Rd Sharma Xi 2018 Solutions for Class 11 Science Math Chapter 4 Measurement Of Angles are provided here with simple step-by-step explanations. These solutions for Measurement Of Angles are extremely popular among Class 11 Science students for Math Measurement Of Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2018 Book of Class 11 Science Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2018 Solutions. All Rd Sharma Xi 2018 Solutions for class Class 11 Science Math are prepared by experts and are 100% accurate.

#### Question 1:

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

$=57°16\text{'}21.81\text{'}\text{'}\phantom{\rule{0ex}{0ex}}\approx 57°16\text{'}22\text{'}\text{'}$

#### 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'

#### Question 3:

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

Given:
Difference between two acute angles of a right-angled triangle = $\frac{2\mathrm{\pi }}{5}$ rad

Now, let one acute angle of the triangle be x$°$.
Therefore, the other acute angle will be 90$°$$-$x$°$.
Now,
$x°-\left(90°-x°\right)=72°\phantom{\rule{0ex}{0ex}}⇒x-90+x=72\phantom{\rule{0ex}{0ex}}⇒2x=162\phantom{\rule{0ex}{0ex}}⇒x=81$
Thus, we have:
x$°$ = 81$°$
And,
90$°$$-$x$°$ = 90$°$$-$81$°$ = 9$°$

#### Question 4:

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

One angle of the triangle =

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

Now,

Thus, the angles are:
.

#### Question 5:

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

(i)

(ii)

(iii)

(iv)

#### Question 6:

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

Let the angles of the quadrilateral be .
We know:
$a-3d+a-d+a+d+a-2d=360\phantom{\rule{0ex}{0ex}}⇒4a=360\phantom{\rule{0ex}{0ex}}⇒a=90\phantom{\rule{0ex}{0ex}}$
We have:
Greatest angle = 120°
Now,
$a+3d=120\phantom{\rule{0ex}{0ex}}⇒90+3d=120\phantom{\rule{0ex}{0ex}}⇒3d=30\phantom{\rule{0ex}{0ex}}⇒d=10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence, are , respectively.

=

#### 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.

Let the angles of the triangle be .
We know:
$a-d+a+a+d=180\phantom{\rule{0ex}{0ex}}⇒3a=180\phantom{\rule{0ex}{0ex}}⇒a=60\phantom{\rule{0ex}{0ex}}$

Hence, the angles are , i.e., .

∴ Angles of the triangle in radians =
=

#### 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.

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 = $\left(\frac{n-2}{n}\right)\mathrm{\pi }$ radian
∴ Angle of the first polygon = $\left(\frac{2x-2}{2x}\right)\mathrm{\pi }=\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)\mathrm{\pi }$ radian

Angle of the second polygon = $\left(\frac{x-2}{x}\right)\mathrm{\pi }$ radian
Thus, we have:
$\frac{\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)\mathrm{\pi }}{\left(\frac{\mathrm{x}-2}{\mathrm{x}}\right)\mathrm{\pi }}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{x-1}{x-2}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}⇒2x-2=3x-6\phantom{\rule{0ex}{0ex}}⇒x=4$
Thus,
Number of sides in the first polygon = 2x = 8
Number of sides in the first polygon = x = 4

#### 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.

Let the angles of the triangle be .
We know:
$a-d+a+a+d=180\phantom{\rule{0ex}{0ex}}⇒3a=180\phantom{\rule{0ex}{0ex}}⇒a=60\phantom{\rule{0ex}{0ex}}$
Given:

Hence, the angles are , i.e., , respectively.

∴ Angles of the triangle in radians =
=

#### 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.

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 = $\left(\frac{n-2}{n}\right)180°$
Thus, we have:
Angle of the first polygon = $\left(\frac{5x-2}{5x}\right)180°$

Angle of the second polygon =
Now,

$\left(\frac{5x-2}{5x}\right)180-\left(\frac{4x-2}{4x}\right)180=9\phantom{\rule{0ex}{0ex}}⇒180\left(\frac{4\left(5x-2\right)-5\left(4x-2\right)}{20x}\right)=9\phantom{\rule{0ex}{0ex}}⇒\frac{20x-8-20x+10}{20x}=\frac{9}{180}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{20x}=\frac{1}{20}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{x}=1\phantom{\rule{0ex}{0ex}}⇒x=2$
Thus, we have:
Number of sides in the first polygon = 5x = 10
Number of sides in the second polygon = 4x = 8

#### 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?

Length of the arc = 40 m

We know:

So, the radius of the track should be 91.64 m.

