Class 11 Maths Ncert Solutions Ex 3.1

Q.1: Calculate the radian measurement of the given degree measurement:

(i).  $25\;^{\circ}$

(ii).  $240\;^{\circ}$

(iii).  $-47\;^{\circ} \; 30′$

(iv).  $520\;^{\circ}$

 

Sol:

(i).  $25\;^{\circ}$

As we know that, $180\;^{\circ}$ = π radian

Therefore, $25\;^{\circ}$ = $\frac{\pi }{180} \times 25$ radian = $\frac{5\pi }{36}$ radian

Hence, $25\;^{\circ}$ = $\frac{5\pi }{36}$ radian

 

(ii).  $240\;^{\circ}$

As we know that $180\;^{\circ}$ = π radian

Therefore, $240\;^{\circ}$ = $\frac{\pi }{180} \times 240$ radian = $\frac{4\pi }{3}$ radian

Hence, $240\;^{\circ}$ = $\frac{4\pi }{3}$ radian

 

(iii).  $-47\;^{\circ} \; 30′$

= $-47\frac{1}{2}$

= $\frac{-95}{2}$ degree

As we know that, $180\;^{\circ}$ = π radian

Therefore, $\frac{-95}{2}$ degree = $\frac{\pi }{180}\times \frac{-95}{2}$ radian = $\frac{-19 }{36\; \times \;2}\pi$ radian = $\frac{-19 }{72}\pi$

Hence, $-47\;^{\circ} \; 30′$ = $\frac{-19 }{72}\pi$

 

(iv).  $520\;^{\circ}$

As we know that, $180\;^{\circ}$ = n radian

Therefore, $520\;^{\circ}$ = $\frac{\pi }{180} \times 520$ radian = $\frac{26\pi }{9}$ radian

Hence, $520\;^{\circ}$ = $\frac{26\pi }{9}$ radian

 

 

Q.2: Calculate the degree measurement of the given degree measurement: [Use π = $\frac{22}{7}$]

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

 

(ii) -4

 

(iii) $\frac{5\pi }{3}$

 

(iv) $\frac{7\pi }{6}$

 

Sol:

(i).  $\frac{11}{16}$:

As we know, that Π Radian = 180°

Therefore, $\frac{11}{16}$ radian = $\frac{180}{\pi }\times \frac{11}{16}$ degree

= $\frac{45\times 11}{\pi \times 4 }$ degree

= $\frac{45 \times 11 \times 7}{22 \times 4 }$ degree

= $\frac{315}{8}$ degree

= $39\;\frac{3}{8}$ degree

= $39\;^{\circ} \;+ \; \frac{3\times 60}{8}$ minute  [$1^{\circ}$ = 60’]

= $39^{\circ} \;+ \; 22′ \;+ \;\frac{1}{2}$ minute

= $39^{\circ} \;+ \; 22′ \;+ \;30”$  [1’ = 60’’]

 

(ii).  -4:

As we know, that Π Radian = 180°

Therefore, -4 radian = $\frac{180}{\pi } \times (-4)$ degree

= $\frac{180 \times 7(-4)}{22}$ degree

= $\frac{-2520}{11}$ degree

= $-229\frac{1}{11}$ degree

= $-229^{\circ} \; + \; \frac{1\times 60}{11}$ minutes     [$1^{\circ}$ = 60’]

= $-229^{\circ} \; +\; 5′ +\;\frac{5}{11}$

= $-229^{\circ} \; +\; 5′ +\;27”$ [1’ = 60’’]

 

(iii).  $\frac{5\pi }{3}$

As we know, that Π Radian = 180°

Therefore, $\frac{5\pi}{3}$ radian = $\frac{180}{\pi }\times \frac{5\pi}{3}$ degree

= $300^{\circ}$

 

(iv).  $\frac{7\pi }{6}$

As we know, that Π Radian = 180°

Therefore, $\frac{7\pi}{6}$ radian = $\frac{180}{\pi }\times \frac{7\pi }{6}$

= $210^{\circ}$

 

 

Q.3: In a minute, wheel makes 360 revolutions. Through how many radians does it turn in 1 second?

