RD Sharma XI 2020 2021 Volume 1 Solutions for Class 12 Commerce Maths Chapter 4 - Measurement Of Angles

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Page No 4.15:

Question 1:

Find the degree measure corresponding to the following radian measures:
(i) $\frac{9 π}{5}$
(ii) $- \frac{5 π}{6}$
(iii) $(\frac{18 π}{5})$
(iv) (−3)^c
(v) 11^c
(vi) 1^c

Answer:

$We have : π rad = 180 ° ∴ 1 rad = {(\frac{180}{π})}^{°}$

$(i) \frac{18 π}{5} = {(\frac{180}{π} \times \frac{9 π}{5})}^{°} = {(36 \times 9)}^{°} = 324^{°}$

$(ii) - \frac{5 π}{6} = {(\frac{180}{π} \times (- \frac{5 π}{6}))}^{°} = - {(30 \times 5)}^{°} = - {(150)}^{°}$

$(iii) {(\frac{18 π}{5})}^{c} = {(\frac{180}{π} \times \frac{18 π}{5})}^{°} = {(36 \times 18)}^{°} = 648^{°}$

$(iv) {(- 3)}^{c} = {(\frac{180}{π} \times - 3)}^{°} = {(\frac{180}{22} \times 7 \times - 3)}^{°} = {(\frac{- 3780}{22})}^{°} = {(- 171 \frac{18}{22})}^{°} = \{- 171 ° {(\frac{18}{22} \times 60)}^{'}\} = \{- 171 ° {(49 \frac{1}{11})}^{'}\} = - \{171 ° 49' {(\frac{1}{11} \times 60)}^{''}\} = - (171 ° 49' 5 . 45^{''}) \approx - (171 ° 49' 5^{''})$

$(v) {(11)}^{c} = {(\frac{180}{π} \times 11)}^{°} = {(\frac{180}{22} \times 7 \times 11)}^{°} = 630^{°}$

$(vi) {(1)}^{c} = {(\frac{180}{π} \times 1)}^{°} = {(\frac{180}{22} \times 7 \times 1)}^{°} = {(\frac{630}{11})}^{°} = {(57 \frac{3}{11})}^{°} = 57 ° {(\frac{3}{11} \times 60)}^{'} = 57 ° {(16 \frac{4}{11})}^{'} = 57 ° 16' {(\frac{4}{11} \times 60)}^{''}$
$= 57 ° 16' 21.81'' \approx 57 ° 16' 22''$

Page No 4.15:

Question 2:

Find the radian measure corresponding to the following degree measures:
(i) 300°
(ii) 35°
(iii) −56°
(iv) 135°
(v) −300°
(vi) 7° 30'
(vii) 125° 30'
(viii) −47° 30'

Answer:

$We have : 180 ° = π rad ∴ 1 ° = \frac{π}{180} rad$

$(i) 300 ° = (300 \times \frac{π}{180}) = \frac{5 π}{3} rad$

$(ii) 35 ° = 35 \times \frac{π}{180} = \frac{7 π}{36} rad$

$(iii) - 56 ° = - (56 \times \frac{π}{180}) = - \frac{14 π}{45} rad$

$(iv) 135 ° = 135 \times \frac{π}{180} = \frac{3 π}{4} rad$

$(v) - 300 ° = - (300 \times \frac{π}{180}) = - \frac{5 π}{3} rad$

$(vi) 30' = {(\frac{1}{2})}^{°} ∴ 7 ° 30' = {(7 \frac{1}{2})}^{°} = {(\frac{15}{2})}^{°} = \frac{15}{2} \times \frac{π}{180} = \frac{π}{24} rad$

$(vii) 30' = {(\frac{1}{2})}^{°} ∴ 125 ° 30' = {(125 \frac{1}{2})}^{°} = {(\frac{251}{2})}^{°} = \frac{251}{2} \times \frac{π}{180} = \frac{251 π}{360} rad$

$(viii) 30' = {(\frac{1}{2})}^{°} ∴ 47 ° 30' = - {(47 \frac{1}{2})}^{°} = - {(\frac{95}{2})}^{°} = - (\frac{95}{2} \times \frac{π}{180}) = - \frac{19 π}{72} rad$

Page No 4.15:

Question 3:

The difference between the two acute angles of a right-angled triangle is $\frac{2 π}{5}$ radians. Express the angles in degrees.

