Bihar Board Class 10th Model Paper 2025 Maths Advanced

BIHAR
BOARD
Model
Paper 2025

DOWNLOAD PDF

SECONDARY SCHOOL EXAMINATION-2025
ek/;fed Ldwy ijh{kk&2025
( ANNUAL / okf’kZd )
Subject:– ADVANCED MATHEMATICS ( OPT. )
fo"k;%& mPp xf.kr ¼ ,sfPNd ½
dqy ç”u : 100+30+8 = 138 fo"k; dksM % 114
Total Questions : 100+30+8 = 138 Subject Code : 114
¼ le; : 3 ?kaVs 15 feuV ½ ¼ iw.kkZad:100 ½
[Time : 3 Hours 15 Minutes] [Full Marks:100]

ijh{kkfFkZ;ksa ds fy, funsZ'k%
Instructions for the candidates:
1. ijh{kkFkhZ 𝑂𝑀𝑅 mÙkj i=d ij viuk ç'u iqfLrdk Øekad ¼10 vadks dk½ vo';
fy[ksaA
Candidates 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 their own words as far as
practicable.
3. nkfguh vksj gkf”k;s ij fn;s gq, vad iw.kkaZd fufnZ"V djrs gSaA
Figures in the right hand margin indicate full marks.
4. ç'uksa dks /;ku iwoZd i<+us ds fy, 15 feuV dk vfrfjDr le; fn;k x;k gSA
An extra time of 15 minutes has been allotted for the candidates to read the
questions carefully.

5. ;g ç'u iqfLrdk nks [k.Mksa esa gS – [k.M&v ,oa [k.M&c A
This question booklet is divided into two sections- SECTION-A and
SECTION-B
6. [k.M&v esa 100 oLrqfu"B ç'u gSa] ftuesa ls dsoy 50 ç'uksa dk gh mÙkj nsuk vfuok;Z
gSA 50 ls vf/kd ç”uksa ds mÙkj nsus ij ÁFke 50 mÙkjksa dk gh ewY;kadu fd;k tk,xkA
Page 1 of 30

çR;sd ç'u ds fy, 1 vad fu/kkZfjr gSA budk lgh mÙkj dks miyC/k djk;s x;s
OMR mÙkj i=d esa fn;s x;s lgh fodYi dks uhys@dkys c‚y isu ls izxk<+ djsA
a
fdlh Hkh çdkj ds OgkbVuj@rjy inkFkZ@CysM@uk[kwu vkfn dk OMR mÙkj&iqfLrdk
esa ç;ksx djuk euk gS] vU;Fkk ijh{kk 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
the 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; ç'u gSa] ftuesa ls fdUgha 15 ç'uksa dk mÙkj nsuk vfuok;Z
gSA çR;sd ç”u ds fy, 2 vad fu/kkZfjr gSaA buds vfrfjDr bl [k.M esa 8 nh?kZ
mÙkjh; ç'u fn;s x;s gSa] ftuesa ls fdUgha 4 ç'uksa dk mÙkj nsuk gSA çR;sd ç”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.

Page 2 of 30

[k.M & v / SECTION-A
oLrqfu"B ç'u / Objective Type Questions

ç'u la[;k 1 ls 100 rd ds ç’u ds lkFk pkj fodYi fn, x, gSa] ftuesa ls ,d lgh
gSA fdUgha 50 ç'uksa ds mÙkj vius }kjk pqus x, lgh fodYi dks 𝑶𝑴𝑹 'khV ij
fpfºur djsAa 𝟓𝟎 × 𝟏 = 𝟓𝟎
Questions 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. 50 × 1 = 50

1. r`rh; ikn esa dks.k dk eku gksrk gS
(A) 90 ls 180 (B) 180 ls 2700
(C) 270 ls 360 (D) 0 ls 90
The value of angle in third quadrant is
(A) 90 𝑡𝑜 180 (B) 180 𝑡𝑜 270
(C) 270 𝑡𝑜 360 (D) 0 𝑡𝑜 90
2. 135 ds lefLFkr dks.k dk eku gS
(A) 225 (B) 315
(C) 405 (D) 495
The value of co-terminal angles of 135 is
(A) 225 (B) 315
(C) 405 (D) 495
3. ledks.k dk eku fMxzh esa gksrk gS
(A) 18 (B) 36
(C) 54 (D) 72
The value of right angle in degree is

(A) 18 (B) 36
(C) 54 (D) 72

4. =

Page 3 of 30

(A) 15 (B) 55
(C) 75 (D) 115
5. ;fn fdlh lecgqHkqt dk ,d cfg’dks.k 45 gks rks Hkqtkvksa dh la[;k gksxh
(A) 5 (B) 6
(C) 7 (D) 8
If 45 be an exterior angle of a regular polygon then the number of sides
will be
(A) 5 (B) 6
(C) 7 (D) 8
6. oxZ ds çR;sd cfg’dks.k dk eku xzsM esa gksrk gS
(A) 50 (B) 100
(C) 90 (D) 120
The value of each and every exterior angle of a square in grade is
(A) 50 (B) 100
(C) 90 (D) 120
7. ;fn 𝑟 f=T;k okys fdlh o`Ùk ds pki }kjk dsUæ ij vUrfjr dks.k 𝜃 gks rks
f=T;[k.M dk {ks=Qy gS
(A) (B)

(C) (D)

If 𝜃 be the angle subtended by an arc at the centre of a circle having
radius 𝑟 then area of the sector is

(A) (B)

(C) (D)

8. 135 dk jsfM;u esa eku gksxk
(A) (B)

Page 4 of 30

(C) (D)

The value of 135 in radian will be
(A) (B)

(C) (D)

9. 75 dk jsfM;e esa eku gksrk gS

(A) (B)

(A) (C)

The value of 75 in radian is
(A) (B)

(C) (D)

10. ;fn 7 lseh f=T;k okys f=T;[k.M ds pki dh yackbZ 11 lseh gks] rks f=T;[k.M
dk dsUæh; dks.k gksxk
(A) 45 (B) 60
(C) 90 (D) 120
If 11𝑐𝑚 be the length of the arc of a sector having radius 7𝑐𝑚 then
central angle of the sector will be
(A) 45 (B) 60
(C) 90 (D) 120
11. fuEukafdr esa dkSu ik;Fkkx‚fj;u f=d ds lkekU; :i dks n”kkZrk gS\
(A) (𝑚, 𝑚 − 1, 𝑚 + 1) (B) (𝑚 , 𝑚 − 1, 𝑚 + 1)
(C) (2𝑚, 𝑚 − 1, 𝑚 + 1) (D) buesa ls dksbZ ugha
Which one of the following represents the general form of Pythagorean
triplet?
(A) (𝑚, 𝑚 − 1, 𝑚 + 1) (B) (𝑚 , 𝑚 − 1, 𝑚 + 1)
(C) (2𝑚, 𝑚 − 1, 𝑚 + 1) (D) None of these

