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Page No 4.15:

Question 1:

Find the degree measure corresponding to the following radian measures:
(i) 9π5
(ii) -5π6
(iii) 18π5
(iv) (−3)c
(v) 11c
(vi) 1c

Answer:

We have:π rad =180°1 rad=180π°

i 18π5=180π×9π5°                  =36×9°                  =324°                                 


ii-5π6=180π×-5π6°                  =-30×5°                  =-150°                                 


iii 18π5c=180π×18π5°                  =36×18°                  =648°                                 


iv -3c=180π×-3°                  =18022×7×-3°                  =-378022°                  =-1711822°                   =-171°1822×60'                   =-171°49111'                   =-171°49'111×60''                   =-171°49'5.45''                                -171°49'5''


(v) 11c=180π×11°                  =18022×7×11°                  =630°               


vi 1c=180π×1°                  =18022×7×1°                  =63011°                  =57311°                   =57°311×60'                   =57°16411'                   =57°16'411×60''                     
                   =57°16'21.81''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 rad1°=π180 rad


i 300°=300×π180=5π3 rad


ii 35°=35×π180=7π36 rad


iii -56°=-56×π180=-14π45 rad


iv 135°=135×π180=3π4 rad

v -300°=-300×π180=-5π3 rad


vi 30'=12°7°30'=712°=152°=152×π180=π24 rad


vii 30'=12°125°30'=12512°=2512°=2512×π180=251π360 rad


viii 30'=12°47°30'=-4712°=-952°=-952×π180=-19π72 rad

Page No 4.15:

Question 3:

The difference between the two acute angles of a right-angled triangle is 2π5 radians. Express the angles in degrees.

Answer:

Given:
Difference between two acute angles of a right-angled triangle = 2π5 rad
1 rad=180π°

2π5 rad=180π×2π5°                  =36×2°                  =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°x-90+x=722x=162x=81
Thus, we have:
x° = 81°
And,
90°-x° = 90°-81° = 9°

Page No 4.15:

Question 4:

One angle of a triangle 23x grades and another is 32x degrees while the third is πx75 radians. Express all the angles in degrees.

Answer:

One angle of the triangle = 23x grad
=23x×910°            1 grad=910°=35x°

Another angle = 32x°

1 radian=180π°
Third angle of the triangle =xπ75 rad                                             =180π×xπ75°                                             =125x°                                                   
Now,
35x+32x+125x=180   (Angle sum property)6x+15x+24x10=18045x10=180x=40


Thus, the angles are:
35x °=24°32x °=60° 12x5°=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-2πNumber of sides in the pentagon=5 Sum of the interior angles of the pentagon =5-2π=3πEach angle of the pentagon =Sum of the interior angles of the polygonNumber of sides=3π5radEach angle of the pentagon=3π5×180π°=108°                 

(ii)

Sum of the interior angles of the polygon=n-2πNumber of sides in the octagon=8  Sum of the interior angles of the octagon =8-2π=6πEach angle of the octagon =Sum of the interior angles of the polygonNumber of sides=6π8 =3π4 radEach angle of octagon=3π4×180π°=135°                 

(iii)

Sum of the interior angles of the polygon=n-2πNumber of sides in the heptagon=7  Sum of the interior angles of the heptagon =7-2π=5πEach angle of the heptagon =Sum of the interior angles of the polygonNumber of sides=5π7 radEach angle of the heptagon=5π7×180π°=9007°=128°34'17''                 

(iv)

Sum of the interior angles of the polygon =n-2πNumber of sides in the duodecagon=12  Sum of the interior angles of the duodecagon =12-2π=10πEach angle of the duodecagon =Sum of the interior angles of the polygonNumber of sides=10π12 =5π6 radEach angle of duodecagon=5π6×180π°=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-3d°, a-d°, a+d° and a+3d°.
We know:
a-3d+a-d+a+d+a-2d=3604a=360a=90
We have:
Greatest angle = 120°
Now,
a+3d=12090+3d=1203d=30d=10

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

Angles of the quadrilateral in radians = 60×π180, 80×π180 , 100×π180 and 120×π180
                                                            =π3, 4π9, 5π9 and 2π3

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=1803a=180a=60

Given:Number of degrees in the least angleNumber of degrees in the mean angle=1120or, a-da=1120or, 60-d60=1120or, 60-d1=12or,120-2d=1or, 2d=119or, 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×π180, 60×π180 and 119.5×π180
                                                        = π360, π3 and 239π360