Answer:

Given:
Difference between two acute angles of a right-angled triangle = $\frac{2 π}{5}$ rad
$∵ 1 rad = {(\frac{180}{π})}^{°}$

$∴ \frac{2 π}{5} rad = {(\frac{180}{π} \times \frac{2 π}{5})}^{°} = {(36 \times 2)}^{°} = 72^{°}$

Now, let one acute angle of the triangle be x $°$ .
Therefore, the other acute angle will be 90 $°$ $-$ x $°$ .
Now,
$x ° - (90 ° - x °) = 72 ° \Rightarrow x - 90 + x = 72 \Rightarrow 2 x = 162 \Rightarrow x = 81$
Thus, we have:
x $°$ = 81 $°$
And,
90 $°$ $-$ x $°$ = 90 $°$ $-$ 81 $°$ = 9 $°$

Page No 4.15:

Question 4:

One angle of a triangle $\frac{2}{3}$ x grades and another is $\frac{3}{2}$ x degrees while the third is $\frac{πx}{75}$ radians. Express all the angles in degrees.

Answer:

One angle of the triangle = $\frac{2}{3} x grad$
$= {(\frac{2}{3} x \times \frac{9}{10})}^{°} [∵ 1 grad = {(\frac{9}{10})}^{°}] = {(\frac{3}{5} x)}^{°}$

Another angle = ${(\frac{3}{2} x)}^{°}$

$∵ 1 radian = {(\frac{180}{π})}^{°}$
$Third angle of the triangle = \frac{x π}{75} rad = {(\frac{180}{π} \times \frac{x π}{75})}^{°} = {(\frac{12}{5} x)}^{°}$
Now,
$\frac{3}{5} x + \frac{3}{2} x + \frac{12}{5} x = 180 (Angle sum property) \Rightarrow \frac{6 x + 15 x + 24 x}{10} = 180 \Rightarrow \frac{45 x}{10} = 180 \Rightarrow x = 40$

Thus, the angles are:
${(\frac{3}{5} x)}^{°} = 24 ° (\frac{3}{2} x)^{°} = 60 ° {(\frac{12 x}{5})}^{°} = 96 °$ .

Page No 4.15:

Question 5:

Find the magnitude, in radians and degrees, of the interior angle of a regular (i) pentagon (ii) octagon (iii) heptagon (iv) duodecagon.

Answer:

(i)

$Sum of the interior angles of the polygon = (n - 2) π Number of sides in the pentagon = 5 ∴ Sum of the interior angles of the pentagon = (5 - 2) π = 3 π Each angle of the pentagon = \frac{Sum of the interior angles of the polygon}{Number of sides} = \frac{3 π}{5} rad Each angle of the pentagon = {(\frac{3 π}{5} \times \frac{180}{π})}^{°} = 108 °$

(ii)

$Sum of the interior angles of the polygon = (n - 2) π Number of sides in the octagon = 8 ∴ Sum of the interior angles of the octagon = (8 - 2) π = 6 π Each angle of the octagon = \frac{Sum of the interior angles of the polygon}{Number of sides} = \frac{6 π}{8} = \frac{3 π}{4} rad Each angle of octagon = {(\frac{3 π}{4} \times \frac{180}{π})}^{°} = 135 °$

(iii)

