 Hello and welcome to the session. I am Deepika here. Let's discuss a question. It says a 1.5 meter tall boy is standing at some distance from a 30 meter tall building. The angle of elevation from his eyes to the top of the building increases from 30 degree to 60 degree as he walks towards the building. Find the distance he walked towards the building. Let us first understand the angle of elevation. In this figure qr is the horizontal line. The angle of elevation p from q angle down by pq with the horizontal line qr. That is the angle of elevation is angle pqr. So this is the key idea behind that question. We will take the help of this key idea to solve the above question. So let's start the solution. First we will draw a simple diagram to represent our problem. Here in this figure ab represents a boy and e represents the tall building. Angle ead and angle efd are the angle of elevation. So we have given ab is equal to 1.5 meter. Therefore ab is equal to cd is equal to 1.5 meter. Building is 30 meter tall that is equal to 30 meter. Now cd is 1.5 meter. Therefore be is equal to e minus cd and this is equal to 30 meter minus 1.5 meter and this is equal to 28.5 meter. Hence be is 28.5 meter. Now according to the question the angle of elevation from a to f increases from 30 degree to 60 degree. Let af is equal to x meter and fb is equal to y meter. Therefore ad is equal to x plus y meter. In right angle ad we have be is equal to 28.5 meter and angle ead is equal to 30 degree. Now we want to find ad. Now we will choose the trigonometric ratio as 1030 or 1030 as this ratio involves both ad and de. So we have de upon ad is equal to 1040 degree. Now de is given to us 28.5 meter. So we have 28.5 over ad the x plus y meter is equal to now 1030 degrees 1 over root 3. So on cross modification we have x plus y is equal to 28.5 into root 3 meter. Now in right angle esp we have angle efd is equal to 60 degree and de is equal to 28.5 meter. We want to find efd again we have de upon efd is equal to 1060 degree. Now de is 28.5 meter so we have 28.5 over y is equal to root 3 as 1060 degree is root 3. This implies y is equal to 28.5 over root 3 meter. Now x plus y is equal to 28.5 into root 3 meter and y is equal to 28.5 over root 3 meter. So we have x is equal to x plus y minus y and this is equal to 28.5 into root 3 minus 28.5 over root 3 or this is equal to 28.5 into 3 minus 28.5 over root 3. And this is equal to 28.5 into 3 minus 1 over root 3. Now multiply the denominator and the numerator by root 3 we have this is equal to 28.5 into 2 root 3 upon 3. And this is again equal to 57.0 root 3 upon 3 on cancellation we have x is equal to 19 root 3 meter. Hence when the angle of acceleration increases from 30 degree to 60 degree avoid what 19 root 3 meter towards a building. Therefore the distance he walked towards a building is 19 root 3 meter and the answer for the above question is 19 root 3 meter. I hope the solution is clear to you. Bye and take care.