Bihar Board Class 10th Model Paper 2023 Maths Advanced

SECONDARY SCHOOL EXAMINATION – 2023 (ANNUAL)
Model Set
Sub. Code – 114
Advanced Mathematics (Optional)
mPp xf.kr ¼,sfPNd½

Total no. of Questions : 100+30+8 = 138 Full Marks - 100

Instructions for the Candidates :

1- ijh{kkFkhZ OMR mÙkj i=d ij viuk iz’u iqfLrdk Øekad ¼10 vadksa dk½ vo’;
fy[ksaA
Candidate must enter his/her Question Booklet Serial No. (10
digits) in the OMR Answer Sheet.
2- ijh{kkFkhZ ;FkklaHko vius 'kCnksa esa gh mÙkj nsaA
Candidates are required to give their answers in own words as far
as practicable.
3- nkfguh vksj gkf’k, ij fn;s gq, vad iw.kkZad fufnZ"V djrs gSaA
Figures in the right hand margin indicate full marks.
4- iz’uksa dks /;kuiwoZd i<+us ds fy, 15 feuV dk vfrfjDr le; fn;k x;k gSA
15 minutes of extra time has been allotted to the candidates to
read the questions carefully.
5- ;g iz’u iqfLrdk nks [k.Mksa esa gS & ,oa A
This question booklet is divided into two sections – Section-A and
Section-B.

1

6- [k.M&v esa 100 oLrqfu"B iz’u gSa] ftuesa ls fdUgha 50 iz’uksa dk mÙkj nsuk
vfuok;Z gSA ipkl ls vf/kd iz’uksa ds mÙkj nsus ij izFke 50 mÙkjksa dk gh
ewY;kadu }kjk fd;k tk,xkA izR;sd iz’u ds fy, 1 vad fu/kkZfjr gSA budk
lgh mÙkj dks miyC/k djk, x;s OMR mÙkj i=d esa fn, x, lgh fodYi dks
uhys@dkys ckWy isu ls izxk<+ djsaA fdlh Hkh izdkj ds âkbVuj @ rjy inkFkZ
@ CysM @ uk[kwu vkfn dk OMR mÙkj i=d esa iz;ksx djuk euk gS] vU;Fkk
ifj.kke vekU; gksxkA
In Section-A there are 100 objective type questions, out of which
any 50 questions are to be answered. First 50 answers will be
evaluated in case more than 50 questions are answered. Each
question carries 1 mark. For answering these darken the circle
with blue / black ball pen against the correct option on OMR
Answer Sheet provided to you. Do not use Whitener / liquid / blade
/ nail etc. on OMR-sheet, otherwise the result will be treated
invalid.
7- [k.M&c esa 30 y?kq mÙkjh; iz’u gSa] ftuesa ls fdUgha 15 iz’uksa dk mÙkj nsuk
vfuok;Z gSA izR;sd iz’u ds fy, 2 vad fu/kkZfjr gSA buds vfrfjDr] bl [k.M
esa 8 nh?kZ mÙkjh; iz’u fn;s x;s gSa] ftuesa ls fdUgha 4 iz’uksa dk mÙkj nsuk
vfuok;Z gSA izR;sd iz’u ds fy, 5 vad fu/kkZfjr gSA
In Section-B, there are 30 short answer type questions, out of
which any 15 questions are to be answered. Each question carries
2 marks. Apart from these, there are 8 long answer type questions,
out of which any 4 questions are to be answered. Each question
carries 5 marks.
8- fdlh izdkj ds bysDVªkWfud midj.k dk iz;ksx iw.kZr;k oftZr gSA
Use of any electronic appliances is strictly prohibited.

2

[k.M & v @ Section - A
oLrqfu"B iz’u @ Objective Type Qestions

iz’u la[;k 1 ls 100 rd ds iz’u ds lkFk pkj fodYi fn, x, gSa ftuesa ls ,d lgh gSA
fdUgha 50 iz’uksa ds mÙkj vius }kjk pqus x, lgh fodYi dks OMR 'khV ij fpfUgr djsaA
50x1=50
Question nos. 1 to 100 have four options, out of which only one is correct.
Answer any 50 questions. You have to mark your selected option on the
OMR-sheet. 50x1=50

1- 5700 fdl prqFkkZa’k esa gS \

(A) izFke (B) f}rh;

(C) r`rh; (D) prqFkZ

In which quadrant does 5700 lie ?

(A) first (B) second

(C) third (D) fourth

2- &3950 fdl prqFkkZa’k esa gS \

(A) izFke (B) f}rh;

(C) r`rh; (D) prqFkZ

In which quadrant does -3950 lie ?

(A) first (B) second

(C) third (D) fourth

3

3- dks.k 800 dk eku jsfM;u esa gS

(A) (B)

(C) (D)

The value of the angle 800 in radian is –

(A) (B)

(C) (D)

4- 2𝜋 jsfM;u dk eku fdrus ledks.k ds cjkcj gksrk gS \

(A) 1 (B) 2

(C) 3 (D) 4

How many right angles is equal to the value of 2𝜋 radian ?