$Sum of the interior angles of the polygon = (n - 2) π Number of sides in the heptagon = 7 ∴ Sum of the interior angles of the heptagon = (7 - 2) π = 5 π Each angle of the heptagon = \frac{Sum of the interior angles of the polygon}{Number of sides} = \frac{5 π}{7} rad Each angle of the heptagon = {(\frac{5 π}{7} \times \frac{180}{π})}^{°} = {(\frac{900}{7})}^{°} = 128 ° 34' 17''$

(iv)

$Sum of the interior angles of the polygon = (n - 2) π Number of sides in the duodecagon = 12 ∴ Sum of the interior angles of the duodecagon = (12 - 2) π = 10 π Each angle of the duodecagon = \frac{Sum of the interior angles of the polygon}{Number of sides} = \frac{10 π}{12} = \frac{5 π}{6} rad Each angle of duodecagon = {(\frac{5 π}{6} \times \frac{180}{π})}^{°} = 150^{°}$

Page No 4.15:

Question 6:

The angle of a quadrilateral are in A.P. and the greatest angle is 120°. Express the angles in radians.

Answer:

Let the angles of the quadrilateral be $(a - 3 d) °, (a - d) °, (a + d) ° and (a + 3 d) °$ .
We know:
$a - 3 d + a - d + a + d + a - 2 d = 360 \Rightarrow 4 a = 360 \Rightarrow a = 90$
We have:
Greatest angle = 120°
Now,
$a + 3 d = 120 \Rightarrow 90 + 3 d = 120 \Rightarrow 3 d = 30 \Rightarrow d = 10$

Hence, $(a - 3 d) °, (a - d) °, (a + d) ° and (a + 3 d) °$ are $60 °, 80 °, 100 ° and 120 °$ , respectively.

Angles of the quadrilateral in radians = $(60 \times \frac{π}{180}), (80 \times \frac{π}{180}), (100 \times \frac{π}{180}) and (120 \times \frac{π}{180})$
= $\frac{π}{3}, \frac{4 π}{9}, \frac{5 π}{9} and \frac{2 π}{3}$

Page No 4.15:

Question 7:

The angles of a triangle are in A.P. and the number of degrees in the least angle is to the number of degrees in the mean angle as 1 : 120. Find the angles in radians.

Answer:

Let the angles of the triangle be $(a - d) °, (a) ° and (a + d) °$ .
We know:
$a - d + a + a + d = 180 \Rightarrow 3 a = 180 \Rightarrow a = 60$

$Given : \frac{Number of degrees in the least angle}{Number of degrees in the mean angle} = \frac{1}{120} or, \frac{a - d}{a} = \frac{1}{120} or, \frac{60 - d}{60} = \frac{1}{120} or, \frac{60 - d}{1} = \frac{1}{2} or, 120 - 2 d = 1 or, 2 d = 119 or, d = 59.5$

Hence, the angles are $(a - d) °, (a) ° and (a + d) °$ , i.e., $0.5 °, 60 ° and 119.5 °$ .

∴ Angles of the triangle in radians = $(0.5 \times \frac{π}{180}), (60 \times \frac{π}{180}) and (119.5 \times \frac{π}{180})$
= $\frac{π}{360}, \frac{π}{3} and \frac{239 π}{360}$

Page No 4.15:

Question 8:

The angle in one regular polygon is to that in another as 3 : 2 and the number of sides in first is twice that in the second. Determine the number of sides of two polygons.

Answer:

Let the number of sides in the first polygon be 2x and the number of sides in the second polygon is x.
We know:
Angle of an n-sided regular polygon = $(\frac{n - 2}{n}) π$ radian
∴ Angle of the first polygon = $(\frac{2 x - 2}{2 x}) π = (\frac{x - 1}{x}) π$ radian

Angle of the second polygon = $(\frac{x - 2}{x}) π$ radian
Thus, we have:
$\frac{(\frac{x - 1}{x}) π}{(\frac{x - 2}{x}) π} = \frac{3}{2} \Rightarrow \frac{x - 1}{x - 2} = \frac{3}{2} \Rightarrow 2 x - 2 = 3 x - 6 \Rightarrow x = 4$
Thus,
Number of sides in the first polygon = 2x = 8
Number of sides in the first polygon = x = 4

Page No 4.15:

Question 9:

The angles of a triangle are in A.P. such that the greatest is 5 times the least. Find the angles in radians.

