#### Provide solution for RD Sharma maths class 12 chapter 12 Derivative as a rate measure exercise Very short answer type question  10

Answer: $\frac{1}{20} \mathrm{rad} / \mathrm{s}$

Hint: Here we use the basic concept of rate of change about bodies and quantities.

Given:

Solution:  let us consider the above ?ABC,right angled B

In above diagram,the right angled is at B and left angle floor and ladder be $\theta$

Let at any time ‘t’ AB=x cm and BC=y cm and we know that AC = 500 cm.

$\sin \theta=\frac{x}{500} \text { and } \cos \theta=\frac{y}{500}$

So,

\begin{aligned} &\Rightarrow x=500 \sin \theta \\ &\Rightarrow y=500 \cos \theta \end{aligned}

Also, it is given that

\begin{aligned} &\frac{d x}{d t}=10 \mathrm{~cm} / \mathrm{sec} \\\\ &\frac{d(500 \sin \theta)}{d t}=10 \mathrm{~cm} / \sec \end{aligned}

\begin{aligned} &\text { for } Y=2 m=200 \mathrm{~cm} \text { (given) } \\\\ &\Rightarrow \frac{d \theta}{d t}=\frac{1}{50 \times \frac{y}{500}}=\frac{10}{y}=\frac{10}{200}=\frac{1}{20} \mathrm{rad} / \sec \end{aligned}