(A) 1 (B) 2

(C) 3 (D) 4

5- 80 xzsM dk eku fMxzh esa gS &

(A) 480 (B) 540

(C) 720 (D) 800

The value of 80 grade in degree is –

(A) 480 (B) 540

(C) 720 (D) 800

4

6- jsfM;u dk eku fMxzh esa gS &

(A) 100 (B) 180

(C) 200 (D) 360

The value of radian in degree is -

(A) 100 (B) 180

(C) 200 (D) 360

7- fdlh dks.k ds 1850 dh fLFkfr esa ifjHkze.k fdj.k fLFkr gksxh &

(A) izFke ikn esa (B) f}rh; ikn esa

(C) r`rh; ikn esa (D) prqFkZ ikn esa

The position of rotating ray in the case of 1850 will be –

(A) in first quadrant (B) in second quadrant

(C) in third quadrant (D) in fourth quadrant

8- 𝜃 fdl ikn esa gksxk rkfd cos𝜃 vkSj tan𝜃 nksuksa _.kkRed gksa \

(A) izFke (B) f}rh;

(C) r`rh; (D) prqFkZ

In which quadrant does 𝜃 lie such that both cos𝜃 and tan𝜃 are

negative ?

(A) first (B) second

(C) third (D) fourth

5

9- leckgq f=Hkqt dk izR;sd dks.k gksrk gS &

(A) jsfM;u (B) jsfM;u

(C) jsfM;u (D) jsfM;u

Each angle of an equilateral triangle is –

(A) radian (B) radian

(C) radian (D) radian

10- 450 dk ledks.k esa eku gksxk &

(A) 1 (B)

(C) (D)

The value of 450 in right angle will be –

(A) 1 (B)

(C) (D)

11- 4%00 cts ?kM+h ds feuV dh lwbZ vkSj ?kaVs dh lwbZ ds chp dk dks.k gksxk &

(A) 600 (B) 900

(C) 1200 (D) 1800

The angle between the minute and hour hands of a clock at 4:00

o’clock will be -

6

(A) 600 (B) 900

(C) 1200 (D) 1800

12- n Hkqtk okys lecgqHkqt dk izR;sd vUr% dks.k gksrk gS &

(A) (2𝑛 − 4) x 900 (B) x 900

(C) x 1800 (D) x 900

Each interior angle of a regular polygon of n sides is equal to –

(A) (2𝑛 − 4) x 900 (B) x 900

(C) x 1800 (D) x 900

13- fuEufyf[kr esa dkSu vlR; gS \

(A) = (B) 𝜃 =

(C) 𝑙 = 𝜃 x r (D) =

Which of the following is false ?

(A) = (B) 𝜃 =

(C) 𝑙 = 𝜃 x r (D) =

14- 1400 dk lefLFkr dks.k gS &

(A) 3600 (B) 5000

(C) 2200 (D) buesa ls dksbZ ugha

The co-terminal angle of 1400 is –
7

(A) 3600 (B) 5000

(C) 2200 (D) none of these

15- lkekU; ladsrksa ds lkFk fdlh o`Ùk esa ;fn r ¾ 28 lseh0 rFkk 𝑙 ¾ 132 lseh0 rks

𝜃 ¾

(A) jsfM;u (B) jsfM;u

(C) jsfM;u (D) jsfM;u

With usual notation in any circle of r = 28 cm and 𝑙 = 132 cm then

𝜃=

(A) radian (B) radian

(C) radian (D) radian

16- ,d lecgqHkqt dk ,d cfg"dks.k jsfM;u gS] rks Hkqtkvksa dh la[;k gksxh &
(A) 2 (B) 4
(C) 6 (D) 8

The exterior angle of a regular polygon is radian then the number of

sides will be -
(A) 2 (B) 4

(C) 6 (D) 8

17- cosA.tanA =
(A) sinA (B) cosA
(C) 1 (D) cotA

8

18- ;fn cos𝜃 = rks sin𝜃 dk eku gksxk &

(A) (B)

(C) (D)

If cos𝜃 = then the value of sin𝜃 will be

(A) (B)

(C) (D)

