Question

A student sitting in a classroom sees a picture on the black board at a height of 1.5m from the horizontal level of sight.The angle of elevation of the picture is 30 degree.As the picture is not clear to him,he moves towards the black board and sees the picture at an angle of elevation 45 degree.Find the distance moved by the student.

Answer 1

I hope this answer will help you

Answer 2

according to the above picture

tan∅ = AB/BC
tanC = 1.5/BC
tan30° = 1.5/BC
1/√3 = 1.5/BC
BC = 1.5×√3
BC = 1.5√3

tan∅ = AB/BD
tan45° = 1.5/BD
1 = 1.5/BD
BD = 1.5

CD = BC - BD
CD = 1.5√3-1.5
CD = 1.5(√3-1) m


so distance moved by the student is 1.5(√3-1) meter