Answer:

Let the angles of the triangle be $(a - d) °, (a) ° and (a + d) °$ .
We know:
$a - d + a + a + d = 180 \Rightarrow 3 a = 180 \Rightarrow a = 60$
Given:
$Greatest angle = 5 \times Least angle or, \frac{Greatest angle}{Least angle} = 5 or, \frac{a + d}{a - d} = 5 or, \frac{60 + d}{60 - d} = 5 or, 60 + d = 300 - 5 d or, 6 d = 240 or, d = 40$

Hence, the angles are $(a - d) °, (a) ° and (a + d) °$ , i.e., $20 °, 60 ° and 100 °$ , respectively.

∴ Angles of the triangle in radians = $(20 \times \frac{π}{180}), (60 \times \frac{π}{180}) and (100 \times \frac{π}{180})$
= $\frac{π}{9}, \frac{π}{3} and \frac{5 π}{9}$

Page No 4.15:

Question 10:

The number of sides of two regular polygons are as 5 : 4 and the difference between their angles is 9°. Find the number of sides of the polygons.

Answer:

Let the number of sides in the first polygon be 5x and the number of sides in the second polygon be 4x.
We know:
Angle of an n-sided regular polygon = $(\frac{n - 2}{n}) 180 °$
Thus, we have:
Angle of the first polygon = $(\frac{5 x - 2}{5 x}) 180 °$

Angle of the second polygon = $(\frac{4 x - 2}{4 x}) 180 °$
Now,

$(\frac{5 x - 2}{5 x}) 180 - (\frac{4 x - 2}{4 x}) 180 = 9 \Rightarrow 180 (\frac{4 (5 x - 2) - 5 (4 x - 2)}{20 x}) = 9 \Rightarrow \frac{20 x - 8 - 20 x + 10}{20 x} = \frac{9}{180} \Rightarrow \frac{2}{20 x} = \frac{1}{20} \Rightarrow \frac{2}{x} = 1 \Rightarrow x = 2$
Thus, we have:
Number of sides in the first polygon = 5x = 10
Number of sides in the second polygon = 4x = 8

Page No 4.15:

Question 11:

A rail road curve is to be laid out on a circle. What radius should be used if the track is to change direction by 25° in a distance of 40 metres?

Answer:

Length of the arc = 40 m
$θ = 25 ° = (25 \times \frac{π}{180}) = \frac{5 π}{36} radian$
We know:
$θ = \frac{Arc}{Radius} \Rightarrow \frac{5 π}{36} = \frac{40}{Radius} \Rightarrow Radius = \frac{40}{\frac{5 π}{36}} = \frac{40 \times 36 \times 7}{5 \times 22} = 91.64 m$
So, the radius of the track should be 91.64 m.

Page No 4.15:

Question 12:

Find the length which at a distance of 5280 m will subtend an angle of 1' at the eye.

Answer:

We have:
Radius = 5280 m
Now,
$θ = 1' = {(\frac{1}{60})}^{°} = (\frac{1}{60} \times \frac{π}{180}) radian$
We know:
$θ = \frac{Arc}{Radius} \Rightarrow \frac{1}{60} \times \frac{π}{180} = \frac{Arc}{5280} \Rightarrow Arc = \frac{5280 \times 22}{60 \times 180 \times 7} = 1.5365 m$

Page No 4.15:

Question 13:

A wheel makes 360 revolutions per minute. Through how many radians does it turn in 1 second?