19- 5cosec2𝜃 – 5cot2𝜃 =

(A) -5 (B) 5

(C) 0 (D) 1

20- sec𝜃 dk eku tan𝜃 ds inksa esa gksxk &

(A) √1 + 𝑡𝑎𝑛 𝜃 (B) √1 − 𝑡𝑎𝑛 𝜃

(C) √𝑡𝑎𝑛 𝜃 − 1 (D) 1 + tan2𝜃

The value of sec𝜃 in terms of tan𝜃 will be –

(A) √1 + 𝑡𝑎𝑛 𝜃 (B) √1 − 𝑡𝑎𝑛 𝜃

(C) √𝑡𝑎𝑛 𝜃 − 1 (D) 1 + tan2𝜃

21- ;fn cosec𝜃 – cot𝜃 = 𝑥 rks cosec𝜃 dk eku gksxk &

(A) 𝑥− (B) 𝑥+

(C) 𝑥 − (D) 𝑥 +
9

If cosec𝜃 – cot𝜃 = 𝑥 then the value of cosec𝜃 will be –

(A) 𝑥− (B) 𝑥+

(C) 𝑥 − (D) 𝑥 +

22- cos2𝜃.sec2𝜃 =

(A) 1 (B) 0

(C) cos𝜃 (D) sec𝜃

23- =

(A) 0 (B) -1

(C) 1 (D)

24- sec4𝜃 – sec2𝜃 =

(A) tan4𝜃 – tan2𝜃 (B) tan2𝜃 – tan4𝜃

(C) 1 (D) tan2𝜃 + tan4𝜃

𝜃
25- ¾
𝜃

(A) cosec𝜃 + cot𝜃 (B) cosec𝜃 – cot𝜃

(C) cosec2𝜃 – cot2𝜃 (D) cosec2𝜃 + cot2𝜃

26- ;fn 16cot𝜃 = 12 rks =

(A) (B)

10

(C) 0 (D)

If 16cot𝜃 = 12 then =

(A) (B)

(C) 0 (D)

27- cosec1200 =

√
(A) (B)
√

(C) 2 (D) -2

28- tan3150 =

(A) 1 (B) -1

(C) √3 (D) - √3

29- cos(900 - 𝜃) =

(A) cos𝜃 (C) - cos𝜃

(C) sin𝜃 (D) - sin𝜃

30- cos10.cos20.cos30 …. Cos1800 dk eku gS &

(A) 1 (B) -1

(C) 2 (D) 0

The value of cos10.cos20.cos30 …. Cos1800 is –

(A) 1 (B) -1

11

(C) 2 (D) 0

31- cot =

(A) - cot (B) cot

(C) tan (D) - tan

32- cosec2330 – sec2570 dk eku gS &

(A) 1 (B) 0

(C) – 1 (D)

The value of cosec2330 – sec2570 is –

(A) 1 (B) 0

(C) – 1 (D)

33- ;fn 2sin𝜃 – 2cos𝜃 = 0 rks 𝜃 dk eku gksxk &

(A) 300 (B) 600

(C) 450 (D) 900

If 2sin𝜃 – 2cos𝜃 = 0 then the value of 𝜃 will be –

(A) 300 (B) 600

(C) 450 (D) 900

34- cos(-x) =

(A) –sinx (B) sinx

(C) -cosx (D) cosx

12

35- ;fn 𝛼 ¾ rks sin𝛼 – cos𝛼 dk eku gksxk &

√
(A) 0 (B)

√ √
(C) (D)
√

If 𝛼 ¾ then the value of sin𝛼 – cos𝛼 will be

√
(A) 0 (B)

√ √
(C) (D)
√

36- fuEufyf[kr esa ls dkSu tan𝜃 ds cjkcj gS \

(A) tan (900 + 𝜃) (B) tan (900 - 𝜃)

(C) cot (1800 - 𝜃) (D) tan (1800 + 𝜃)

Which of the following is equal to tan𝜃 ?

(A) tan (900 + 𝜃) (B) tan (900 - 𝜃)

(C) cot (1800 - 𝜃) (D) tan (1800 + 𝜃)

37- cos (A + B) =

(A) cosA.CosB – sinA.sinB (B) cosA.cosB + sinA.sinB

(C) cosA.sinB – sinA.cosB (D) sinA.sinB – cosA.cosB

38- sin320.cos280 + cos320.sin280 dk eku gksxk

√
(A) 0 (B)

13

(C) (D)
√

The value of sin320.cos280 + cos320.sin280 will be –

√
(A) 0 (B)

(C) (D)
√

39- sin150 =

√ √
(A) (B)

√ √
(C) (D)
√ √

40- 3sin𝜃 + 4cos𝜃 dk egÙke eku gS &

(A) 7 (B) 5

(C) 4 (D) 3

The maximum value of 3sin𝜃 + 4cos𝜃 is

(A) 7 (B) 5

(C) 4 (D) 3

41- cot(𝑥 – 𝑦) =

. .
(A) (B)

. .
(C) (D)

42- cos(A + B).cos(A – B) =

(A) cos2B – sin2A (B) sin2A – cos2B
14

(C) cos2B + sin2A (D) cos2B – cos2A

43- sin( + A).sin( - A)