#### Question 12:

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

We have:
Now,

We know:

#### Question 13:

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

#### 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.

We know:

(i)
Length of the arc = 10 cm
Now,

(ii)
Length of the arc = 15 cm
Now,

(iii)
Length of the arc = 21 cm
Now,

#### 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.

Let AB be the chord and O be the centre of the circle.
Here,
AO = BO = AB = 30 cm
Therefore,
Now, #### 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?

Time = 10 seconds
Speed =

Now,
Radius of the curve = 1500 m

So, the train will turn $\frac{11}{90}$ radian in 10 seconds.

#### 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'.

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, #### 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.

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 = #### 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.

Let the angles subtended at the centres by the arcs and radii of the first and second circles be , respectively.
Thus, we have:

${\mathrm{\theta }}_{1}=\frac{l}{{r}_{1}}$
$⇒{r}_{1}=\frac{l}{\left(65×\frac{\mathrm{\pi }}{180}\right)}$

${\mathrm{\theta }}_{2}=\frac{l}{{r}_{2}}$
$⇒{r}_{2}=\frac{l}{\left(110×\frac{\mathrm{\pi }}{180}\right)}$

$⇒\frac{{r}_{1}}{{r}_{2}}=\frac{\frac{l}{\left(65×\frac{\mathrm{\pi }}{180}\right)}}{\frac{l}{\left(110×\frac{\mathrm{\pi }}{180}\right)}}=\frac{110}{65}=\frac{22}{13}\phantom{\rule{0ex}{0ex}}$

r1:r2 = 22:13

#### 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.

Length of the arc = 22 cm
Now,

∴ Angle subtended at the centre by the arc = ${\left(\frac{11}{50}×\frac{180}{\mathrm{\pi }}\right)}^{°}={\left(\frac{11}{5}×\frac{18}{22}×7\right)}^{°}={\left(\frac{63}{5}\right)}^{°}=12°36\text{'}$

#### Question 1:

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

(c) $\frac{D}{90}=\frac{G}{100}=\frac{2R}{\mathrm{\pi }}$

#### Question 2:

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

(a) $\frac{\mathrm{\pi }}{6}$

(b) $\frac{\mathrm{\pi }}{3}$

(c) $\frac{\mathrm{\pi }}{2}$

(d) $\frac{2\mathrm{\pi }}{3}$

(b) $\frac{\mathrm{\pi }}{3}$

Let the angles of the triangle be .
Thus, we have:
$a-d+a+a+d=180\phantom{\rule{0ex}{0ex}}⇒3a=180\phantom{\rule{0ex}{0ex}}⇒a=60\phantom{\rule{0ex}{0ex}}$
Hence, the angles are , i.e., .
60° is the only angle which is independent of d.
∴ One of the angles of the triangle (in radians) == $\frac{\mathrm{\pi }}{3}$

#### 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°

(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°

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°

∴ Required angle between the two hands of the clock = $255°-180°=75°$

#### Question 4:

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

(b) $\frac{7{\mathrm{\pi }}^{\mathrm{c}}}{12}$

(c) $\frac{13{\mathrm{\pi }}_{\mathrm{c}}}{18}$

(d) $\frac{13{\mathrm{\pi }}_{\mathrm{c}}}{4}$

(c) ${\left(\frac{13\mathrm{\pi }}{18}\right)}^{c}$
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,

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,

∴ Required angle between two hands = $240°-110°=130°$
And,
Value of the angle (in radians) between the two hands of the clock = ${\left(130×\frac{\mathrm{\pi }}{180}\right)}^{c}={\left(\frac{13\mathrm{\pi }}{18}\right)}^{c}=\frac{13{\mathrm{\pi }}^{c}}{18}$

#### 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

(a) 22:13

Let the angles subtended at the centres by the arcs and radii of the first and second circles be respectively.
We have:

${\mathrm{\theta }}_{1}=\frac{l}{{r}_{1}}$
$⇒{r}_{1}=\frac{l}{\left(65×\frac{\mathrm{\pi }}{180}\right)}$

${\mathrm{\theta }}_{2}=\frac{l}{{r}_{2}}$
$⇒{r}_{2}=\frac{l}{\left(110×\frac{\mathrm{\pi }}{180}\right)}$

$⇒\frac{{r}_{1}}{{r}_{2}}=\frac{\frac{l}{\left(65×\frac{\mathrm{\pi }}{180}\right)}}{\frac{l}{\left(110×\frac{\mathrm{\pi }}{180}\right)}}=\frac{110}{65}=\frac{22}{13}\phantom{\rule{0ex}{0ex}}$
$⇒{r}_{1}:{r}_{2}=22:13$

#### 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

(d) 8π

#### 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°

(b) 210°

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

#### Question 8:

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