KC Sinha Mathematics Solution Class 11 Chapter 4 कोण और उनके माप (Angles and their measures) Exercise 4.1

Question 1
[Find the radian measure corresponding to the following degree measures]:
Note: रेडियन माप=$\dfrac{\pi}{180^{\circ}}\times $डिग्री माप

(i) 105°
Sol :
रेडियन माप=$\dfrac{\pi}{180^{\circ}}\times 105^{\circ}$

$=\dfrac{7\pi}{12}$

(ii) 25°
Sol :
रेडियन माप=$\dfrac{\pi}{180^{\circ}}\times 25^{\circ}$

$=\dfrac{5\pi}{36}$


(iii) -56°
Sol :
रेडियन माप=$\dfrac{\pi}{180^{\circ}}\times (-56^{\circ})$

$=-\dfrac{14\pi}{45}$

(iv) 7° 30'
Sol :
Note: 1'=$\left(\dfrac{1}{60}\right)^{\circ}$

=$7^{\circ}\times \left(\dfrac{30}{60}\right)^{\circ}$

=$ \left(7\dfrac{1}{2}\right)^{\circ}=\left(\dfrac{15}{2}\right)^{\circ}$

रेडियन माप=$\dfrac{\pi}{180^{\circ}}\times \left(\dfrac{15}{2}\right)^{\circ}$

$=\dfrac{\pi}{24}$


(v) 40° 20'
Sol :

Question 2
निम्नलिखित रेडियन माप के संगत माप ज्ञात कीजिए ($\pi=\frac{22}{7}$ का प्रयोग करे ):
[Find the degree measures corresponding to the following radian measures]
(i)$\frac{7 \pi}{6}$
Sol :
डिग्री माप=$=\dfrac{180^{\circ}}{\pi} \times \dfrac{7 \pi}{6}=210^{\circ}$

(ii) $\frac{5 \pi}{3}$
(iii)$-\frac{5 \pi}{24}$
(iv)$\frac{-2 \pi}{3}$

(v) 6
Sol :
डिग्री माप$=\dfrac{18 0^{\circ}}{\pi} \times 6=7$

$=\left(\dfrac{3780}{11}\right)^{\circ}$

$=\left(343 \dfrac{7}{11}\right)^{0}$

$=343^{\circ}\left(\dfrac{7}{11} \times 60\right)'$

$=343^{\circ}\left(\dfrac{420}{11}\right)^{\prime}$

$=343^{\circ}\left(38 \dfrac{2}{11}\right)^{\prime}$

$=343^{\circ} 38^{\prime}\left(\dfrac{2}{\pi} \times 60\right)^{\prime \prime}$

$=343^{\circ} 38^{\prime} 11^{\prime \prime}$ approx

Question 3
$18^{\circ} 33^{\prime} 45^{\prime \prime}$ को वृत्तीय माप में लिखे ।
[Express $18^{\circ} 33^{\prime} 45^{\prime \prime}$ in circular measure]
Sol :
$18^{\circ} 33^{\prime} 45^{\prime \prime}=18^{\circ} 33^{\prime} \dfrac{45}{60}^{\prime}$

$=18^{\circ}\left(\frac{135}{4}\right)^{\prime}$

$=18^{\circ}\left(\frac{135}{4 \times 60}\right)^{\circ}$

$=18^{\circ}\left(\frac{9}{16}\right)^{\circ}$

$=\left(\frac{297}{16}\right)^{\circ}$

$=\frac{297}{16} \times \frac{\pi}{180}=\dfrac{33\pi}{320}$

Question 4
किसी त्रिभुज के कोण 1:3:5 की निष्पत्ति में है , तो उनका मान रेडियन में निकाले ।
[The angles of a triangle are in the ratio 1:3:5 , find them in radian]
Sol :
$\begin{aligned} Let & \angle A=x \\ & \angle B=3 x \\ & \angle C=5 x \end{aligned}$