(A) 1 + sin2A (B) 1 - sin2A

(C) - sin2A (D) + sin2A

44- sin(x + y) – sin(x – y) =

(A) 2sinx siny (B) 2sinx cosy

(C) 2cosx siny (D) 2cosx cosy

45- ;fn sin800 + sin200 = Ksin500 rks K dk eku gksxk &

(A) √3 (B) √2

(C) 1 (D) 2

If sin800 + sin200 = Ksin500 then the value of K will be –

(A) √3 (B) √2

(C) 1 (D) 2

46- ;fn cos500 + cos400 = √2cosA rks A dk eku gksxk &

(A) 50 (B) 100

(C) 200 (D) 250

If cos500 + cos400 = √2cosA then the value of A will be –

(A) 50 (B) 100

(C) 200 (D) 250

15

47- sin750 – sin150 =

(A) – (B)
√ √

(C) 0 (D) 1

48- sin450 + cosec450 =

(A) (B) √2

(C) (D) 3√2
√

49- 2sin .cos =

√ √
(A) (B)

√
(C) (D)

50- cos( - 𝜃) + cos( + 𝜃) =

(A) √2cos𝜃 (B) 2sin𝜃

(C) 2cos𝜃 (D) √2sin𝜃

51- =

(A) cos600 (B) tan600

(C) sin 600 (D) sin300

52- sec2 – tan2 =

(A) 1 (B) 0

16

(C) -1 (D)

53- =

(A) tan( ) (B) tan(𝑥 + y)

(C) tan( ) (D) –tan(𝑥 – y)

54- tan2𝐴 =

(A) (B)

(C) (D)

55- 3sin𝜃 – 4sin3𝜃 =

(A) 2cos3𝜃 (B) sin3𝜃

(C) cos3𝜃 (D) 2sin3𝜃

56- ;fn cosx = rks cos3x dk eku gksxk &

(A) (B)

(C) - (D)

If cosx = then the value of cos3x will be –

(A) (B)

(C) - (D)

57- 1 – 2sin2450 =

17

(A) 0 (B) 1

(C) (D) - 1

58- ;fn sinx = rFkk cosx = rks sin2x dk eku gksxk &

(A) (B)

(C) (D)

If sinx = and cosx = then the value of sin2x will be -

(A) (B)

(C) (D)

59- 2sin5x cos5x =

(A) sin5x (B) sin10x

(C) cos5x (D) cos210x

60- ;fn sin𝜃 = cos𝜑 rks 𝜑 dk eku gS &

(A) 900 - 𝜃 (B) 900 + 𝜃

(C) 1800 - 𝜃 (D) 1800 + 𝜃

If sin𝜃 = cos𝜑 then the value of 𝜑 is -

(A) 900 - 𝜃 (B) 900 + 𝜃

(C) 1800 - 𝜃 (D) 1800 + 𝜃

61- =

18

(A) tan2 (B) tan

(C) cot (D) cot2

62- 2cos222½0 – 1 =

(A) 0 (B) 1

(C) (D)
√

63- 1 – cos800 =

(A) 2cos2400 (B) 2sin2400

(C) cos2400 (D) 2cos2200

64- 4cos3100 – 3cos100 =

(A) (B)
√

√
(C) (D) 0

65- ;fn 𝜃 = 1400 rks sin – cos dk fpà gksxk

(A) /kukRed (B) _.kkRed

(C) (A) rFkk (B) nksuksa (D) buesa ls dksbZ ugha

If 𝜃 = 1400 then sign of sin – cos will be

(A) positive (B) negative

(C) both (A) and (B) (D) none of these

66- =

19

(A) tanx (B) cotx

(C) sinx (D) cosx

67- ;fn A + B + C = 𝜋 rks cos(B + C) dk eku gS &

(A) cosA (B) – cosA

(C) sinA (D) - sinA

If A + B + C = 𝜋 then the value of cos(B + C) is –

(A) cosA (B) – cosA

(C) sinA (D) - sinA

68- ;fn A + B + C + D = 2𝜋 rks sin dk eku gS &

(A) cos( ) (B) –cos( )

(C) –sin( ) (D) sin( )

If A + B + C + D = 2𝜋 then the value of sin is –

(A) cos( ) (B) –cos( )

(C) –sin( ) (D) sin( )

69- cos2 – sin2 =

√
(A) (B) 0

(C) -1 (D)