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 = n-2nπ radian
∴ Angle of the first polygon = 2x-22xπ=x-1xπ radian

 Angle of the second polygon = x-2xπ radian
Thus, we have:
x-1xπx-2xπ=32x-1x-2=322x-2=3x-6x=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=1803a=180a=60
Given:
Greatest angle=5×Least angleor, Greatest angleLeast angle=5or, a+da-d=5or, 60+d60-d=5or, 60+d=300-5dor, 6d=240or, 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×π180, 60×π180 and 100×π180
                                                        =π9, π3 and 5π9

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 = n-2n180°
Thus, we have:
Angle of the first polygon = 5x-25x180°

Angle of the second polygon = 4x-24x180° 
Now,

5x-25x180-4x-24x180=91804(5x-2)-5(4x-2)20x=920x-8-20x+1020x=9180220x=1202x=1x=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×π180 =5π36 radian
We know:
θ=ArcRadius5π36=40RadiusRadius=405π36=40×36×75×22=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'=160°=160×π180 radian
We know:
θ=ArcRadius160×π180=Arc5280Arc=5280×2260×180×7=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 = 360Number of revolutions taken by the wheel in 1 second=36060=6We know:1 revolution = 2π radians Number of radians the wheel will turn in 1 second=6×2π=12π 

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=ArcRadius=1075=215 radian

(ii)
Length of the arc = 15 cm
Now,
θ=ArcRadius=1575=15 radian

(iii)
Length of the arc = 21 cm
Now,
θ=ArcRadius=2175=725 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, AOB is an equilateral triangle. 
Now,
Radius = 30 cm
θ=60°=60×π180 =π3 radian
θ=ArcRadiusπ3=Arc30Arc=30π3=10π 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 =66×10003600m/s

We know:Speed = Distance Time66×10003600=DistanceTimeDistance =66×10003600×10=11006 m

Now,
Radius of the curve = 1500 m
 θ=ArcRadius       =110061500       =11001500×6=1190 radian 
So, the train will turn 1190 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'=3160°=3160×π180 radians
θ=ArcRadius3160×π180=2RadiusRadius=180×60×2×731×22                 =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 = 91×106 km
θ=32'=3260°=3260×π180 radians
θ=ArcRadius3260×π180=d91×106d=32×91×106×2260×180×7=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 θ1 and r1 and θ2 and r2, respectively.
Thus, we have:
θ1=65°=65×π180 radian
θ2=65°=110×π180 radian
θ1=lr1
r1=l65×π180

θ2=lr2
r2=l110×π180

r1r2=l65×π180l110×π180=11065=2213

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=ArcRadius=22100=1150 radian
∴ Angle subtended at the centre by the arc = 1150×180π°=115×1822×7°=635°=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) D100=G90=2Rπ
(b) D90=G100=Rπ
(c) D100=G100=2Rπ
(d) D90=G100=R2π

Answer:

(c) D90=G100=2Rπ

It is the relation between degree, grade and radian.

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) π6

(b) π3

(c) π2

(d) 2π3

Answer:

(b) π3

Let the angles of the triangle be a-d°,a° and a+d°.
Thus, we have:
a-d+a+a+d=1803a=180a=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×π180 = π3

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., 172 = 36012×172°=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 = 36060×30°=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) 2πc3

(b) 7πc12

(c) 13πc18

(d) 13πc4

Answer:

(c) 13π18c
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., 113 = 36012×113°=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 = 36060×40°=240°
∴ Required angle between two hands = 240°-110°=130°
And,
Value of the angle (in radians) between the two hands of the clock = 130×π180c=13π18c=13πc18

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 θ1 and r1 and θ2 and r2, respectively.
We have:
θ1=65°=65×π180 radian
θ2=65°=110×π180 radian
θ1=lr1
r1=l65×π180

θ2=lr2
r2=l110×π180

r1r2=l65×π180l110×π180=11065=2213
r1:r2=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 =DistanceTime=4 revolutions1 second=4×2π1       1 revolution = 2π radians=8π 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 = 2πr=14π
Now,
Angle subtended by the arc = ArcRadius=14π12=14π12×180π°=210°

Page No 4.17:

Question 8:

The radius of the circle whose arc of length 15 π cm makes an angle of 3π4 radian at the centre is
(a) 10 cm
(b) 20 cm
(c) 1114cm
(d) 2212cm

Answer:

(b) 20 cm

θ=ArcRadius3π4=15πRadiusRadius=603=20 cm



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