Q10 of 61 Page 215

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


Given, AD = 1.5 m and distance moved = BC = ?


In triangle ABD,


∠ABD = 30°


We know,





⇒ BD = 1.5 x √3


⇒ BD = 1.5√3


Now, in triangle ACD,


∠ACD = 45°


We know,





⇒ CD = 1.5


⇒ BC = BD – CD


⇒ BC = 1.5 √3 – 1.5


⇒ BC = 1.5 (√3 – 1)


⇒ BC = 1.5(1.732 – 1)


⇒ BC = 1.5 (0.732)


⇒ BC = 1.098 m


∴ the distance moved by the student is 1.098 m.


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