20

70- =

(A) -1 (B) 2

(C) 0 (D) 1

71- cot430 =

(A) tan430 (B) tan470

(C) cos470 (D) cosec470

72- =

(A) 1 (B) 0

(C) (D) -1

73- ;fn tan6𝜃 = tan1200 rks 𝜃 dk eku gksxk &

(A) 100 (B) 200

(C) 300 (D) 600

If tan6𝜃 = tan1200 then the value of 𝜃 will be –

(A) 100 (B) 200

(C) 300 (D) 600

74- ;fn =150 rks – sin2A ¾

(A) 0 (B) 1

(C) -1 (D) -

21

If =150 then – sin2A =

(A) 0 (B) 1

(C) -1 (D) -

75- ;fn cos𝜃 = -1 ¼tgk¡ 00≤ 𝜃 ≤1800½ rks 𝜃 dk eku gS

(A) 450 (B) 600

(C) 1500 (D) 1800

If cos𝜃 = -1 (where 00≤ 𝜃 ≤1800) then the value of 𝜃 is

(A) 450 (B) 600

(C) 1500 (D) 1800

76- ;fn √2sinA = 1 ¼tgk¡ 00≤ 𝐴 ≤1800½ rks A dk eku gS &

(A) 600, 1200 (B) 450,1350

(C) 450,1800 (D) 300, 1500

If √2sinA = 1 (where 00≤ 𝐴 ≤1800) then the value of A is –

(A) 600, 1200 (B) 450,1350

(C) 450,1800 (D) 300, 1500

77- ;fn √3sin𝜃 – cos𝜃 = 0 ¼tgk¡ 00≤ 𝜃 ≤3600½ rks 𝜃 dk eku gS &

(A) 900, 2100 (B) 600,1800

(C) 450,1800 (D) 300, 2100

If √3sin𝜃 – cos𝜃 = 0 (where 00≤ 𝜃 ≤3600) then the value of 𝜃 is –

22

(A) 900, 2100 (B) 600,1800

(C) 450,1800 (D) 300, 2100

78- 2sin2𝜃 + 3cos2𝜃 dk U;wure eku gS &

(A) 5 (B) 3

(C) 2 (D) buesa ls dksbZ ugha

The least value of 2sin2𝜃 + 3cos2𝜃 is –

(A) 5 (B) 3

(C) 2 (D) None of these

79- fdlh ∆ABC esa cosB dk eku gksrk gS &

(A) (B)

(C) (D)

In any ∆ABC, the value of cosB is –

(A) (B)

(C) (D)

80- fdlh ∆ABC esa tan .tan =

(A) (B)

(C) (D)

In any ∆ABC tan .tan =
23

(A) (B)

(C) (D)

81- ;fn ∆ABC esa b = 8 lseh0] c = 2 lseh0 rFkk A = 600 rks ∆ABC dk {ks=Qy

gksxk &

(A) 16√3 lseh02 (B) 4√3 lseh02

(C) 16 lseh02 (D) 32 lseh02

In ∆ABC if b = 8cm, c = 2cm and A = 600 then the area of ∆ABC will

be –

(A) 16√3 cm2 (B) 4√3cm2

(C) 16cm2 (D) 32cm2

82- fuEufyf[kr esa ls dkSu lgh gS \

( ) ( )
(A) cos = (B) cos =

( ) ( )( )
(C) cos = (D) cos =
( )

Which of the following is true ?

( ) ( )
(A) cos = (B) cos =

( ) ( )( )
(C) cos = (D) cos =
( )

24

( )
83- ∆ABC esa
( )
=

(A) (B)

(C) (D)

( )
In ∆ABC =
( )

(A) (B)

(C) (D)

84- ;fn ∆ABC esa a = 3, b = 5 rFkk c = 7 rks ∆ABC dk lcls NksVk dks.k gksxk &

(A) A (B) B

(C) C (D) buesa ls dksbZ ugha

In ∆ABC if a = 3, b = 5 and c = 7 then the smallest angle of ∆ABC will

be-

(A) A (B) B

(C) C (D) none of these

85- ,d lh<+h nhokj ds lkFk yxus ij tehu ds lkFk 600 dk dks.k cukrh gSA ;fn

lh<+h dk ikn nhokj ls 2-5 eh0 nwj gS] rks lh<+h dh yackbZ gS &

(A) 2-5 eh0 (B) 4-5 eh0

(C) 10 eh0 (D) 5 eh0

25

A ladder makes an angle of 600 with the ground when placed with a

wall. If the foot of the ladder is 2.5 m away from the wall then the

length of the ladder is –

(A) 2.5 m (B) 4.5 m

(C) 10 m (D) 5 m

86- fcUnq A¼&6] &11½ dh dksfV gS &

(A) -6 (B) -11

(C) -17 (D) 5

The ordinate of the point A(-6, -11) is –

(A) -6 (B) -11

(C) -17 (D) 5

87- fcUnqvksa (a + b, a – b) vkSj (2a + 2b, 2a – 2b) ds chp dh nwjh gS &

(A) √𝑎 + 𝑏 bdkbZ (B) (𝑎 + 𝑏 ) bdkbZ

(C) 2(𝑎 + 𝑏 ) bdkbZ (D) 2(𝑎 − 𝑏 ) bdkbZ

The distance between the points (a + b, a – b) and (2a + 2b, 2a – 2b)

is –

(A) √𝑎 + 𝑏 units (B) (𝑎 + 𝑏 ) units

(C) 2(𝑎 + 𝑏 ) units (D) 2(𝑎 − 𝑏 ) units

88- x &v{k ls fcUnq A(6, 9) dh nwjh gS &

26

(A) 6 bdkbZ (B) 15 bdkbZ

(C) 3 bdkbZ (D) 9 bdkbZ

The distance of the point A(6, 9) from the x-axis is –

(A) 6 units (B) 15 units

(C) 3 units (D) 9 units

89- ;fn fcUnq,¡ (p, o), (o, q) vkSj (2, 2) lajs[k gSa] rks + =

(A) 1 (B)