$\angle A+\angle B+\angle C=180^{\circ}$

$x+3 x+5 x=180^{\circ}$

$9 x=180^{\circ}$

$x=\dfrac{18 0^\circ}{9}=20^{\circ}$

∠A=x=20° $=20 \times \frac{\pi}{180}=\frac{\pi}{9}$

∠B=3x=3×20° $=60^{\circ} \times \frac{\pi}{180}=\frac{\pi}{3}$

∠C=5x=5×20° =100° $=100\times \dfrac{\pi}{180}=\dfrac{5\pi}{9}$


Question 5
Sol :
Angle of assumed triangle are
a-d , a , a+d

The sum of angle of triangle is 180°

a-d+a+a+d=180°

3a=180°

$a=\frac{180^{\circ}}{3}$

a=60°

$\frac{a+d}{a-d}=\frac{\pi}{60^{\circ}} \times \frac{180^{\circ}}{\pi}$

$\frac{60+d}{60-d}=3$

60+d=180-3d

3d+d=180-60

4d=120

$d=\frac{120}{4}=30$

पहला कोण=a-d=60-30=30°

दूसरा कोण=a=60°

तीसरा कोण=a+d=60+30=90°


Question 6
Sol :

Angle between two needle

$=\frac{360^{\circ}}{12} \times 2 \frac{1}{2}$

$=30^{\circ} \times \frac{5}{2}$

=75°

$=75^{\circ} \times \frac{\pi}{180^{\circ}}$

$=\frac{5 \pi}{12}$


Question 7
Sol :

Angle between two needle
$=\frac{360}{12} \times 4$

$=120^{\circ} \times \frac{\pi}{180^{\circ}}$

$=\frac{2 \pi}{3}$


Question 8
Sol :
L=37.4 cm

$\theta=60^{\circ}=60^{\circ} \times \frac{\pi}{180^{\circ}}$

$\theta=\frac{\pi}{3}$

$r=\frac{\text{L}}{\theta}$

$=\frac{37 \cdot 4}{\frac{\pi}{3}}=\frac{37 \cdot 4}{\frac{22}{7 \times 3}}$

$=\frac{37 \cdot 4}{22} \times 7 \times 3$

=35.7 cm


Question 9
Sol :
The radius of both the circle is r₁ and r₂
And the angles are θ₁ and θ₂ both circle have their length l
θ₁=65° θ₂=210°

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

$r_{1}=\frac{l}{65^{\circ}}$..(i)

$r_{2}=\frac{l}{110^{\circ}}$..(ii)

On dividing (i) by (ii)

$\frac{r_{1}}{r_{2}}=\dfrac{\frac{l}{65^{\circ}}}{\frac{l}{110^{\circ}}}=\frac{110^{\circ}}{65^{\circ}}$

=22:13


Question 10
Sol :

r=1.5 cm

$\theta=\frac{360^{\circ}}{12} \times 8=240^{\circ}$

$\theta=240^{\circ}=240^{\circ} \times \frac{\pi}{180^{\circ}}$

$\theta=\frac{4 \pi}{3}$


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

$\frac{4 \pi}{3}=\frac{l}{1 \cdot 5}$

$\frac{4 \times 3.14}{3} \times 1.5=l$

6.28 cm =l


Question 11
Sol :

θ=30°$=\left(\frac{30^{\circ}}{60^{\circ}}\right)=\frac{1}{2}^{\circ}$

$=\frac{1}{2} \times \frac{\pi}{180^{\circ}}=\frac{\pi}{360^{\circ}}$

l=1 cm

$r=\frac{l}{\theta}=\frac{1 \mathrm{cm}}{\frac{\pi}{360^{\circ}}}$

$=\frac{1 \times 360^{\circ}}{\pi} \mathrm{cm}$

$=\frac{360^{\circ}}{22} \times 7$

$=\frac{1260}{11} \mathrm{cm}$

$=114 \frac{6}{11} \mathrm{cm}$


$=\operatorname{1m} 14 \frac{6}{11} \mathrm{cm}$


Question 12
Sol :

The diameter of the moon is d

$\theta=31^{\prime}=\frac{31}{60} \times \frac{\pi}{180^{\circ}}$

r=38400 km

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

l=θ×r

$=\frac{31}{60} \times \frac{\pi}{180^{\circ}} \times 38400 \mathrm{km}$

$=\frac{31}{9} \times \frac{22}{7} \times 32 \mathrm{km}$

$=\frac{21824}{63} \mathrm{km}$

$=346 \frac{26}{63} \mathrm{km}$


Question 13
Sol :
d=4 feet

Speed of train=𝜋×4×6 feet/r

=24𝜋 feet/r