Answer:

$Number of revolutions taken by the wheel in 1 minute = 360 Number of revolutions taken by the wheel in 1 second = \frac{360}{60} = 6 We know : 1 revolution = 2 π radians ∴ Number of radians the wheel will turn in 1 second = 6 \times 2 π = 12 π$

Page No 4.15:

Question 14:

Find the angle in radians through which a pendulum swings if its length is 75 cm and the tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 21 cm.

Answer:

We know:
Radius = 75 cm

(i)
Length of the arc = 10 cm
Now,
$θ = \frac{Arc}{Radius} = \frac{10}{75} = \frac{2}{15} radian$

(ii)
Length of the arc = 15 cm
Now,
$θ = \frac{Arc}{Radius} = \frac{15}{75} = \frac{1}{5} radian$

(iii)
Length of the arc = 21 cm
Now,
$θ = \frac{Arc}{Radius} = \frac{21}{75} = \frac{7}{25} radian$

Page No 4.15:

Question 15:

The radius of a circle is 30 cm. Find the length of an arc of this circle, if the length of the chord of the arc is 30 cm.

Answer:

Let AB be the chord and O be the centre of the circle.
Here,
AO = BO = AB = 30 cm
Therefore, $∆ A O B is an equilateral triangle .$
Now,
Radius = 30 cm
$θ = 60 ° = (60 \times \frac{π}{180}) = \frac{π}{3} radian$
$θ = \frac{Arc}{Radius} \Rightarrow \frac{π}{3} = \frac{Arc}{30} \Rightarrow Arc = \frac{30 π}{3} = 10 π cm$

Page No 4.16:

Question 16:

A railway train is travelling on a circular curve of 1500 metres radius at the rate of 66 km/hr. Through what angle has it turned in 10 seconds?

Answer:

Time = 10 seconds
Speed = $66 km / h = \frac{66 \times 1000}{3600} m / s$

$We know : Speed = \frac{Distance}{Time} \Rightarrow \frac{66 \times 1000}{3600} = \frac{Distance}{Time} \Rightarrow Distance = \frac{66 \times 1000}{3600} \times 10 = \frac{1100}{6} m$

Now,
Radius of the curve = 1500 m
$∴ θ = \frac{Arc}{Radius} = \frac{\frac{1100}{6}}{1500} = \frac{1100}{1500 \times 6} = \frac{11}{90} radian$
So, the train will turn $\frac{11}{90}$ radian in 10 seconds.

Page No 4.16:

Question 17:

Find the distance from the eye at which a coin of 2 cm diameter should be held so as to conceal the full moon whose angular diameter is 31'.

Answer:

Let PQ be the diameter of the coin and E be the eye of the observer.
Also, let the coin be kept at a distance r from the eye of the observer to hide the moon completely.
Now,
$θ = 31' = {(\frac{31}{60})}^{°} = (\frac{31}{60} \times \frac{π}{180}) radians$
$θ = \frac{Arc}{Radius} \Rightarrow \frac{31}{60} \times \frac{π}{180} = \frac{2}{Radius} \Rightarrow Radius = \frac{180 \times 60 \times 2 \times 7}{31 \times 22} = 221.7 cm or 2.217 m$

Page No 4.16:

Question 18:

Find the diameter of the sun in km supposing that it subtends an angle of 32' at the eye of an observer. Given that the distance of the sun is 91 × 10⁶ km.

Answer:

Let PQ be the diameter of the Sun and E be the eye of the observer.
Because the distance between the Sun and the Earth is quite large, we will take PQ as arc PQ.
Now,
r = $91 \times 10^{6} km$
$θ = 32' = {(\frac{32}{60})}^{°} = (\frac{32}{60} \times \frac{π}{180}) radians$
$θ = \frac{Arc}{Radius} \Rightarrow \frac{32}{60} \times \frac{π}{180} = \frac{d}{91 \times 10^{6}} \Rightarrow d = \frac{32 \times 91 \times 10^{6} \times 22}{60 \times 180 \times 7} = 847407.4 km$

Page No 4.16:

Question 19:

If the arcs of the same length in two circles subtend angles 65° and 110° at the centre, find the ratio of their radii.