(C) 2 (D) -

If points (p, o), (o, q) and (2, 2) are collinear then + =

(A) 1 (B)

(C) 2 (D) -

90- ewy fcUnq ls fcUnq P(sin𝜃,cos𝜃) dh nwjh gS &

(A) √2 bdkbZ (B) 2 bdkbZ

(C) 1 bdkbZ (D) buesa ls dksbZ ugha

The distance of the point P(sin𝜃,cos𝜃) from the origin is –

(A) √2 units (B) 2 units
(C) 1 unit (D) none of these
91- fcUnq P(0, 8) vkSj Q(4, 8) dks feykus okyh js[kk PQ ds fy, fuEufyf[kr esa

dkSu&lk lR; gS \

(A) PQ II y-v{k (B) PQ II x-v{k

27

(C) PQ  x-v{k (D) buesa ls dksbZ ugha

Which of the following is true for the line PQ joining the point P(0, 8)

and Q(4, 8) ?

(A) PQ II y-axis (B) PQ II x-axis

(C) PQ  x-axis (D) none of these

92- fcUnq P(10, 6) fLFkr gksxk &

(A) izFke ikn esa (B) prqFkZ ikn esa

(C) f}rh; ikn esa (D) r`rh; ikn esa

The point P(10, 6) lies

(A) in the first quadrant (B) in fourth quadrant

(C) in the second quadrant (D) in the third quadrant

93- fuEufyf[kr esa ls dkSu lk fcUnq r`rh; prqFkkZa’k esa fLFkr gksxk \

(A) (-6, -9) (B) (7, 9)

(C) (8, -6) (D) (-3, 5)

Which of the following points lies in the third quadrant ?

(A) (-6, -9) (B) (7, 9)

(C) (8, -6) (D) (-3, 5)

94- js[kk x = -7 dk vkys[k fuEu esa ls fdl fcUnq ls gksdj xqtjsxh &

(A) (-7, 4) (B) (4, -7)

(C) (0, -7) (D) buesa ls dksbZ ugha

28

The graph of the line x = -7 passes through which of the following

point ?

(A) (-7, 4) (B) (4, -7)

(C) (0, -7) (D) none of these

95- fuEufyf[kr esa ls dkSu&lk fcUnq nwljs prqFkkZa’k esa gSa \

(A) (3, 7) (B) (-8, 10)

(C) (5, 0) (D) (-7, -8)

Which of the following points lies in second quadrant ?

(A) (3, 7) (B) (-8, 10)

(C) (5, 0) (D) (-7, -8)

96- fcUnq A(0, 10) vkSj B(6, 0).dks feykus okyh js[kk[k.M ds e/; fcUnq ds fu;ked

gSa &

(A) (6, 10) (B) (0, 0)

(C) (3, 5) (D) (0, 4)

The coordinates of the midpoint of the line segment joining the points

A(0, 10) and B(6, 0) are

(A) (6, 10) (B) (0, 0)

(C) (3, 5) (D) (0, 4)

97- ;fn ∆PQR ds 'kh"kZ P(4, 9), Q(2, 3) rFkk R(6, 5) gSa rks 'kh"kZ R ls [khaph x;h

ekf/;dk dh yackbZ gS &

29

(A) 5 (B) 10

(C) 25 (D) √10

If P(4, 9), Q(2, 3) and R(6, 5) are the vertices of ∆PQR then the

length of the median drawn through R is –

(A) 5 (B) 10

(C) 25 (D) √10

98- ;fn ∆ABC ds 'kh"kZ A(2, 3), B(5, 6) rFkk C(8, 6) gSa rks bldk dsUnzd dk

fu;ked gksxsa &

(A) (2, 3) (B) (5, 6)

(C) ( , ) (D) (5, 5)

If vertices of the ∆ABC are A(2, 3), B(5, 6) and C(8, 6) then

coordinates of its centroid are

(A) (2, 3) (B) (5, 6)

(C) ( , ) (D) (5, 5)

99- fcUnq P(-1, 3) vkSj Q(4, -7) dks feykus okys js[kk[k.M dks 3 % 4 ds vuqikr esa

vUr% foHkkftr djus okys fcUnq dk x &fu;ked gS &
( ) ( )
(A) (B)

( ) ( )
(C) (D)

The x- co-ordinate of a point which divides the line segment joining
30

P(-1, 3) and Q(4, -7) in the ratio 3 : 4 internally is –

( ) ( )
(A) (B)

( ) ( )
(C) (D)

100- fcUnqvksa (0, 9), (7, 13) vkSj (-15, -3) ls cus f=Hkqt dk {ks=Qy gS &

(A) 24 (B) 12

(C) 8 (D) 4

The area of the triangle formed by points (0, 9), (7, 13) and

(-15, -3) is –

(A) 24 (B) 12

(C) 8 (D) 4

[k.M&c @ Section-B

y?kq mÙkjh; iz’u @ Short Answer Type Questions.

iz'u la[;k 1 ls 30 y?kq mÙkjh; iz’u gSaA buesa ls fdUgha 15 iz’uksa ds mÙkj nsaA izR;sd ds

fy, 2 vad fu/kkZfjr gSA 15x2=30

Question Nos. 1 to 30 are short Answer Type. Answer any 15 questions.

