Let's do this question and the other is inclined to avoid making an angle of 45 degrees, a man is climbing
the ladder at the rate of one meter per second.
How fast is he approaching the wall?
All right.
Pause the video, give it a try and then let's do it together.
All right.
We're back.
I hope you've given it a try.
Let's do it together.
Now, over here, you have a wall and a ladder, and this angle is forty five degrees right now.
Let me complete the triangle by drawing the ground as well.
Now, this angle is 90 degrees, right?
Because the wall and the ground make a 90 degree angle.
Now, it's given that this angle is 40 feet.
So this angle also will be 40 feet.
Right.
Because this angle, which is 90 plus forty five plus whatever you get over here, should add up to
180 degrees.
Right now, over here, it's mentioned that he's climbing the wall at one meter per second.
All right.
And we need to find how fast is approaching the wall.
Let me name these points as A, B and C.
All right.
Now let's find out cos 40 feet, because basically we need to find out B C, right.
Because the ratio between AB and BC will be constant.
Now, if this is moving in one meter per second, we need to find how much B C is changing in one second.
All right.
Cos forty five is equal to B.C. divided by EBR.
Right.
So that's equal to one by rule because we know that's the value for forty five degrees.
Right Corzo.
Forty five is one or two B.C. by ABS equal one by Rotu therefore B C divided by one is equal to one
by Rotu.
Because we know that AB is equal to one right over here we have one, we double second.
So we are taking this one now.
In this case we get BKS equal to one Bierut two meters per second.
So the answer is one by two meters per second.
That is, if we start climbing the wall at the speed of one meter per second, then he's getting closer
to the wall at the rate of one by two meters per second.