Answer:

Let the angles subtended at the centres by the arcs and radii of the first and second circles be $θ_{1} and r_{1} and θ_{2} and r_{2}$ , respectively.
Thus, we have:
$θ_{1} = 65 ° = (65 \times \frac{π}{180}) radian$
$θ_{2} = 65 ° = (110 \times \frac{π}{180}) radian$
$θ_{1} = \frac{l}{r_{1}}$
$\Rightarrow r_{1} = \frac{l}{(65 \times \frac{π}{180})}$

$θ_{2} = \frac{l}{r_{2}}$
$\Rightarrow r_{2} = \frac{l}{(110 \times \frac{π}{180})}$

$\Rightarrow \frac{r_{1}}{r_{2}} = \frac{\frac{l}{(65 \times \frac{π}{180})}}{\frac{l}{(110 \times \frac{π}{180})}} = \frac{110}{65} = \frac{22}{13}$

⇒ r₁:r₂ = 22:13

Page No 4.16:

Question 20:

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.

Answer:

Length of the arc = 22 cm
Radius = 100 cm
Now,
$θ = \frac{Arc}{Radius} = \frac{22}{100} = \frac{11}{50} radian$
∴ Angle subtended at the centre by the arc = ${(\frac{11}{50} \times \frac{180}{π})}^{°} = {(\frac{11}{5} \times \frac{18}{22} \times 7)}^{°} = {(\frac{63}{5})}^{°} = 12 ° 36'$

Page No 4.17:

Question 1:

If D, G and R denote respectively the number of degrees, grades and radians in an angle, then
(a) $\frac{D}{100} = \frac{G}{90} = \frac{2 R}{π}$
(b) $\frac{D}{90} = \frac{G}{100} = \frac{R}{π}$
(c) $\frac{D}{100} = \frac{G}{100} = \frac{2 R}{π}$
(d) $\frac{D}{90} = \frac{G}{100} = \frac{R}{2 π}$

Answer:

Page No 4.17:

Question 2:

If the angles of a triangle are in A.P., then the measures of one of the angles in radians is

(a) $\frac{π}{6}$

(b) $\frac{π}{3}$

(c) $\frac{π}{2}$

(d) $\frac{2 π}{3}$

Answer:

(b) $\frac{π}{3}$

Let the angles of the triangle be $(a - d) °, (a) ° and (a + d) °$ .
Thus, we have:
$a - d + a + a + d = 180 \Rightarrow 3 a = 180 \Rightarrow a = 60$
Hence, the angles are $(a - d) °, (a) ° and (a + d) °$ , i.e., $(60 - d) °, 60 ° and (60 + d) °$ .
60° is the only angle which is independent of d.
∴ One of the angles of the triangle (in radians) = $(60 \times \frac{π}{180})$ = $\frac{π}{3}$

Page No 4.17:

Question 3:

The angle between the minute and hour hands of a clock at 8:30 is
(a) 80°
(b) 75°
(c) 60°
(d) 105°

Answer:

(b) 75°
We know that the hour hand of a clock completes one rotation in 12 hours.
∴ Angle traced by the hour hand in 12 hours = 360°
$Now, Angle traced by the hour hand in 8 hours 30 minutes, i . e ., \frac{17}{2} = {(\frac{360}{12} \times \frac{17}{2})}^{°} = 255 °$
We also know that the minute hand of a clock completes one rotation in 60 minutes.
∴ Angle traced by the minute hand in 60 minutes = 360°
$Now, Angle traced by the minute hand in 30 minutes = {(\frac{360}{60} \times 30)}^{°} = 180 °$
∴ Required angle between the two hands of the clock = $255 ° - 180 ° = 75 °$

Page No 4.17:

Question 4:

At 3:40, the hour and minute hands of a clock are inclined at
(a) $\frac{2 π^{c}}{3}$