Each question carries 2 marks. 15x2=30

1- 15033^ dks o`Ùkh; eki esa fy[ksaA

Express 15033’ in circular measure.

31

2- ,d f=Hkqt dk ,d dks.k vkSj nwljk dks.k gSa rks rhljs dks.k dh eki

fMxzh esa Kkr djsaA

One angle of a triangle is and the second angle is . Find the

third angle in degree mesure.

3- ;fn fdlh dks.k dh eki fMxzh rFkk xzsM esa Øe’k% D vkSj G gSa] rks fl) djsa

fd G – D = .

If the measure of an angle is D and G in degree and grade

respectively then prove that G – D = .

4- ,d o`Ùk dh f=T;k 10 lseh0 gSA ml pki dh yackbZ Kkr djsa tks dsUnz ij 450

dk dks.k cukrh gSA

The radius of a circle is 10 cm. Find the arc of the circle which

makes an angle of 450 at the centre.

5- ?kM+h esa ?kaVs dh lwbZ }kjk 45 feuV esa cuk,¡ x;s dks.k dks jsfM;u esa fudkysaA

Find the angle in radian made by the hour hand of a clock in 45

minutes.

6- ;fn cot𝜃 = vkSj 𝜃 r`rh; ikn esa gks rks cos𝜃 rFkk cosec𝜃 dk eku Kkr

djsaA

32

If cot𝜃 = and 𝜃 is in third quadrant then find the value of cos𝜃

and cosec𝜃.

7- ;fn tan2A + cot2A = a2 vkSj tanA – cotA = b gks rks

fl) djsa fd a2 – b2=2.

If tan2A + cot2A = a2 and tanA – cotA = b then prove that a2 – b2=2.

8- fl) djsa fd cosec2𝜃 + sec2𝜃 = cosec2𝜃.sec2𝜃.

Prove that cosec2𝜃 + sec2𝜃 = cosec2𝜃.sec2𝜃.

9- f=dks.kferh; vuqikrksa sin(-8700), cos(-8700) rFkk tan(-8700) dks ?kVrs Øe

esa fy[ksaA

Write the trigonometrical ratios sin(-8700), cos(-8700) and

tan(-8700) in descending order.

10- eku Kkr djsa %

Sin(2700 – A).sin(900 – A) – cos(2700 – A)cos(900 + A)

Find the value :

Sin(2700 – A).sin(900 – A) – cos(2700 – A)cos(900 + A)

11- sin1200 – cos1500 + tan1350dk eku Kkr djsaA

Find the value of sin1200 – cos1500 + tan1350.

12- ;fn rhu fcUnq (2, 4), (5, 9) rFkk (x, y) lajs[k gks rks

fl) djsa fd 5x – 3y + 2 = 0.

33

If three points (2, 4), (5, 9) and (x, y) are collinear then prove that

5x – 3y + 2 = 0.

13- ml fcUnq ds fu;ked Kkr djsa tks fcUnqvksa ¼4] 6½ ,oa ¼8] 10½ dks feykus okys

js[kk[kaM dks vkarfjd :Ik ls 2 % 3 ds vuqikr esa foHkkftr djrk gSA

Find the co-ordinates of the point which divides the line segment joining the

points (4, 6) and (8, 10) in the ratio 2 : 3 internally.

14- fl) djsa fd %&

√
sin400.cos200 + cos400 . sin200 =

Prove that :-

√
sin400.cos200 + cos400 . sin200 =

15- fl) djsa fd

cos550 + cos650 – cos50 = 0

Prove that :

cos550 + cos650 – cos50 = 0

16- ;fn A vkSj B U;wudks.k gksa rFkk cotA = vkSj cotB = gks rks cot (A + B)

dk eku Kkr djsaA

If A and B are acute angles and cotA = , cotB = then find the

value of cot (A + B).

17- fl) djsa fd %&

34

cos650.cos250 = cos400.

Prove that : -

cos650.cos250 = cos400.

18- fl) djsa fd : tan540 =

Prove that : tan540 =

19- fl) djsa fd % = 𝑠𝑖𝑛𝑥

Prove that : = 𝑠𝑖𝑛𝑥

20- fl) djsa fd % = cot2𝜃.