Page 5 of 30

12. 𝑡𝑎𝑛𝜃 =
(A) (B)

(C) (D)

13. 𝑡𝑎𝑛𝜃 dk O;qRØe gS
(A) 𝑠𝑒𝑐𝜃 (B) 𝑐𝑜𝑠𝑒𝑐𝜃
(C) 𝑠𝑖𝑛𝜃 (D) 𝑐𝑜𝑡𝜃
Reciprocal of 𝑡𝑎𝑛𝜃 is
(A) 𝑠𝑒𝑐𝜃 (B) 𝑐𝑜𝑠𝑒𝑐𝜃
(C) 𝑠𝑖𝑛𝜃 (D) 𝑐𝑜𝑡𝜃
14. ;fn 𝑡𝑎𝑛𝜃 = rks 𝑠𝑖𝑛𝜃 dk eku gS&

(A) (B)

(C) (D)

If 𝑡𝑎𝑛𝜃 = then the value of 𝑠𝑖𝑛𝜃 is

(A) (B)

(C) (D)

15. ;fn 𝑡𝑎𝑛𝜃 = 𝑐𝑜𝑡𝜃 ] rks θ dk eku xzsM esa gS
(A) 25 (B) 50
(C) 75 (D) 100
If 𝑡𝑎𝑛𝜃 = 𝑐𝑜𝑡𝜃 then the value of 𝜃 in grade is
(A) 25 (B) 50
(C) 75 (D) 100

16. =

(A) ∞ (B) 0
(C) −1 (D) 1

Page 6 of 30

17.fuEukafdr esa 𝑡𝑎𝑛𝜃 ∙ 𝑐𝑜𝑡𝜃 dk eku fdlds cjkcj ugha gS\
(A) sec 𝜃 − tan 𝜃 (B) 𝑐𝑜𝑠𝑒𝑐 𝜃 − cot 𝜃
(C) cos 𝜃 − sin 𝜃 (D) sin 𝜃 + cos 𝜃
Which one of the following is not equal to 𝑡𝑎𝑛𝜃 ∙ 𝑐𝑜𝑡𝜃 ?
(A) sec 𝜃 − tan 𝜃 (B) 𝑐𝑜𝑠𝑒𝑐 𝜃 − cot 𝜃
(C) cos 𝜃 − sin 𝜃 (D) sin 𝜃 + cos 𝜃

18. + dk eku gS
(A) sin 𝜃 𝑐𝑜𝑠𝜃 (B) 𝑠𝑖𝑛𝜃 𝑠𝑒𝑐𝜃
(C) 𝑐𝑜𝑠𝑒𝑐𝜃 cos 𝜃 (D) 𝑠𝑒𝑐𝜃 𝑐𝑜𝑠𝑒𝑐𝑒𝜃

The value of + is

(A) sin 𝜃 𝑐𝑜𝑠𝜃 (B) 𝑠𝑖𝑛𝜃 𝑠𝑒𝑐𝜃
(C) 𝑐𝑜𝑠𝑒𝑐𝜃 cos 𝜃 (D) 𝑠𝑒𝑐𝜃 𝑐𝑜𝑠𝑒𝑐𝑒𝜃

19.fuEukafdr esa 𝑐𝑜𝑠𝑒𝑐 𝜃 − cot 𝜃 ds cjkcj ugha gS
(A) 𝑐𝑜𝑠𝑒𝑐 𝜃 + cot 𝜃 (B) 1 + 2 cot 𝜃
(C) 1 − 2𝑐𝑜𝑠𝑒𝑐 𝜃 (D) 2 𝑐𝑜𝑠𝑒𝑐 − 1
Which one of the following is not equal to 𝑐𝑜𝑠𝑒𝑐 𝜃 − cot 𝜃?
(A) 𝑐𝑜𝑠𝑒𝑐 𝜃 + cot 𝜃 (B) 1 + 2 cot 𝜃
(C) 1 − 2𝑐𝑜𝑠𝑒𝑐 𝜃 (D) 2 𝑐𝑜𝑠𝑒𝑐 − 1

20. fdlh f=Hkqt ds dks.k 3 ∶ 4 ∶ 5 ds vuqikr esa gS]a rks lcls cM+s dks.k dk eku
jsfM;u esa gS
(A) (B)

(C) (D)

Page 7 of 30

The angles of a triangle are in the ratio of 3 ∶ 4 ∶ 5 then the value of the
greatest angle in radian is
(A) (B)

(C) (D)

21.𝑡𝑎𝑛75 =
(A) −2 − √3 (B) −2 + √3
(C) 2 − √3 (D) 2 + √3
22.𝑡𝑎𝑛2𝜃 =

(A) (B)

(C) (D)

23.tan0 tan1 tan2 … tan89 =
(A) –1 (B) 1
(C) ±1 (D) 0
24.𝑠𝑖𝑛𝜃 ∙ 𝑐𝑜𝑠𝜃 ∙ 𝑡𝑎𝑛𝜃 ∙ 𝑐𝑜𝑠𝑒𝑐𝜃 ∙ 𝑠𝑒𝑐𝜃 ∙ 𝑐𝑜𝑡𝜃 =
(A) 0 (B) 1
(C) 2 (D) 3
25.cos(𝜋 + 𝜃) =
(A) −𝑠𝑖𝑛𝜃 (B) 𝑠𝑖𝑛𝜃
(C) −𝑐𝑜𝑠𝜃 (D) 𝑐𝑜𝑠𝜃
26.tan(𝐴 + 𝐵 + 𝐶) =
(A)

(B)

(C)

(D)

27.cos(𝐴 + 𝐵) + cos(𝐴 − 𝐵) =

Page 8 of 30

(A) 2𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵 (B) 2𝑐𝑜𝑠𝐴 𝑠𝑖𝑛𝐵
(C) 2 𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐵 (D) 2𝑐𝑜𝑠𝐴 𝑐𝑜𝑠𝐵

28.tan =

(A) (B) −2

(C) −
√ (D) −
√

29.(𝑡𝑎𝑛𝐴 + 𝑐𝑜𝑡𝐴) =
(A) tan 𝐴 + cot 𝐴 (B) tan 𝐴 − cot 𝐴
(C) sec 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴 (D) 1
30.𝑠𝑒𝑐30 𝑐𝑜𝑠𝑒𝑐45 𝑐𝑜𝑡60
√ √
(A) (B)
√