(b) $\frac{7 π^{c}}{12}$

(c) $\frac{13 π_{c}}{18}$

(d) $\frac{13 π_{c}}{4}$

Answer:

(c) ${(\frac{13 π}{18})}^{c}$
We know that the hour hand of a clock completes one rotation in 12 hours.
∴ Angle traced by the hour hand in 12 hours = 360°
Now,
$Angle traced by the hour hand in 3 hours 40 minutes, i . e ., \frac{11}{3} = {(\frac{360}{12} \times \frac{11}{3})}^{°} = 110 °$
We also know that the minute hand of a clock completes one rotation in 60 minutes.
∴ Angle traced by the minute hand in 60 minutes = 360°
Now,
$Angle traced by the minute hand in 40 minutes = {(\frac{360}{60} \times 40)}^{°} = 240 °$
∴ Required angle between two hands = $240 ° - 110 ° = 130 °$
And,
Value of the angle (in radians) between the two hands of the clock = ${(130 \times \frac{π}{180})}^{c} = {(\frac{13 π}{18})}^{c} = \frac{13 π^{c}}{18}$

Page No 4.17:

Question 5:

If the arcs of the same length in two circles subtend angles 65° and 110° at the centre, than the ratio of the radii of the circles is
(a) 22 : 13
(b) 11 : 13
(c) 22 : 15
(d) 21 : 13

Answer:

(a) 22:13

Let the angles subtended at the centres by the arcs and radii of the first and second circles be $θ_{1} and r_{1} and θ_{2} and r_{2},$ respectively.
We have:
$θ_{1} = 65 ° = (65 \times \frac{π}{180}) radian$
$θ_{2} = 65 ° = (110 \times \frac{π}{180}) radian$
$θ_{1} = \frac{l}{r_{1}}$
$\Rightarrow r_{1} = \frac{l}{(65 \times \frac{π}{180})}$

$θ_{2} = \frac{l}{r_{2}}$
$\Rightarrow r_{2} = \frac{l}{(110 \times \frac{π}{180})}$

$\Rightarrow \frac{r_{1}}{r_{2}} = \frac{\frac{l}{(65 \times \frac{π}{180})}}{\frac{l}{(110 \times \frac{π}{180})}} = \frac{110}{65} = \frac{22}{13}$
$\Rightarrow r_{1} : r_{2} = 22 : 13$

Page No 4.17:

Question 6:

If OP makes 4 revolutions in one second, the angular velocity in radians per second is
(a) π
(b) 2 π
(c) 4 π
(d) 8 π

Answer:

(d) 8π

$Angular velocity = \frac{Distance}{Time} = \frac{4 revolutions}{1 second} = \frac{4 \times 2 π}{1} (∵ 1 r e v o l u t i o n = 2 π radians) = 8 π radians per second$

Page No 4.17:

Question 7:

A circular wire of radius 7 cm is cut and bent again into an arc of a circle of radius 12 cm. The angle subtended by the arc at the centre is
(a) 50°
(b) 210°
(c) 100°
(d) 60°
(e) 195°

Answer:

(b) 210°

Length of the arc of radius = Circumference of the circle of radius 7 cm = $2 π r = 14 π$
Now,
Angle subtended by the arc = $\frac{Arc}{Radius} = \frac{14 π}{12} = {(\frac{14 π}{12} \times \frac{180}{π})}^{°} = 210 °$

Page No 4.17:

Question 8:

The radius of the circle whose arc of length 15 π cm makes an angle of $\frac{3 π}{4}$ radian at the centre is
(a) 10 cm
(b) 20 cm
(c) $11 \frac{1}{4} cm$
(d) $22 \frac{1}{2} cm$

Answer:

(b) 20 cm

$θ = \frac{Arc}{Radius} \Rightarrow \frac{3 π}{4} = \frac{15 π}{Radius} \Rightarrow Radius = \frac{60}{3} = 20 cm$

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