Prove that : = cot2𝜃.

21- ;fn tan𝜃 = rks cos2𝜃 dk eku fudkysaA

If tan𝜃 = then find the value of cos2𝜃.

22- fl) djsa fd % tan + cot = 2cosecA.

Prove that : tan + cot = 2cosecA.

23- tan22 dk eku fudkysaA

.
Find the value of tan22

24- ;fn A+B+C = 𝜋 rks fl) djsa fd %&

35

cotB.cotC + cotC.cotA + cotA.cotB = 1

If A+B+C = 𝜋 then prove that

cotB.cotC + cotC.cotA + cotA.cotB = 1

25- gy djsa % cosecx + √2 = 0 tgk¡ 00 ≤ x ≤ 3600.

Solve : cosecx + √2 = 0 where 00 ≤ x ≤ 3600.

26- ;fn fdlh f=Hkqt ABC esa dks.k 3 % 4 % 5 ds vuqikr esa gks rks a : b : c Kkr

djsaA

If angles of any triangle ABC are in the ratio 3 : 4 : 5 then find

a : b : c.

27- fdlh ∆ABC esa] fl) djsa fd = .

In any ∆ABC, prove that = .

28- x dk eku Kkr djsa ftlds fy, fcUnq P(x, 4) rFkk Q(9, 10) ds chp dh nwjh

10 bdkbZ gSA

Find the value of x for which the distance between the points

P(x, 4) and Q(9, 10) is 10 units.

29- y-v{k ij fcUnq Kkr djsa tks fcUnq ¼&5] &2½ rFkk ¼3] 2½ ls lenwjLFk gksA

Find the points on the y-axis which is equidistant from the points

(-5, -2) and (3, 2).

36

30- fl) djsa fcUnq,¡ ¼&2] 3½] ¼4] 0½ rFkk ¼1] &3½ ,d lef}ckgq f=Hkqt ds 'kh"kZ gSaA

Prove that the points (-2, 3), (4, 0) and (1, &3) are the vertices of

an isosceles triangle.

Long Answer Type Questions.

iz'u la[;k 31 ls 38 nh?kZ mÙkjh; iz’u gSaA buesa ls fdUgha 4 iz’uksa ds mÙkj nsaA izR;sd

iz’u ds fy, 5 vad fu/kkZfjr gSA 4x5=20

Question Nos 31 to 38 are Long Answer Type. Answer any 4 questions.

Each question carries 5 marks. 4x5=20

31- T;kferh; fof/k ls fl) djsa fd] sin(A + B) = sinA.cosB + cosA.sinB.

Prove geometrically, sin(A + B) = sinA.cosB + cosA.sinB.

32- fl) djsa fd cos . cos .cos .cos =

Prove that cos . cos .cos .cos =

33- ;fn sin𝜃 + sin𝜑 = p vkSj cos𝜃 + cos𝜑 = q

rks fl) djsa fd sin(𝜃 +𝜑) = .

If sin𝜃 + sin𝜑 = p and cos𝜃 + cos𝜑 = q

then prove that sin(𝜃 +𝜑) = .

34- lehdj.k sec𝛼 + cos𝛼 = ¼tgk¡ 00 ≤ 𝛼 ≤ 2𝜋½ dks gy djsaA

Solve the equation sec𝛼 + cos𝛼 = (where 00 ≤ 𝛼 ≤ 2𝜋).

37

35- ;fn fdlh ∆ABC esa A + B + C = 𝜋]

rks fl) djsa fd cos2A + cos2B + cos2C = -4cosAcosBcosC – 1

In any ∆ABC if A + B + C = 𝜋

then prove that cos2A + cos2B + cos2C = -4cosAcosBcosC – 1.

36- fdlh f=Hkqt ∆ABC esa fl) djsa fd (a + b – c) cot = (a – b + c)cot .

In any ∆ABC Prove that

(a + b – c) cot = (a – b + c)cot .

37- ml fcUnq dk fu;ked fudkysa tks fcUnq ¼1] 2½ vkSj ¼11] 9½ dks feykusokyh

js[kk[kaM dks lef=Hkkftr djrh gSA

Find the co-ordinates of the point which trisects the line segment

joining the points (1, 2) and (11, 9).

38- 28 eh0 pkSM+h lM+d ds nksuksa vksj leku Å¡pkbZ ds nks [kEHks gSaA lM+d ds fdlh

fcUnq ij] tks nksuksa [kEHkksa ds chp esa gS] [kEHkkssa ds 'kh"kZ dk mUu;u dks.k 600 vkSj

300 gSaA [kEHkksa dh Å¡pkbZ rFkk ml fcUnq dh fLFkfr Kkr djsaA

Two pillars of equal height stand on either side of a roadway which

is 28m wide. At a point in the road between the pillars, the angles

of elevations of the tops of the pillars are 600 and 300. Find the

height of the pillar and position of the point.

38