(C)
√ (D)
√ √

31.𝑡𝑎𝑛2𝐴 =

(A) (B)

(C) (D)

32. 𝑠𝑖𝑛36 =
√ √
(A) (B)

√ √
(C) (D)

33.𝑐𝑜𝑠𝐶 =

(A) (B)

(C) (D)

34. =

(A) 2 (B) 1
(C) 0 (D) –1

Page 9 of 30

35.𝑡𝑎𝑛3𝐴 =

(A) (B)

(C) (D)

36.𝑡𝑎𝑛22 =

(A) 2 + √2 (B) √2 − 1

(C) √2 + 1 (D) 2 − √2

37.𝑡𝑎𝑛𝐴 =

(A) (B)

(C) (D)

38. =

(A) 0 (B) –1
(C) 1 (D) ±1

39.;fn = 45 rks 𝑠𝑖𝑛𝜃 + 𝑐𝑜𝑠𝜃 dk eku gS

(A) 1 (B) 2
(C) –1 (D) 0

If = 45 then the value of 𝑠𝑖𝑛𝜃 + 𝑐𝑜𝑠𝜃 is

(A) 1 (B) 2
(C) –1 (D) 0
40.sec 60 + 𝑐𝑜𝑠𝑒𝑐 30 =
(A) 2 (B) 4
(C) 8 (D) 16

41. =

(A) 𝑠𝑒𝑐 𝜃 (B) 𝑐𝑜𝑠𝑒𝑐 𝜃

Page 10 of 30

(C) cot 𝜃 (D) tan 𝜃
42.;fn 𝑐𝑜𝑠𝑒𝑐 𝜃 = 2 rks tan 𝜃 + cot 𝜃 dk eku gS

(A)
√ (B)
√

(C) 2 (D) 0
If 𝑐𝑜𝑠𝑒𝑐 𝜃 = 2 then the value of tan 𝜃 + cot 𝜃 is

(A)
√ (B)
√

(C) 2 (D) 0
43.;fn 13 sin 𝜃 = 12 rks 𝑐𝑜𝑡𝜃 dk eku gS

(A) (B)

(C) (D)

If 13 sin 𝜃 = 12 then the value of cot 𝜃 is
(A) (B)

(C) (D)

44.𝑐𝑜𝑠73 𝑐𝑜𝑠17 − 𝑠𝑖𝑛73 𝑠𝑖𝑛17 dk eku gksxk
(A) –1 (B) 0
(C) 1 (D) ±1
The value of 𝑐𝑜𝑠73 𝑐𝑜𝑠17 − 𝑠𝑖𝑛73 𝑠𝑖𝑛17 will be

(A) –1 (B) 0
(C) 1 (D) ±1
45.𝑡𝑎𝑛150 =

(A) (B) −
√ √

(C) √3 (D) −√ 3
46.;fn 𝑡𝑎𝑛𝜆 = cot Ψ rks 𝜆 dk eku gS&
(A) 90 − Ψ (B) Ψ − 90
(C) 90 + Ψ (D) buesa ls dksbZ ugha

Page 11 of 30

If 𝑡𝑎𝑛𝜆 = cot Ψ then the value of 𝜆 is

(A) 90 − Ψ (B) Ψ − 90
(C) 90 + Ψ (D) None of these

47.𝑐𝑜𝑠40 + 𝑐𝑜𝑠20 =
(A) √3 𝑠𝑖𝑛10 (B) √3 𝑐𝑜𝑠10
(C) √2 𝑠𝑖𝑛10 (D) √2 𝑐𝑜𝑠10
48.f=Hkqt dh ekf/;dk,¡ ,d nwljs dks fdl vuqikr esa foHkkftr djrh gSa\
(A) 1∶2 (B) 2∶1
(C) 2∶3 (D) 3∶2
In which ratio does intersect the medians of a triangle to each other?

(A) 1:2 (B) 2:1
(C) 2:3 (D) 3:2
49.tan =

( )( ) ( )( )
(A) (B)
( )

( ) ( )
(C) (D)
( )( )

50.;fn 𝛾 = 𝛿 = 45 gks rks 𝑡𝑎𝑛𝛾 + 𝑡𝑎𝑛𝛿 eku gksxk&
(A) 0 (B) 1
(C) 2 (D)

If 𝛾 = 𝛿 = 45 then the value of 𝑡𝑎𝑛𝛾 + 𝑡𝑎𝑛𝛿 will be
(A) 0 (B) 1
(C) 2 (D)

51.f=Hkqt dh ekf/;dkvksa dk dVku fcUnq dgykrk gS
(A) vr%dsUæ (B) yEcdsUæ
(C) e/;dsUæ (D) ifjdsUæ
The point of concurrency of the medians of a triangle is known as
Page 12 of 30

(A) incentre (B) orthocentre
(C) centroid (D) circumcentre
52.;fn ,d Å/okZ/kj ehy ds iRFkj dh Å¡pkbZ lrg ij iM+us okyh mldh Nk;k

dh yackbZ ds xquh gks] rks lw;Z dk mUu;u dks.k gksxk
√

(A) 30 (B) 45
(C) 60 (D) 75

If the height of a vertical mile stone is times of the length of its shadow
√

on the ground, the angle of elevation of the sun is

(A) 30 (B) 45
(C) 60 (D) 75
53.;fn 𝐴 + 𝐵 + 𝐶 = 𝜋 rks cos(𝐶 + 𝐴) =
(A) 𝑐𝑜𝑠𝐵 (B) −𝑐𝑜𝑠𝐵
(C) 𝑠𝑖𝑛𝐵 (D) −𝑠𝑖𝑛𝐵
If 𝐴 + 𝐵 + 𝐶 = 𝜋 then cos(𝐶 + 𝐴) =
(A) 𝑐𝑜𝑠𝐵 (B) −𝑐𝑜𝑠𝐵
(C) 𝑠𝑖𝑛𝐵 (D) −𝑠𝑖𝑛𝐵
54. ;fn ∆ 𝐴𝐵𝐶 esa 𝑏 = 5√3 lseh] 𝑐 = 5 lseh rFkk 𝐵 = 120 rks 𝐶 dk eku gS
(A) 15 (B) 30
(C) 60 (D) 75
If in ∆ 𝐴𝐵𝐶, 𝑏 = 5√3𝑐𝑚, 𝑐 = 5𝑐𝑚 and 𝐵 = 120 then the value of 𝐶 is
(A) 15 (B) 30
(C) 60 (D) 75
55.cos(270 + 𝜃) =
(A) − cos 𝜃 (B) cos 𝜃
(C) − sin 𝜃 (D) sin 𝜃
56.cos(𝐴 − 𝐵) − cos(𝐴 + 𝐵) =

