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10 months ago

Class 11th Trigonometric Functions

5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

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Payal Gupta

Contributor-Level 10

5. Given, diameter of circle = 40 cm

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 *
$\frac{π}{180}$ radian = $\frac{π}{3}$ radian

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

l = 20 * $\frac{π}{3}$ cm

l = $\frac{2 0π}{3}$ cm.

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Class 11th Trigonometric Functions

4. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm $(U s e π = \frac{22}{7}) .$

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Payal Gupta

Contributor-Level 10

4. Here l = 22cm.

r =100cm.

Ø =?

Hence by r = $\frac{1}{Ø}$

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

= $\frac{22}{100}$ * $180°$ /π

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

= ${(\frac{63}{5})}^{0}$

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

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Class 11th Trigonometric Functions

3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

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Payal Gupta

Contributor-Level 10

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.

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10 months ago

Class 11th Trigonometric Functions

2. Find the degree measures corresponding to the following radian measures $(U s e π = \frac{22}{7}) .$

(i) $\frac{11}{16}$ (ii)-4 (iii) $\frac{5 π}{3}$ (iv) $\frac{7 π}{6}$

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Payal Gupta

Contributor-Level 10

2. (i) $\frac{11}{16}$

We know that radian= 180°,

Hence, $\frac{11}{16}$ radian= $\frac{11}{16}$ *
$\frac{180°}{π}$ = $\frac{11}{16}$ * $\frac{180 °}{22 / 7}$ = $\frac{11}{16}$ * $\frac{7}{22}$ *180 $°$

= ${(\frac{315}{8})}^{0}$

=39 0 $\frac{3 °}{8}$

= 39 $°$ + $\frac{3 * 60}{8}$ minute (as 1 $°$ =60′)

=39°+22′+ ${(\frac{1}{2})}^{'}$

=39°+22′+ ${(\frac{60}{2})}^{''}$ (as 1′=60”)

=39°+22′+30”.

=39° 22′ 30”.

(ii) -4

We know that radian = 180°.

Hence: -4 radian = -4* $(\frac{180°}{π})$ = 4* $\frac{180 °}{\frac{22}{7}}$ = 4*180°* $\frac{7}{22}$ .

= - ${(\frac{2520}{11})}^{0}$

=229 0 $\frac{1 °}{11}$

=229+ $\frac{1 * 60'}{11}$

=229+5′+ ${(\frac{5}{11})}^{'}$ .

=229°+5′+27″

=229° 5′27″

(iii) $\frac{5π}{3} .$ .

Solution: We know that, π radian= 180°.

Here $\frac{5π}{3}$ radian = $\frac{5π}{3}$ * $\frac{180°}{π}$

=300°

(iv) $\frac{7π}{6}$

Solution: We know that radian =180° .

Here, $\frac{7π}{6}$ &n

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2. (i) $\frac{11}{16}$

We know that radian= 180°,

Hence, $\frac{11}{16}$ radian= $\frac{11}{16}$ *
$\frac{180°}{π}$ = $\frac{11}{16}$ * $\frac{180 °}{22 / 7}$ = $\frac{11}{16}$ * $\frac{7}{22}$ *180 $°$

= ${(\frac{315}{8})}^{0}$

=39 0 $\frac{3 °}{8}$

= 39 $°$ + $\frac{3 * 60}{8}$ minute (as 1 $°$ =60′)

=39°+22′+ ${(\frac{1}{2})}^{'}$

=39°+22′+ ${(\frac{60}{2})}^{''}$ (as 1′=60”)

=39°+22′+30”.

=39° 22′ 30”.

(ii) -4

We know that radian = 180°.

Hence: -4 radian = -4* $(\frac{180°}{π})$ = 4* $\frac{180 °}{\frac{22}{7}}$ = 4*180°* $\frac{7}{22}$ .

= - ${(\frac{2520}{11})}^{0}$

=229 0 $\frac{1 °}{11}$

=229+ $\frac{1 * 60'}{11}$

=229+5′+ ${(\frac{5}{11})}^{'}$ .

=229°+5′+27″

=229° 5′27″

(iii) $\frac{5π}{3} .$ .

Solution: We know that, π radian= 180°.

Here $\frac{5π}{3}$ radian = $\frac{5π}{3}$ * $\frac{180°}{π}$

=300°

(iv) $\frac{7π}{6}$

Solution: We know that radian =180° .

Here, $\frac{7π}{6}$ radian = $\frac{7π}{6}$ * $\frac{180°}{π}$ =210°

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Class 11th Trigonometric Functions

Class 11 Chapter 3 Trigonometric Functions Ex 3.1 NCERT Solution

1. Find the radian measures corresponding to the following degree measures:

(i) 25° (ii) – 47°30′ (iii) 240° (iv) 520°

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Payal Gupta

Contributor-Level 10

1. (i) 25°

Solution:We know that 180° = π radian.

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

(ii) 47°30′

Solution: We know that 180° = π radian,

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

= $- \frac{1 9π}{72}$ radians

(iii) 240°

Solution:We know that, 180°= radian.

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

= $\frac{4π}{3}$ radian.

(iv) 520°

Solution: We know that, 180= radian.

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

= $\frac{2 6π}{9}$ radian.

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10 months ago

Class 11th Science

What should I do if I am not able to decide which stream to choose in class 11th?

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Aashi Saxena

Contributor-Level 6

Choosing the right stream for students who are in class 11 is a crucial decision. This will set the foundation for them to make further decisions related to their academic goals and professional careers.

One of the first things to focus on while choosing the right stream is to calculate the student's passion and interest. Make sure you select subjects that are of interest and determine your level of commitment, interest and motivation. For those who love problem-solving and numbers, there is a high chance that they will be more interested in Mathematics, Science stream or computer science. On the other side, those who are interested in

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