Ncert Solutions Maths class 11th

So, radius, r = $\frac{40}{2}$ cm = 20 cm

Length of chord (AB) = 20cm

In OAB

OA = OB=AB=20 cm

Hence, AOAB is equilateral triangle and end of the angle is 60°

:. Ø =60° = 60 *
$\frac{\mathrm{}\mathrm{\pi }}{180}$ radian =$\frac{\mathrm{\pi }}{3}$ radian

Hence, length of minor are of the chord, l=rØ.

l = 20 * $\frac{\mathrm{\pi }}{3}$ cm

l = $\frac{2\mathrm{0\pi }}{3}$ cm.

4. Here l = 22cm.

r =100cm.

Ø =?

Hence by r = $\frac{1}{\mathrm{Ø}}$

= Ø = $\frac{l}{r}$ = $\frac{22}{100}$ radian

= $\frac{22}{100}$ * $\mathrm{180°}$/π

$\frac{\mathrm{}\mathrm{}}{}$$\frac{}{}$= $\frac{22}{100}$ * 180° * $\frac{7}{22}$

= ${\left(\frac{63}{5}\right)}^0$

=12 $\frac{3°}{5}$ = 12° $\frac{3*60\text{'}}{5}=12°36\text{'}$

3. Given that a wheel makes 360 revolutions in one minute

Then, number of revolutions in one second = $\frac{360}{60}$ =6.

In 1 complete revolution the wheel turns 360°= 2π radian.

So, In 6 revolution, the wheel will turns 6*2π radian = 12π radian.

Hence, in one second the wheel will turn an angle of 12π radian.

1. (i) 25°

Solution:We know that 180° = π radian.

Hence, 25° = $\frac{\pi }{180}$ 25 radian= $\frac{\mathrm{5\pi }}{36}$ radians.

(ii) 47°30′

Solution: We know that 180° = π radian,

Hence, -47°30′= -47 * $\frac{1}{2}$ degree= $\frac{-47}{2}$ * $\frac{\pi }{\mathrm{180°}}$ radians.

= $-\frac{1\mathrm{9\pi }}{72}$ radians

(iii) 240°

Solution:We know that, 180°= radian.

Hence, 240°= 240* $\frac{\pi }{\mathrm{180 }}$ radian.

= $\frac{\mathrm{4\pi }}{3}$ radian.

(iv) 520°

Solution: We know that, 180= radian.

Hence, 520°= 520°* $\frac{\pi }{\mathrm{180 }}$radian.

= $\frac{2\mathrm{6\pi }}{9}$ radian.

36. For the given inequality, x> –3.

The equation of the line is x= –3.

This line divides the xy-plane into planer I and II. We take a point (0,0) to check the correctness of the inequality.

So, 0> –3 which is true.

So, the solution of the region is I which includes the origin.

The dotted line indicates that any point on the line does not satisfy the inequality.

35. For the given equation inequality y< 2,  the equation of line is y= 2

This line devides the xy-plane into two planer I and II. We take a point (0,0) to check the correctness of the inequality.

So, 0< 2

0< 2 which is false.

So, the solution of the region is II which does not include the origin.

The dotted line indicates that any point on the line does not satisfy the inequality.

34. For the inequality 3y – 5x<30, the equation of line is 3y 5x=30. We consider the table below to plot 3y 5x =30.

$\begin{array}{l}x\\ y\end{array}|\begin{array}{l}0\\ 10\end{array}|\begin{array}{l}-6\\ 0\end{array}|$

This line divides the xy-plane into two planer I and II. We select a point (0,0) and check the correctness the inequality.

3 * 0 – 5 * 0 < 30

0 < 30 which is true.

So, the solution region is I which includes the origin. The dotted line indicates that any point on the line will not satisfy the given inequality.

33.For the inequality –3x+2y≥ –6 the equation of line is – 3x+2y=6.

We consider the table below to plot – 3x+2y= –6.

$\begin{array}{l}x\\ y\end{array}|\begin{array}{l}0\\ -3\end{array}|\begin{array}{l}2\\ 0\end{array}|$

This line divides the xy-plane into two planer I and II. We select a point (0,0) and check the correctness the inequality,

–3 * 0+2 * 0 ≥ –6

0 ≥ –6 which is true.

So, the solution region is I which includes the origin. The continuous line also indicates that any point on the line also satisfy the given inequality.