Page 13 of 30

(A) 2𝑐𝑜𝑠𝐴 𝑐𝑜𝑠𝐵 (B) 2𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐵
(C) 2𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵 (D) 2𝑐𝑜𝑠𝐴 𝑠𝑖𝑛𝐵
57.200 fdl prqFkkaZ'k esa fLFkr gS\
(A) çFke (B) f}rh;
(C) r`rh; (D) prqFkZ
In which quadrant does 200 lie?
(A) first (B) second
(C) third (D) fourth
58.sec(−𝜃) =
(A) − sec 𝜃 (B) sec 𝜃
(C) 𝑐𝑜𝑠𝑒𝑐 𝜃 (D) −𝑐𝑜𝑠𝑒𝑐 𝜃
59.;fn 𝑠𝑖𝑛𝜃 = − tgk¡ 𝜃, 180 vkSj 360 ds chp gS] rks 𝜃 dk eku gksxk
(A) 210 , 330 (B) 150 , 330
(C) 210 , 150 (D) buesa ls dksbZ ugha

If 𝑠𝑖𝑛𝜃 = − where 𝜃 lies between 180 and 360 , then the value of 𝜃

will be

(A) 210 , 330 (B) 150 , 330
(C) 210 , 150 (D) None of these
60. ;fn 𝑠𝑖𝑛3𝜃 = 𝑐𝑜𝑠3𝜃 gks rks 𝜃 dk eku gS

(A) 7 (B) 15

(C) 22 (D) 30

If 𝑠𝑖𝑛3𝜃 = 𝑐𝑜𝑠3𝜃 then the value of 𝜃 is

(A) 7 (B) 15

(C) 22 (D) 30

Page 14 of 30

61.fcUnq (𝑥, 0) rFkk (0, 𝑦) ds chp dh nwjh gS&

(A) 𝑥 −𝑦 (B) 𝑥+𝑦

(C) 𝑥 +𝑦 (D) 𝑥−𝑦

Distance between the points (𝑥, 0) and (0, 𝑦) is

(A) 𝑥 −𝑦 (B) 𝑥+𝑦

(C) 𝑥 +𝑦 (D) 𝑥−𝑦

62. fcUnq (13, −22) fdl pFkqrkaZ'k esa fLFkr gS\
(A) çFke (B) f}rh;
(C) r`rh; (D) prqFkZ
In which quadrant does the point (13, −22) lie?
(A) first (B) second
(C) third (D) fourth
63.fcUnq ¼&2]&5½ esa dksfV dk eku gS\
(A) –2 (B) 2
(C) –5 (D) 5

The value of ordinate in the point (−2, −5) is

(A) –2 (B) 2
(C) –5 (D) 5

64.fcUnq (𝑙, 𝑚) esa Hkqt dk eku gS\
(A) √𝑙 (B) 𝑙

(C) √𝑚 (D) 𝑚

Value of abscissa in the point (𝑙, 𝑚) is

(A) √𝑙 (B) 𝑙

(C) √𝑚 (D) 𝑚

Page 15 of 30

65.ewy fcUnq ls fcUnq ¼3]4½ dh nwjh gS
(A) 3 (B) 4
(C) 7 (D) 5

Distance of the point (3, 4) from the origin is

(A) 3 (B) 4
(C) 7 (D) 5

66.rhu fcUnq,¡ ,djSf[kd dgykrs gS ;fn muls cus f=Hkqt dk {ks=Qy gksxk
(A) _.kkRed (B) “kwU;
(C) /kukRed (D) buesa ls dksbZ ugha
Three points are said to be collinear if the area of the triangle so formed
will be
(A) negative (B) zero
(C) positive (D) none of these
67. 'kh"kksaZ (12, 0), (0, 0)rFkk (0, 35) ls cus f=Hkqt dh ifjfefr gS
(A) 47 bdkbZ (B) 37 bdkbZ
(C) 84 bdkbZ (D) 94 bdkbZ
Perimeter of the triangle formed having vertices (12, 0), (0, 0) and
(0, 35) is
(A) 47 𝑢𝑛𝑖𝑡𝑠 (B) 37 𝑢𝑛𝑖𝑡𝑠
(C) 84 𝑢𝑛𝑖𝑡𝑠 (D) 94 𝑢𝑛𝑖𝑡𝑠
68.'kh"kksZa (2, 1), (4, 5) rFkk (0, 3) ls cus f=Hkqt dk {ks=Qy gS&
(A) 4 oxZ bdkbZ (B) 6 oxZ bdkbZ
(C) 9 oxZ bdkbZ (D) 12 oxZ bdkbZ

Area of the triangle formed having vertices (2, 1), (4, 5) and (0, 3) is

(A) 4 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠 (B) 6 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
(C) 9 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠 (D) 12 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠

Page 16 of 30

69.𝑦-v{k ij fLFkr og fcUnq tks fcUnqvksa ¼6] 5½ rFkk ¼&4] 3½ ls lenwjLFk gS
(A) (9, 0) (B) (0, 9)
(C) (0, 3) (D) (0, 6)
The point on 𝑦-axis which is equidistant from the points (6, 5) and
(−4, 3) is
(A) (9, 0) (B) (0, 9)
(C) (0, 3) (D) (0,6)
70.fdlh o`Ùk ds O;kl ds fljksa ds funsZ'kkad ¼2]3½ rFkk ¼6]7½ gSa] rks o`Ùk ds ds Uæ ds
funsZ'kkad gksx
a s
(A) (5, 13) (B) (8, 10)
(C) (4, 5) (D) (5, 4)
The co-ordinates of the ends of diameter of a circle be (2, 3) and (6, 7)
then the co-ordinates of centre of the circle will be
(A) (5, 13) (B) (8, 10)
(C) (4, 5) (D) (5, 4)
71. fcUnqvksa ¼4] &3½ rFkk ¼9] 7½ dks feykusokyh js[kk[k.M dks 3%2 ds vuqikr esa
vUrfoHkZDr djus okyh fcUnq ds funsZ”kkad gSa
(A) (7, 3) (B) (3, 7)
(C) (5, 7) (D) (7, 5)
Co-ordinates of the point dividing internally the line segment joining the
points (4, −3) and (9, 7) in the ratio 3 ∶ 2 are
(A) (7, 3) (B) (3, 7)
(C) (5, 7) (D) (7, 5)

72.fcUnqvksa ¼&5] 4½ vkSj ¼7]&8½ dks feykusokyh js[kk[k.M ds e/; fcUnq ds funsZ'kkad
gSa
(A) (−1, −2) (B) (−1, 2)
(C) (1, −2) (D) (1, 2)

Page 17 of 30

Co-ordinates of the mid point of the line segment joining the points
(−5, 4) and (7, −8) are
(A) (−1, −2) (B) (−1, 2)
(C) (1, −2) (D) (1, 2)
73.;fn fcUnqvksa (2, 3), (5, 𝑘) rFkk (6, 7) ,djSf[kd gksa rks 𝑘 eku gksxk
(A) 6 (B) 3
(C) 2 (D) 4

If the points (2, 3), (5, 𝑘) and (6, 7) are collinear then the value of 𝑘 will
be

(A) 6 (B) 3
(C) 2 (D) 4

74.fcUnqvksa ¼2] 3½ rFkk ¼7] 8½ dks feykusokyh js[kk[k.M dks fcUnq ¼4] 5½ fdl
vuqikr esa foHkDr djrh gS\
(A) 3∶2 (B) 2∶3
(C) 1∶2 (D) 3∶4
In which ratio does the point (4, 5) intersect the line segment joining the
points (2, 3) and (7, 8)?
(A) 3∶2 (B) 2∶3
(C) 1∶2 (D) 3∶4
75.;fn fcUnq,¡ (1, 2), (0, 0) rFkk (𝑥, 𝑦) ,djSf[kd gks rks
(A) 𝑥=𝑦 (B) 2𝑥 = 𝑦
(C) 𝑦 = 2𝑥 (D) 𝑥𝑦 = 10
If the points (1, 2), (0, 0) and (𝑥, 𝑦) are collinear then
(A) 𝑥=𝑦 (B) 2𝑥 = 𝑦
(C) 𝑦 = 2𝑥 (D) 𝑥𝑦 = 10
76.'kh"kksZa ¼1]2½] ¼3]4½ rFkk ¼2]1½ ls cuk f=Hkqt fdl çdkj dk gS\
(A) leckgq (B) lef}ckgq

Page 18 of 30

(C) fo"keckgq (D) ledks.k
Which kind of the triangle is formed having vertices (1, 2), (3, 4) and
(2, 1)?
(A) equilateral (B) isosceles
(C) scalene (D) right angled
77.fcUnq,¡ (𝑘, 1), (1, −1) rFkk (11, 4) laj[s k gSa] rks 𝑘 dk eku gksxk
(A) 5 (B) 4
(C) 3 (D) 2
Points (𝑘, 1), (1, −1) and (11, 4) are collinear then the value of 𝑘 will be
(A) 5 (B) 4
(C) 3 (D) 2
78.“kh’kksZa 1, √3 , (0, 0) rFkk (2, 0) ls cus f=Hkqt ds vUr%dsUæ ds funsZ”kkad gSa
√
(A) 1, (B) ,
√

√
(C) , (D) 1,
√

Co-ordinates of incentre of the triangle formed having vertices 1, √3 ,
(0, 0) and (2, 0) are
√
(A) 1, (B) ,
√

√
(C) , (D) 1,
√

79. fcUnqvksa (𝑎, 𝑏) rFkk , ds chp dh nwjh gS

(A) √𝑎 + 𝑏 √
(B)

√ √
(C) (D)

Distance between the points (𝑎, 𝑏) and , is

(A) √𝑎 + 𝑏 √
(B)

Page 19 of 30

√ √
(C) (D)

80. fcUnqvksa (𝑥 , 𝑦 ) rFkk (𝑥 , 𝑦 ) dks feykus okyh js[kk[k.M dks 𝑚 ∶ 𝑛 ds
vuqikr esa cfgxZr :i ls foHkDr djus okyh fcUnq dh dksfV gS&
(A) (B)

(C) (D)

Ordinate of the point of the line segment joining the points (𝑥 , 𝑦 ) and
(𝑥 , 𝑦 ) intersecting externally in the ration 𝑚: 𝑛 is

(A) (B)

(C) (D)

81.𝑐𝑜𝑠5𝜃 − 𝑐𝑜𝑠9𝜃 =
(A) 2𝑐𝑜𝑠7𝜃 𝑐𝑜𝑠2𝜃 (B) 2𝑠𝑖𝑛7𝜃 𝑠𝑖𝑛2𝜃
(C) 2𝑐𝑜𝑠7𝜃 𝑠𝑖𝑛2𝜃 (D) 2𝑠𝑖𝑛7𝜃 𝑐𝑜𝑠2𝜃
82.fuEukafdr esa fdldk eku bdkbZ ds cjkcj ugha gS\
(A) 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝑒𝑐𝜃 (B) 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃
(C) 𝑐𝑜𝑠𝜃 𝑠𝑒𝑐𝜃 (D) 𝑡𝑎𝑛𝜃 𝑐𝑜𝑡𝜃

Which of the following is not equal to unity?

(A) 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝑒𝑐𝜃 (B) 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃
(C) 𝑐𝑜𝑠𝜃 𝑠𝑒𝑐𝜃 (D) 𝑡𝑎𝑛𝜃 𝑐𝑜𝑡𝜃
83.𝑐𝑜𝑡𝜃 =
√ √
(A) (B)
√ √
(C) √1 − cos 𝜃 (D) √1 − sin 𝜃

Page 20 of 30

84.ledks.k f=Hkqt esa] vk/kkj vkSj Hkqtk ftlds lkFk lanfHkZr dks.k θ cukrh gS]
dgykrh gS\
(A) lef}Hkktd (B) Ekkf/;dk
(C) d.kZ (D) yEc
In a right angled triangle, base and the side together with which form
reference angle 𝜃, is said to be
(A) bisector (B) median
(C) hypotenuse (D) perpendicular
85. fdlh ledks.k f=Hkqt es]a ;fn yEc] vk/kkj rFkk d.kZ Øe'k% 𝑝, 𝑏 rFkk ℎ gksa rks
ℎ¾

(A) 𝑝 −𝑏 (B) 𝑏 −𝑝

(C) 𝑝 +𝑏 (D) (𝑝 + 𝑏)

In a right angled triangle, if perpendicular, base and hypotenuse are 𝑝, 𝑏
and ℎ respectively then ℎ =

(A) 𝑝 −𝑏 (B) 𝑏 −𝑝

(C) 𝑝 +𝑏 (D) (𝑝 + 𝑏)

86.;fn 𝑡𝑎𝑛𝜃 = rks 𝑡𝑎𝑛2𝜃 dk eku gS

(A) (B) −

(C) (D) −

If 𝑡𝑎𝑛𝜃 = then the value of 𝑡𝑎𝑛2𝜃 is

(A) (B) −

(C) (D) −

87.59 ds iwjd dks.k dk eku gS
(A) 149 (B) 121
(C) 301 (D) 31

Page 21 of 30

The value of complementary angle of 59 is

(A) 149 (B) 121
(C) 301 (D) 31

88.fdlh f=Hkqt ds Hkqtkvksa dh yEckbZ Øe'k% 18] 24 rFkk 30 lseh gks rks mldk
{ks=Qy gksxk
(A) 216 lseh2 (B) 256 lseh2

(C) 196 lseh2 (D) 212 lseh2
The length of the sides of a triangle are 18, 24 and 30 cm respectively
then its area will be
(A) 216 𝑐𝑚 (B) 256 𝑐𝑚
(C) 196 𝑐𝑚 (D) 212 𝑐𝑚

89.;fn 𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃 + 1 = 0 rks 𝑠𝑒𝑐𝜃 − 𝑡𝑎𝑛𝜃 =
(A) –1 (B) 1
(C) 0 (D) 2

If 𝑠𝑒𝑐𝜃 + 𝑡𝑎𝑛𝜃 + 1 = 0 then 𝑠𝑒𝑐𝜃 − 𝑡𝑎𝑛𝜃 =

(A) –1 (B) 1
(C) 0 (D) 2

90. vkdk'k esa mM+rs gq, irax dh Mksjh dh x.kuk fdl :i esa gksrh gS\
(A) d.kZ (B) vk/kkj
(C) yEc (D) buesa ls dksbZ ugha
The concept of kite string flying in the sky is considered to be in which
form?
(A) Hypotenuse (B) Base
(C) Perpendicular (D) None of these

Page 22 of 30

91.9%00 cts ?kM+h ds feuV dh lwbZ rFkk ?kaVs dh lwbZ ds chp dk dks.k gksxk
(A) 360 (B) 180
(C) 90 (D) 270

The angle between the minute and hour hands of a clock 9 𝑜′𝑐𝑙𝑜𝑐𝑘 will
be

(A) 360 (B) 180
(C) 90 (D) 270

92. 𝑡𝑎𝑛𝜃 + 𝑐𝑜𝑡𝜃 =
(A) 𝑠𝑒𝑐𝜃 − 𝑐𝑜𝑠𝑒𝑐𝜃 (B) 𝑠𝑒𝑐𝜃 + 𝑐𝑜𝑠𝑒𝑐𝜃
(C) 𝑠𝑒𝑐𝜃 𝑐𝑜𝑠𝑒𝑐𝜃 (D)

93. f=Hkqt ∆𝐴𝐵𝐶 esa c }kjk dkSu lh Hkqtk fu:fir gksxh\
(A) 𝐶𝐴 (B) 𝐵𝐶
(C) 𝐴𝐵 (D) 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐴

In ∆𝐴𝐵𝐶 which side is represented by c?

(A) 𝐶𝐴 (B) 𝐵𝐶
(C) 𝐴𝐵 (D) 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐴

94.tan =

(A) cot (B) cot

(C) cot (D) cot

95.;fn 𝑥𝑐𝑜𝑠 𝜃 = 𝑎 rFkk 𝑦 = 𝑎 𝑡𝑎𝑛𝜃 rks 𝑥 − 𝑦 =
(A) 𝑎 (B) 𝑎
(C) (D)

If 𝑥𝑐𝑜𝑠 𝜃 = 𝑎 and 𝑦 = 𝑎 𝑡𝑎𝑛𝜃 then 𝑥 − 𝑦 =

(A) 𝑎 (B) 𝑎

Page 23 of 30

(C) (D)

96.𝑠𝑒𝑐69 − 𝑐𝑜𝑠𝑒𝑐21 =
(A) –1 (B) 0
(C) 1 (D) 2
97.;fn = rks dEiksusUMks vkSj fMohMsUMks fu’ifr ds QyLo:i gksxk

(A) = (B) =

(C) = (D) None of these

If = then after applying componendo and dividendo it reduces to

(A) = (B) =

(C) = (D) None of these

98.;fn 𝐴 + 𝐵 + 𝐶 = 𝜋 rks 𝑡𝑎𝑛𝐴 + 𝑡𝑎𝑛𝐵 + 𝑡𝑎𝑛𝐶 =

(A) (B) 𝑡𝑎𝑛𝐴 − 𝑡𝑎𝑛𝐵 − 𝑡𝑎𝑛𝐶

(C) 𝑡𝑎𝑛𝐴 𝑡𝑎𝑛𝐵 𝑡𝑎𝑛𝐶 (D) 𝑡𝑎𝑛𝐴 𝑡𝑎𝑛𝐵 + 𝑡𝑎𝑛𝐵 𝑡𝑎𝑛𝐶 + 𝑡𝑎𝑛𝐶 𝑡𝑎𝑛𝐴

If 𝐴 + 𝐵 + 𝐶 = 𝜋 then 𝑡𝑎𝑛𝐴 + 𝑡𝑎𝑛𝐵 + 𝑡𝑎𝑛𝐶 =

(A) (B) 𝑡𝑎𝑛𝐴 − 𝑡𝑎𝑛𝐵 − 𝑡𝑎𝑛𝐶

(C) 𝑡𝑎𝑛𝐴 𝑡𝑎𝑛𝐵 𝑡𝑎𝑛𝐶 (D) 𝑡𝑎𝑛𝐴 𝑡𝑎𝑛𝐵 + 𝑡𝑎𝑛𝐵 𝑡𝑎𝑛𝐶 + 𝑡𝑎𝑛𝐶 𝑡𝑎𝑛𝐴

99. θ dk og eku tks lehdj.k tan 4𝜃 = 𝑐𝑜𝑡2𝜃 dks larq"V djrk gS] gksxk

(A) 7 (B) 15

(C) 18 (D) 22

The value of 𝜃 satisfying the equation tan 4𝜃 = 𝑐𝑜𝑡2𝜃 will be

(A) 7 (B) 15

(C) 18

Page 24 of 30

(D) 22

100. =

(A) tan (B) cot

(C) sec (D) 𝑐𝑜𝑠𝑒𝑐

[k.M&c @ SECTION-B
y?kq mÙkjh; ç'u @Short Answer Type Questions

ç'u la[;k 1 ls 30 rd y?kq mÙkjh; gSa A buesa ls fdUgha 15 ç'uksa ds mÙkj nsaA çR;sd
ç’u ds fy, 2 vad fu/kkZfjr gSA 𝟏𝟓 × 𝟐 = 𝟑𝟎
Question Nos. 1 to 30 are Short Answer Type. Answer any 15 questions. Each
question carries 2 marks. 15 × 2 = 30

1. fl) djsa fd 420 , 1140 rFkk −1380 lefLFkr dks.k gaSA
Prove that 420 , 1140 and −1380 are co-terminal angles.

2. ,d f=Hkqt dk ,d dks.k rFkk nwljk dks.k 80 gks rks rhljs dks.k dk
eku fMxzh esa Kkr djsaA

An angle of a triangle is and another angle be 80 then find the

value of third angle in degree.
3. fdlh dks.k ds iwjd dks.k dk eku mldk vk/kk gks rks mldk eku fudkysaA
The value of complementary angle of an angle is half of the angle then
find its value.
4. ;fn 𝑠𝑒𝑐𝜃 = rks 𝑐𝑜𝑡𝜃 dk eku Kkr djsaA

If 𝑠𝑒𝑐𝜃 = then find the value of 𝑐𝑜𝑡𝜃.

5. ;fn leprqHkqt
Z ds fod.kksaZ dh yackbZ 12 lseh vkSj 16 lseh gks rks blds
çR;sd Hkqtk dh yackbZ Kkr djsaA

Page 25 of 30

If the length of the diagonals of a rhombus are 12 𝑐𝑚 and 16 𝑐𝑚 then
find the length of its each side.

6. ,d lecgqHkqt dk ,d cfg’dks.k gks rks Hkqtkvksa dh la[;k Kkr djsaA

Exterior angle of a regular polygon be then find the number of

sides.
7. 72 fdehå@?kaVk dks ehVj@lsdaM esa cnysaA
Convert 72𝑘𝑚/ℎ𝑟 into 𝑚/𝑠𝑒𝑐.
8. lef}ckgq f=Hkqt dk 'kh"kZ dks.k 96 gks rks vk/kkj ij ds dks.k dk eku Kkr
djsasA
Vertex angle of an isosceles triangle be 96 then find the value of the
base angles.

9. fl) djsa fd = tan

Prove that = tan

10.fl) djsa fd = tan

Prove that = tan

11. cosec 150 dk eku Kkr djasA
Find the value of cosec 150 .

12.,d f=Hkqt dk ,d dks.k gS rFkk nwljk dks.k 75 gS rks rhljs dks.k dk
eki xzsM esa crkosaA
One angle of a triangle be and another angle is 75 then find the

measure of third angle in grade.
13.;fn 𝑥 = 𝑎𝑠𝑖𝑛𝜃 rFkk 𝑦 = 𝑏𝑐𝑜𝑠𝜃 rks 𝑥, 𝑦, 𝑎 rFkk 𝑏 esa lac/a k LFkkfir djsaA
If 𝑥 = 𝑎𝑠𝑖𝑛𝜃 and 𝑦 = 𝑏𝑐𝑜𝑠𝜃 then establish a relation among 𝑥, 𝑦, 𝑎
and 𝑏.
14. Ekku fudkysa (1 − 𝑠𝑖𝑛𝜃)(1 + 𝑠𝑖𝑛𝜃)(1 + tan 𝜃)

Page 26 of 30

Evaluate (1 − 𝑠𝑖𝑛𝜃)(1 + 𝑠𝑖𝑛𝜃)(1 + tan 𝜃).
15.fdlh f=Hkqt dh nks Hkqtkvksa dh yEckbZ Øe'k% 5 lseh rFkk 12 lseh gks rFkk
muds chp dk dks.k ,d ledks.k gks rks ml f=Hkqt dk {ks=Qy Kkr djsa A
Length of two sides of a triangle are 5 𝑐𝑚 and 12 𝑐𝑚 respectively and
the angle between them is a right angle then find the area of the
triangle.
16. ;fn ∆𝐴𝐵𝐶 esa 𝑎 = 1 lseh, 𝑏 = √3 lseh rFkk 𝐴 = 30 rks 𝐵 dk eku
fudkysaA
If in ∆𝐴𝐵𝐶 𝑎 = 1 𝑐𝑚, 𝑏 = √3 𝑐𝑚 and 𝐴 = 30 then find value of 𝐵.
17. 𝑠𝑖𝑛𝜃 + 𝑐𝑜𝑠𝜃 dk egÙke eku fudkysaA
Find the maximum value of 𝑠𝑖𝑛𝜃 + 𝑐𝑜𝑠𝜃.

18.fl) djas fd = − 1.

Prove that = − 1.
19. 𝑡𝑎𝑛3𝜃 dks 𝑡𝑎𝑛𝜃 inksa esa O;Dr djsaA
Express 𝑡𝑎𝑛3𝜃 in terms of 𝑡𝑎𝑛𝜃.
20.fdlh f=Hkqt ds dks.k 2 ∶ 3 ∶ 4 ds vuqikr esa gSa] rks chp okys dks.k dk eku
jsfM;u esa Kkr djsAa
The angles of a triangle are in the ratio of 2 ∶ 3 ∶ 4 then find the value
of the middle angle in radian.
21.;fn ,d 12 ehå Å¡ps [kEHks dh lrg ij iM+us okyh Nk;k dh yEckbZ 4√3
ehå gks] rks lw;Z dk mUu;u dks.k Kkr djsaA
If a pole 12 m high casts a shadow 4√3 𝑚 long on the ground then
find the sun’s elevation.
22.;fn 𝑠𝑖𝑛𝐴 + sin 𝐴 = 1 rks cos 𝐴 + cos 𝐴 dk eku fudkysaA
If 𝑠𝑖𝑛𝐴 + sin 𝐴 = 1 then find the value of cos 𝐴 + cos 𝐴.

Page 27 of 30

23.;fn 𝑥 = asin 𝜃 + 𝑏𝑐𝑜𝑠𝜃 rFkk 𝑦 = 𝑎𝑐𝑜𝑠𝜃 − 𝑏𝑠𝑖𝑛𝜃 rks 𝑥 + 𝑦 dk
eku fudkysAa
If 𝑥 = asin 𝜃 + 𝑏𝑐𝑜𝑠𝜃 and 𝑦 = 𝑎𝑐𝑜𝑠𝜃 − 𝑏𝑠𝑖𝑛𝜃 then find the value of
𝑥 +𝑦 .
24.15 ehå yEch ,d lh<+h nhokj ds lkFk 60 dk dks.k cukrh gS] rks lrg ls
ml fcUnq dh Å¡pkbZ Kkr djs]a tgk¡ lh<+h nhokj dks Li”kZ djrh gSA
A ladder 15 𝑚 long makes an angle of 60 with the wall then find the
height of the point, where the ladder touches the wall.
25.'kh"kksZa ¼8] 0½] ¼5] 0½ vkSj ¼8] 4½ okys f=Hkqt dk {ks=Qy fudkysaA
Find the area of a triangle having vertices (8, 0), (5, 0) and (8, 4).
26.;fn fcUnqvksa ¼&2] 3½ vkSj ¼𝑥, 𝑦½ dks feykus okys js[kk[k.M ds e/; fcUnq ds

funsZ'kkad ¼0] 4½ gSa rks vHkh’V fcUnq ¼𝑥, 𝑦½ Kkr djsaA
If the co-ordinates of mid point of the line segment joining the points
(−2, 3) and (𝑥, 𝑦) be (0, 4) then find the required point (𝑥, 𝑦).
27.eqy fcUnq ls fcUnq ¼&5]&12½ dh nwjh Kkr djsaA
Find the distance of the point (−5, −12) from the origin.
28.fl) djsa fd fcUnq,¡ ¼7]10½] ¼&2] 5½ rFkk ¼3]&4½ ledks.k lef}ckgq f=Hkqt
ds 'kh"kZ gSaA
Show that the points (7, 10), (−2, 5) and (3, −4) are the vertices of a
right isosceles triangle.
29.fcUnq ¼&2]7½ vkSj ¼4] 5½ dks feykus okyh js[kk[k.M o`Ùk dk O;kl gks] rks o`Ùk
ds dsUæ dk fu;ked Kkr djsaA
Line segment joining the points (−2, 7) and (4, 5) be the diameter of
the circle then find the co-ordinates of the centre of the circle.
30. 𝑥 dk eku Kkr djsa ftlls fd fcUnq,¡ ¼&3]12½] ¼7]6½ vkSj ¼𝑥]𝑎½ ,d jSf[kd
gSaA

Page 28 of 30

Find the value of 𝑥 so that the points (−3, 12), (7, 6) and (𝑥, 𝑎) are
collinear.

nh?kZ mÙkjh; ç'u@Long Answer Type Questions

ç'u la[;k 31 ls 38 nh?kZ mÙkjh; ç’u gSa A buesa ls fdUgha 4 ç'uksa ds mÙkj nsa A
çR;sd ç'u ds fy, 4 vad fu/kkZfjr gSA 𝟒 × 𝟓 = 𝟐𝟎

Question Nos. 31 to 38 are Long Answer Type. Answer any 4 questions. Each
question carries 5 marks. 4 × 5 = 20

31.θ dh leLr f=dks.kfefr; fu’ifÙk;ksa dks 𝑐𝑜𝑠𝑒𝑐𝜃 ds :i esa O;Dr djsaA
Express each of the trigonometrical ratios of 𝜃 in the form of 𝑐𝑜𝑠𝑒𝑐𝜃.
32.f=dks.kfefr; O;atd 5 𝑠𝑖𝑛𝜃 + 12𝑐𝑜𝑠𝜃 dk U;qure ,oa egÙke eku
fudkyaAs
Find the minimum and maximum value of the trigonometrical
expression 5 𝑠𝑖𝑛𝜃 + 12𝑐𝑜𝑠𝜃.
33.;fn fdlh f=Hkqt 𝐴𝐵𝐶 esa 𝐴 ∶ 𝐵 ∶ 𝐶 = 1 ∶ 2 ∶ 3 rks 𝑎 ∶ 𝑏 ∶ 𝑐 Kkr djsaA
If in triangle 𝐴𝐵𝐶, 𝐴 ∶ 𝐵 ∶ 𝐶 = 1 ∶ 2 ∶ 3 then find 𝑎 ∶ 𝑏 ∶ 𝑐.
√
34.fl) djsa fd sin 20 𝑠𝑖𝑛 30 sin 40 sin 80 =
√
Prove that sin 20 𝑠𝑖𝑛 30 sin 40 sin 80 = .

35. sin 5𝜃 dks 𝑠𝑖𝑛𝜃 ds inksa esa O;Dr dhft,A
Express sin 5𝜃 𝑖𝑛 terms of 𝑠𝑖𝑛𝜃.
36.,d igkM+h dh f”k[kj ls nks yxkrkj fdyksehVj n”kkZus okys iRFkj dk
voueu dks.k Øe'k% 45 vkSj 30 gks rks igkM+h dh Å¡pkbZ Kkr djsAa
(tgk¡ √3 = 1.732)

Page 29 of 30

From the top of a hill, the angles of depression of two consecutive
kilometre stones are found to be 45 and 30 respectively then find
the height of the hill. ( use √3 = 1.732)
37. ;fn fcUnq,¡ 𝐴(1, −2), 𝐵(2, 3), 𝐶(−3, 2) rFkk 𝐷(−4, −3) lekUrj
prqHkZt 𝐴𝐵𝐶𝐷 ds “kh’kZ gkas rks 𝐴𝐵 dks vk/kkj ekudj lekUrj prqHkZt dh
Å¡pkbZ Kkr djsaA
If the points 𝐴(1, −2), 𝐵(2, 3), 𝐶(−3, 2) and 𝐷(−4, −3) are the
vertices of a parallelogram 𝐴𝐵𝐶𝐷 then considering 𝐴𝐵 as the base
find the height of the parallelogram.
38.;fn fcUnq,¡ (3, −2), (−3, 1) rFkk (4, −3) f=Hkqt 𝐴𝐵𝐶 dh Hkqtk,¡ 𝐵𝐶, 𝐶𝐴
rFkk 𝐴𝐵 ds Øe”k% e/; fcUnq,¡ gksa rks f=Hkqt 𝐴𝐵𝐶 ds “kh’kksZa ds funsZ”kkad Kkr
djsaA
If the points (3, −2), (−3, 1) and (4, −3) be the mid points of the
sides 𝐵𝐶, 𝐶𝐴 and 𝐴𝐵 respectively then find the co-ordinates of the
vertices of triangle 𝐴𝐵𝐶.

Page 30 of 30

Sample Papers

CBSE Sample Papers
ICSE / ISC Board Sample Papers
AP Board Sample Papers
Assam Board Sample Papers
Bihar Board Sample Papers
Chhattisgarh Board Sample Papers
Goa Board Sample Papers
Gujarat Board Sample Papers
Haryana Board Sample Papers
HP Board Sample Papers
J&K State Board Sample Papers
Jharkhand Board Sample Papers
Karnataka Board Sample Papers
Kerala Board Sample Papers

Sample Papers

MP Board Sample Papers
Maharashtra Board Sample Papers
Manipur Board Sample Papers
Mizoram Board Sample Papers
Orissa Board Sample Papers
Punjab Board Sample Papers
Rajasthan Board Sample Papers
Tamil Nadu Board Sample Papers
Telangana State Board Sample Papers
Tripura Board Sample Papers
UP Board Sample Papers
Uttarakhand Board Sample Papers
West Bengal Board Sample Papers