Answer:

No. of revolutions made in a minute = 360 revolutions

Therefore, no. of revolutions made in a second = $\frac{360}{60}$ = 6

In one revolution, the wheel rotates an angle of 2π radian.

Therefore, in 6 revolutions, it will turn an angle of 12π radian.

 

 

Q.4: Calculate the degree measurement of the angle subtended at the centre of a circle of radius 100 m by an arc of length 22 m.

[Use π = $\frac{22}{7}$]

Answer:

As we know,

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then, $\theta = \frac{l}{r}$

Given:

r = 100 m and L = 22 m

We have,

$\theta = \frac{22}{100}$ radian

= $\frac{180}{\pi }\times \frac{22}{100}$ degree

= $\frac{180 \;\times \;7 \;\times \;22}{22 \; \times 100 }$ degree

= $\frac{126}{10 }$ degree

= $12\frac{3}{5}$ degree

= $12^{\circ} \; 36′$ [$1^{\circ}$ = 60’]

Therefore, the req angle is $12^{\circ} \; 36′$

 

 

Q.5: In a circle of diameter 40 m, the length of the chord 20 m. Find the length of minor arc of chord.

Answer:

Diameter of circle = 40 m

Radius of circle = $\frac{40}{20}$ m = 20 m

Let XY be the chord (length = 20 m) of the circle.

In $\Delta OXY$, OX = OY  = radius of the circle = 20 m

XY = 20 m

Therefore,

$\Delta OXY$ is an equilateral triangle.

$\theta \;= \;60^{\circ}$ = $\frac{\pi }{3}$ radian

As we know,

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then, θ = $\frac{l}{r}$

$\frac{\pi }{3}$ = $\frac{arc\;(AB\;)}{20}$

$\frac{arc\;(AB\;)}{20}$ = $\frac{20\pi }{3}$ m

The length of the minor arc of the chord is $\frac{20\pi }{3}$ m.

 

 

Q.6: In two circles, arcs which has same length subtended at an angle of $60^{\circ}$ and $75^{\circ}$ at the center. Calculate the ratio of their radii.

 Sol:

Let, the radii of two circles be $r_{1}$ and $r_{2}$.

Let, an arc of length l subtend an angle of $60^{\circ}$ at the center of the circle of radius $r_{1}$, while let an arc of length l subtend an angle of $75^{\circ}$ at the center of the circle of radius $r_{2}$.

$\\60^{\circ}$ = $\frac{\pi }{3}$ radian

$75^{\circ}$ = $\frac{5\pi }{12}$ radian

In a circle of radius r unit, if an arc of length l unit subtends an angle θ then:

θ = $\frac{l}{r}$

l = r θ

l = $\frac{r_{1}\;\pi }{3}$

l = $\frac{r_{2}\;5\pi }{12}$

$\frac{r_{1}\;\pi }{3}$ = $\frac{r_{2}\;5\pi }{12}$

$r_{1}$ = $\frac{r_{2}5}{4}$

$\frac{r_{1}}{r_{2}}$ = $\frac{5}{4}$

Therefore, the ratio of radii is 5: 4

 

Q.7: Calculate the angle in radian through which a pendulum swings if the length is 75 cm and the tip describes an arc of length

(i) 10 cm

 

(ii) 15 cm

 

(iii) 21 cm

 

Sol:

As we know that, in a circle of radius ‘r’ unit, if an arc of length ‘l’ unit subtends an angle $\theta$ radian at the center, then:

$\theta = \frac{l}{r}\\$

Here, r = 75 cm

(i).  10 cm:

$\theta$ = $\frac{10}{75}$ radian = $\frac{2}{15}$ radian

 

(ii).  15 cm:

$\theta$ = $\frac{15}{75}$ radian = $\frac{1}{5}$ radian

 

(iii).  21 cm:

$\theta$ = $\frac{21}{75}$ radian = $\frac{7}{25}$ radian

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