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Let's do this question and the other is inclined to avoid making an angle of 45 degrees, a man is climbing

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the ladder at the rate of one meter per second.

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How fast is he approaching the wall?

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All right.

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Pause the video, give it a try and then let's do it together.

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All right.

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We're back.

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I hope you've given it a try.

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Let's do it together.

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Now, over here, you have a wall and a ladder, and this angle is forty five degrees right now.

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Let me complete the triangle by drawing the ground as well.

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Now, this angle is 90 degrees, right?

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Because the wall and the ground make a 90 degree angle.

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Now, it's given that this angle is 40 feet.

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So this angle also will be 40 feet.

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Right.

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Because this angle, which is 90 plus forty five plus whatever you get over here, should add up to

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180 degrees.

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Right now, over here, it's mentioned that he's climbing the wall at one meter per second.

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All right.

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And we need to find how fast is approaching the wall.

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Let me name these points as A, B and C.

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All right.

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Now let's find out cos 40 feet, because basically we need to find out B C, right.

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Because the ratio between AB and BC will be constant.

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Now, if this is moving in one meter per second, we need to find how much B C is changing in one second.

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All right.

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Cos forty five is equal to B.C. divided by EBR.

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Right.

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So that's equal to one by rule because we know that's the value for forty five degrees.

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Right Corzo.

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Forty five is one or two B.C. by ABS equal one by Rotu therefore B C divided by one is equal to one

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by Rotu.

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Because we know that AB is equal to one right over here we have one, we double second.

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So we are taking this one now.

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In this case we get BKS equal to one Bierut two meters per second.

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So the answer is one by two meters per second.

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That is, if we start climbing the wall at the speed of one meter per second, then he's getting closer

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to the wall at the rate of one by two meters per second.