# The shadow of a tower, when the angle of elevation of the sun is 45°, is found to be 10 m longer than when it was 60°. Find the height of the tower.

Let the height of the tower = h (m) Let the point of 60° elevation is (m) away from the foot of the tower.

In ∆ABC,

tan 45° = 1 = 1 = h = 10+ ----(1)

In ∆ABD,

tan 60° = √3 = h = √3  = ---------(2) From eqn. (2) in eqn. (1)

h = h - = 10 = 10 = 10√3  h= h= 15+53 23.66 m.

Therefore height of the tower is 23.66 m.

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 