# Class 11 Maths Ncert Solutions Ex 3.1

## Class 11 Maths Ncert Solutions Chapter 3 Ex 3.1

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

(i).  25$25\;^{\circ}$

(ii).  240$240\;^{\circ}$

(iii).  4730'$-47\;^{\circ} \; 30′$

(iv).  520$520\;^{\circ}$

Sol:

(i).  25$25\;^{\circ}$

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

Therefore, 25$25\;^{\circ}$ = π180×25$\frac{\pi }{180} \times 25$ radian = 5π36$\frac{5\pi }{36}$ radian

Hence, 25$25\;^{\circ}$ = 5π36$\frac{5\pi }{36}$ radian

(ii).  240$240\;^{\circ}$

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

Therefore, 240$240\;^{\circ}$ = π180×240$\frac{\pi }{180} \times 240$ radian = 4π3$\frac{4\pi }{3}$ radian

Hence, 240$240\;^{\circ}$ = 4π3$\frac{4\pi }{3}$ radian

(iii).  4730'$-47\;^{\circ} \; 30′$

= 4712$-47\frac{1}{2}$

= 952$\frac{-95}{2}$ degree

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

Therefore, 952$\frac{-95}{2}$ degree = π180×952$\frac{\pi }{180}\times \frac{-95}{2}$ radian = 1936×2π$\frac{-19 }{36\; \times \;2}\pi$ radian = 1972π$\frac{-19 }{72}\pi$

Hence, 4730'$-47\;^{\circ} \; 30′$ = 1972π$\frac{-19 }{72}\pi$

(iv).  520$520\;^{\circ}$

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

Therefore, 520$520\;^{\circ}$ = π180×520$\frac{\pi }{180} \times 520$ radian = 26π9$\frac{26\pi }{9}$ radian

Hence, 520$520\;^{\circ}$ = 26π9$\frac{26\pi }{9}$ radian

Q.2: Calculate the degree measurement of the given degree measurement: [Use π = 227$\frac{22}{7}$]

(i) 1116$\frac{11}{16}$

(ii) -4

(iii) 5π3$\frac{5\pi }{3}$

(iv) 7π6$\frac{7\pi }{6}$

Sol:

(i).  1116$\frac{11}{16}$:

As we know, that Π Radian = 180°

Therefore, 1116$\frac{11}{16}$ radian = 180π×1116$\frac{180}{\pi }\times \frac{11}{16}$ degree

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

= 45×11×722×4$\frac{45 \times 11 \times 7}{22 \times 4 }$ degree

= 3158$\frac{315}{8}$ degree

= 3938$39\;\frac{3}{8}$ degree

= 39+3×608$39\;^{\circ} \;+ \; \frac{3\times 60}{8}$ minute  [1$1^{\circ}$ = 60’]

= 39+22'+12$39^{\circ} \;+ \; 22′ \;+ \;\frac{1}{2}$ minute

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

(ii).  -4:

As we know, that Π Radian = 180°

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

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

= 252011$\frac{-2520}{11}$ degree

= 229111$-229\frac{1}{11}$ degree

= 229+1×6011$-229^{\circ} \; + \; \frac{1\times 60}{11}$ minutes     [1$1^{\circ}$ = 60’]

= 229+5'+511$-229^{\circ} \; +\; 5′ +\;\frac{5}{11}$

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

(iii).  5π3$\frac{5\pi }{3}$

As we know, that Π Radian = 180°

Therefore, 5π3$\frac{5\pi}{3}$ radian = 180π×5π3$\frac{180}{\pi }\times \frac{5\pi}{3}$ degree

= 300$300^{\circ}$

(iv).  7π6$\frac{7\pi }{6}$

As we know, that Π Radian = 180°

Therefore, 7π6$\frac{7\pi}{6}$ radian = 180π×7π6$\frac{180}{\pi }\times \frac{7\pi }{6}$

= 210$210^{\circ}$

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

No. of revolutions made in a minute = 360 revolutions

Therefore, no. of revolutions made in a second = 36060$\frac{360}{60}$ = 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 = 227$\frac{22}{7}$]

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

Given:

r = 100 m and L = 22 m

We have,

θ=22100$\theta = \frac{22}{100}$ radian

= 180π×22100$\frac{180}{\pi }\times \frac{22}{100}$ degree

= 180×7×2222×100$\frac{180 \;\times \;7 \;\times \;22}{22 \; \times 100 }$ degree

= 12610$\frac{126}{10 }$ degree

= 1235$12\frac{3}{5}$ degree

= 1236'$12^{\circ} \; 36′$ [1$1^{\circ}$ = 60’]

Therefore, the req angle is 1236'$12^{\circ} \; 36′$

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

Diameter of circle = 40 m

Radius of circle = 4020$\frac{40}{20}$ m = 20 m

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

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

XY = 20 m

Therefore,

ΔOXY$\Delta OXY$ is an equilateral triangle.

θ=60$\theta \;= \;60^{\circ}$ = π3$\frac{\pi }{3}$ 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 = lr$\frac{l}{r}$

π3$\frac{\pi }{3}$ = arc(AB)20$\frac{arc\;(AB\;)}{20}$

arc(AB)20$\frac{arc\;(AB\;)}{20}$ = 20π3$\frac{20\pi }{3}$ m

The length of the minor arc of the chord is 20π3$\frac{20\pi }{3}$ m.

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

Sol:

Let, the radii of two circles be r1$r_{1}$ and r2$r_{2}$.

Let, an arc of length l subtend an angle of 60$60^{\circ}$ at the center of the circle of radius r1$r_{1}$, while let an arc of length l subtend an angle of 75$75^{\circ}$ at the center of the circle of radius r2$r_{2}$.

60$\\60^{\circ}$ = π3$\frac{\pi }{3}$ radian

75$75^{\circ}$ = 5π12$\frac{5\pi }{12}$ radian

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

θ = lr$\frac{l}{r}$

l = r θ

l = r1π3$\frac{r_{1}\;\pi }{3}$

l = r25π12$\frac{r_{2}\;5\pi }{12}$

r1π3$\frac{r_{1}\;\pi }{3}$ = r25π12$\frac{r_{2}\;5\pi }{12}$

r1$r_{1}$ = r254$\frac{r_{2}5}{4}$

r1r2$\frac{r_{1}}{r_{2}}$ = 54$\frac{5}{4}$

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 θ$\theta$ radian at the center, then:

θ=lr$\theta = \frac{l}{r}\\$

Here, r = 75 cm

(i).  10 cm:

θ$\theta$ = 1075$\frac{10}{75}$ radian = 215$\frac{2}{15}$ radian

(ii).  15 cm:

θ$\theta$ = 1575$\frac{15}{75}$ radian = 15$\frac{1}{5}$ radian

(iii).  21 cm:

θ$\theta$ = 2175$\frac{21}{75}$ radian = 725$\frac{7}{25}$ radian