ∆ABC

tan 45° =

1 =

Y = 3000m ------(2)

tan 30° =

=

x + y = 3000

Using equation (1) to replace value of y, we get

x = 3000√3 – 3000

3000(√3-1)

2196m

Since the distance travelled by aeroplane in 15 seconds is 2196m. Therefore distance travelled by aeroplane in 1 hour =

527.04 km/hr

Therefore speed of aeroplane is 527.04 km/hr

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Trick to learn all Trigonometric Formulae28 mins
Champ Quiz | Trigger on Trigonometry47 mins
Fundas of Trigonometry50 mins
Champ Quiz | Trigonometry Important Questions33 mins
NCERT | Trigonometric Identities52 mins
Champ Quiz | NTSE Trigonometry50 mins
Testing the T- Ratios of Specified Angles57 mins
Foundation | Cracking Previous Year IMO QuestionsFREE Class
NCERT | Imp. Qs. on Trigonometry42 mins
Quiz | Trail of Mixed Questions on Trigonometry59 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses