Bihar Board Class 10th Model Paper 2026 Maths Advance

iz'u iqfLrdk Øekad / Question Booklet Serial No. :

SECONDARY SCHOOL EXAMINATION – 2026
ek/;fed Ldwy ijh{kk & 2026
¼ANNUAL@okf"kZd ½
fo"k; dksM % MODEL QUESTION PAPER Question Booklet Set

Subject Code :
114 ADV. MATHEMATICS (OPT.)
Code

mPp xf.kr ¼,sfPNd½
dqy iz'u % 100 $ 30 $ 8 ¾ 138 dqy eqfnzr i`"B % 35
Total Questions : 100 + 30 +8 = 138 Total Printed Pages : 35
¼le; % 3 ?kaVs 15 feuV½ ¼iw.kkZad % 100½
[Time : 3 Hours 15 Minutes] [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
Candidates must enter their 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 answers in their own words as far as practicable.

3- nkfguh vksj gkf'k;s 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 vfrfjä le; fn;k x;k gSA
An extra time of 15 minutes have been allotted for the candidate to read the questions carefully.

5- ;g iz'u i= nks [k.Mksa esa foHkä gS & [k.M Þvß ,oa [k.M ÞcßA
This question booklet is divided into two sections – Section “A” and Section “B”.

6- [k.M Þvß esa 100 oLrqfu"B iz'u gSa ftuesa 50 iz'u vfuok;Z gSaA ;fn ijh{kkFkhZ 50 ls vf/kd iz'uksa ds mÙkj
nsr@
s nsrh gSa] rks izFke 50 mÙkjksa dk gh ewY;kadu fd;k tk,xkA izR;sd iz'u ds fy, 1 vad fu/kkZfjr gSA
buds mÙkj nsus gsrq OMR mÙkj&i=d ij lgh fodYi dks dkys@uhys ckWy isu ls izxk<+ djsAa fdlh Hkh
izdkj ds g~okbVuj@ rjy inkFkZ@ CysM@uk[kwu vkfn dk OMR mÙkj i=d esa iz;ksx ugha djsa vU;Fkk
ijh{kk ifj.kke vekU; gksxkA
In Section – A there are 100 objective type question out of which 50 questions are compulsory. If
the candidates answer more than 50 questions, the first 50 questions only will be evaluated. Each

Page 1 of 35

question carries 1 mark. For answering these darken the correct circle on the OMR answer sheet
with black/blue ball pen. Do not use whitener/liquid material/ blade/ nail etc on the OMR answer
sheet otherwise the result will be invalid.

7- [k.M Þcß esa 30 y?kq mÙkjh; iz'u gSAa ftuesa ls fdUgha 15 iz'uksa ds mÙkj nsuk vfuok;Z gSA izR;sd iz'u ds
fy, 2 vad fu/kkZfjr gSAa buds vfrfjä] bl [k.M esa 8 nh?kZ mÙkjh; iz'u fn;s x;s gSa] ftuesa ls fdUgha 4
iz'uksa dk mÙkj nsuk gSA izR;sd iz'u ds fy, 5 vad fu/kkZfjr gSAa
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 35

[k.M & v @ Section-A
oLrqfu"B iz'u @ Objective Type Questions
iz'u la[;k 1 ls 100 rd ds izR;sd 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 fpfàr djsAa
50 x 1 = 50
Question Nos. 1 to 100 have four options, out of which only one is correct, you have to mark your
selected option on the OMR sheet. Answer any 50 questions. 50 x 1 = 50

1- fuEufyf[kr esa dkSu izFke prqFkkZ'a k esa fLFkr gksxk \

(A) -2800 (B) -1850

(C) -800 (D) -1000

Which of the following will lie in first quadrant?

(A) -2800 (B) -1850

(C) -800 (D) -1000

2- 15800 fdl prqFkkZ'a k esa gS \

(A) izFke (B) f}rh;

(C) r`rh; (D) prqFkZ

In which quadrant does 15800 lie?

(A) First (B) Second

(C) Third (D) Fourth

3- 200 ¾

(A) (B)

(C) (D)

Page 3 of 35

4- =

(A) 3080 (B) 3060

(C) 2750 (D) 2880

5- fuEufyf[kr esa dkSu xyr gS \

(A) 1 ledks.k ¾ 900 (B) 1 xzsM ¾

(C) π jsfM;u ¾ 100 xzsM (D) 10 ¾ ( ) jsfM;u

Which of the following is false?

(A) 1 right angle = 900 (B) 1 grade =

(C) π radian = 100 grade (D) 10 = (𝜋/180) radian

6- 900 dk eku xzsM esa gksxk &

(A) 100 (B) 200

(C) 110 (D) 300

The value of 900 in grade will be

(A) 100 (B) 200

(C) 110 (D) 300

7- 𝜃 fdl ikn esa gksxk rkfd 𝑡𝑎𝑛𝜃 /kukRed rFkk 𝑠𝑒𝑐𝜃 _.kkRed gks \

(A) izFke (B) f}rh;

(C) r`rh; (D) prqFkZ

In which quadrant does 𝜃 lie such that 𝑡𝑎𝑛𝜃 is positive and 𝑠𝑒𝑐𝜃 is negative?

Page 4 of 35

(A) First (B) Second

(C) Third (D) Fourth

8- iapHkqt ds var% dks.kksa dk ;ksx gksrk gS &

(A) 8400 (B) 5400

(C) 7000 (D) 5000

The sum of the interior angles of a pentagon is

(A) 8400 (B) 5400

(C) 7000 (D) 5000

9- n Hkqtk okys lecgqHkqt dk izR;sd cfg"dks.k gksrk gS &

(A) (B)

(C) (D)

Each exterior angle of a regular polygon of n sides is

(A) (B)

(C) (D)

10- 1500 dk lefLFkr dks.k gS

(A) -2100 (B) 5100

(C) -5100 (D) buesa ls dksbZ ugha

The Co-terminal angle of 1500 is
(A) -2100 (B) 5100

(C) -5100 (D) None of these

Page 5 of 35

11- fdlh Hkh o`Ùk esa lkekU; ladsru ds lkFk fuEufyf[kr esa dkSu lR; gS \

(A) 𝜃 = (B) 𝜃 =

(C) 𝜃 = 𝑙 × 𝑟 (D) buesa ls dksbZ ugha

With usual notation in any circle which of the following is true?

(A) 𝜃 = (B) 𝜃 =

(C) 𝜃 = 𝑙 × 𝑟 (D) None of these

12- 6 % 30 cts ?kM+h ds feuV vkSj ?kaVs dh lwbZ ds chp dk dks.k gksxk

(A) 300 (B) 450

(C) 600 (D) 150

The angle between the minute and hour hands of a Clock at 6:30 O'clock will be

(A) 300 (B) 450

(C) 600 (D) 150

13- fdlh o`Ùk esa ;fn 𝑙 = 12 lsehŒ rFkk 𝑟 = 3 lsehŒ gks] rks θ dk eku ¼jsfM;u es½a gksxk

(A) 4 (B) 1/4

(C) 36 (D) 8

In any circle if 𝑙 = 12 cm and 𝑟 = 3 cm. then the value of θ (in radian) will be

(A) 4 (B) 1/4

(C) 36 (D) 8

14- ;fn 𝑠𝑒𝑐𝐴 = rks 7(𝑐𝑜𝑠𝑒𝑐 𝐴 − cot 𝐴) =

(A) 0 (B) 25

(C) 7 (D) -7

Page 6 of 35

If 𝑠𝑒𝑐𝐴 = then 7(𝑐𝑜𝑠𝑒𝑐 𝐴 − cot 𝐴) =

(A) 0 (B) 25

(C) 7 (D) -7

15- 3 =

(A) 1 (B) 3

(C) -1 (D) -3

16- 7 cos 𝐴 × 7 sec 𝐴 =

(A) 7 (B) 1

(C) 14 (D) 49

17- =

(A) 1 (B) 0

(C) -1 (D) −1/2

18- ;fn 𝑠𝑒𝑐𝐴 = √𝑥 rks tan 𝐴 =
(A) (𝑥 − 1) (B) √1 − 𝑥

(C) 1 − 𝑥 (D) 1 − 𝑥

If 𝑠𝑒𝑐𝐴 = √𝑥 𝑡ℎ𝑒𝑛 tan 𝐴 =

(A) (𝑥 − 1) (B) √1 − 𝑥

(C) 1 − 𝑥 (D) 1 − 𝑥

Page 7 of 35

19- ;fn 𝑐𝑜𝑠𝑒𝑐𝜃 − 𝑐𝑜𝑡𝜃 = rks 𝑐𝑜𝑡𝜃 =

(A) (B)

(C) (D)

If 𝑐𝑜𝑠𝑒𝑐𝜃 − 𝑐𝑜𝑡𝜃 = then 𝑐𝑜𝑡𝜃 =

(A) (B)

(C) (D)

20- 4(1 − Sin 𝐴) − 4(𝐶𝑜𝑠 𝐴 − 1) =

(A) 8 (B) 4

(C) 0 (D) 12

21- ;fn 𝑐𝑜𝑠𝑒𝑐𝐴 = rks 𝑐𝑜𝑠𝐴 =
√

(A) (B) 2

√
(C) (D) s
√

If 𝑐𝑜𝑠𝑒𝑐𝐴 = then 𝑐𝑜𝑠𝐴 =
√

(A) (B) 2

√
(C) (D) s
√

Page 8 of 35

22- ;fn 𝑥 = 𝑝𝑠𝑖𝑛 ∝ rFkk 𝑦 = 𝑞 𝑡𝑎𝑛 ∝ rks − =

(A) 0 (B) 1

(C) −1 (D) 2

If 𝑥 = 𝑝𝑠𝑖𝑛 ∝ and 𝑦 = 𝑞 𝑡𝑎𝑛 ∝ then − =

(A) 0 (B) 1

(C) −1 (D) 2

23- =

(A) 𝑠𝑒𝑐𝐴 + 𝑡𝑎𝑛𝐴 (B) 𝑠𝑒𝑐𝐴 − 𝑡𝑎𝑛𝐴

(C) 1 (D) 𝑠𝑒𝑐 𝐴 + tan 𝐴

24- ;fn ∆ABC esa, ∠𝐵 = 90 , AB=24 lsehŒ rFkk BC = 7 lsehŒ rks sinA =

(A) (B)

(C) (D)

If in ∆ABC, ∠𝐵 = 90 , AB=24 cm. and BC = 7cm. then sinA =

(A) (B)

(C) (D)

25- 𝑆𝑖𝑛 (270 + 𝜃) =

(A) 𝑆𝑖𝑛𝜃 (B) 𝐶𝑜𝑠𝜃

(C) −𝑆𝑖𝑛𝜃 (D) −𝐶𝑜𝑠𝜃

Page 9 of 35

26- 𝑐𝑜𝑠𝑒𝑐(−𝜃) × 𝑐𝑜𝑠(− 𝜃) =

(A) 𝑡𝑎𝑛𝜃 (B) tan(−𝜃)

(C) cot(−𝜃) (D) 𝑐𝑜𝑡𝜃

27- ;fn 𝑥 = rks 𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠 𝑥 =

(A) 0 (B) 1

(C) −1 (D) 2

If 𝑥 = then 𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠2 𝑥 =

(A) 0 (B) 1

(C) −1 (D) 2

28- 𝑡𝑎𝑛150 =

(A) √3 (B) −√3

1
(C) (D) −
√ √3

29- 𝑠𝑖𝑛(−1215 ) =

(A) − (B)
√ √

1
(C) (D) −
2

30- sec =

(A) (B)
√ √

√ √
(C) (D)

Page 10 of 35

31- 𝑐𝑜𝑡4𝜋 =

(A) 0 (B) 1

(C) -1 (D) vifjHkkf"kr

𝑐𝑜𝑡4𝜋 =

(A) 0 (B) 1

(C) -1 (D) undefined

32- ;fn 𝑥 = 𝑠𝑖𝑛24 rFkk 𝑦 = 𝑐𝑜𝑠240 rks fuEufyf[kr esa dkSu lgh gS \

(A) 𝑥 > 𝑦 (B) 𝑥 < 𝑦

(C) 𝑥 = 𝑦 (D) buesa ls dksbZ ugha

If 𝑥 = 𝑠𝑖𝑛24 and 𝑦 = 𝑐𝑜𝑠24 then which of the following is true?

(A) 𝑥 > 𝑦 (B) 𝑥 < 𝑦

(C) 𝑥 = 𝑦 (D) none of these

33- 𝑐𝑜𝑠(2𝜋 − ∅) =

(A) 𝑐𝑜𝑠∅ (B) −𝑐𝑜𝑠∅

(C) 𝑠𝑖𝑛∅ (D) −𝑠𝑖𝑛∅

34- 2 × 𝑡𝑎𝑛1 × 𝑡𝑎𝑛2 × 𝑡𝑎𝑛3 . . . . . 𝑡𝑎𝑛89 =

(A) 0 (B) 1

(C) −1 (D) 2

35- 𝑐𝑜𝑠(𝜃 + Φ) =

(A) 𝑐𝑜𝑠𝜃. 𝑐𝑜𝑠Φ − 𝑠𝑖𝑛𝜃. 𝑠𝑖𝑛Φ (B) 𝑐𝑜𝑠𝜃. 𝑐𝑜𝑠Φ + 𝑠𝑖𝑛𝜃. 𝑠𝑖𝑛Φ

(C) 𝑐𝑜𝑠𝜃. 𝑠𝑖𝑛Φ + 𝑠𝑖𝑛𝜃. 𝑐𝑜𝑠Φ (D) 𝑠𝑖𝑛θ. 𝑐𝑜𝑠𝜙 − 𝑐𝑜𝑠θ. 𝑠𝑖𝑛𝜙

Page 11 of 35

36- 𝑠𝑖𝑛27 . 𝑐𝑜𝑠9 − 𝑐𝑜𝑠9 . 𝑠𝑖𝑛27 =

(A) 𝑐𝑜𝑠18 (B) 𝑠𝑖𝑛18

(C) 𝑠𝑖𝑛36 (D) 𝑐𝑜𝑠36

37- 𝑐𝑜𝑠22 . 𝑐𝑜𝑠8 − 𝑠𝑖𝑛22 . 𝑠𝑖𝑛8 =

√
(A) (B)

(C) (D) 1
√

38- 8𝑠𝑖𝑛𝐴 + 15𝑐𝑜𝑠𝐴 dk U;wure eku gS
(A) −17 (B) 17

(C) 8 (D) 7

The minimum value of 8𝑠𝑖𝑛𝐴 + 15𝑐𝑜𝑠𝐴 is
(A) −17 (B) 17

(C) 8 (D) 7

39- 𝑐𝑜𝑠 𝐵 − 𝑐𝑜𝑠 𝐴 =

(A) 𝑐𝑜𝑠(𝐴 + 𝐵). 𝑐𝑜𝑠(𝐴 − 𝐵) (B) 𝑠𝑖𝑛(𝐴 + 𝐵). 𝑠𝑖𝑛(𝐴 − 𝐵)

(C) 𝑐𝑜𝑠(𝐴 + 𝐵). 𝑠𝑖𝑛(𝐴 − 𝐵) (D) 𝑐𝑜𝑠(𝐴 − 𝐵). 𝑠𝑖𝑛(𝐴 + 𝐵)

40- 𝑐𝑜𝑡 −𝜃 =

(A) (B)

(C) (D)

Page 12 of 35

41- 𝑠𝑖𝑛 . 𝑐𝑜𝑠 + 𝑐𝑜𝑠 . 𝑠𝑖𝑛 =

(A) 𝑠𝑖𝑛 (B) 𝑠𝑖𝑛

(C) 𝑐𝑜𝑠 (D) 𝑐𝑜𝑠

42- 𝑐𝑜𝑠15𝜋 =

(A) -1 (B) 0

(C) 1 (D)

43- 2𝑐𝑜𝑠𝐴. 𝑠𝑖𝑛𝐵 =

(A) 𝑠𝑖𝑛(𝐴 + 𝐵) − 𝑠𝑖𝑛(𝐴 − 𝐵) (B) 𝑐𝑜𝑠(𝐴 + 𝐵) − 𝑐𝑜𝑠(𝐴 − 𝐵)

(C) 𝑠𝑖𝑛(𝐴 + 𝐵) + 𝑠𝑖𝑛(𝐴 − 𝐵) (D) 𝑐𝑜𝑠(𝐴 + 𝐵) + 𝑐𝑜𝑠(𝐴 − 𝐵)

44- 𝑐𝑜𝑠𝐴 + 𝑐𝑜𝑠𝐵 =

(A) 2𝑐𝑜𝑠 . 𝑐𝑜𝑠 (B) 2𝑠𝑖𝑛 . 𝑠𝑖𝑛

(C) 2𝑠𝑖𝑛 . 𝑐𝑜𝑠 (D) 2𝑐𝑜𝑠 . 𝑠𝑖𝑛

45- ;fn 𝑠𝑖𝑛65 + 𝑠𝑖𝑛 25 = 𝐾 𝑐𝑜𝑠20 rks K dk eku gksxk &

(A) √2 (B) − 2

(C) √3 (D) − 3

If 𝑠𝑖𝑛65 + 𝑠𝑖𝑛 25 = 𝐾 𝑐𝑜𝑠20 then the value of K will be -

(A) √2 (B) − 2

(C) √3 (D) − 3

Page 13 of 35

46- 2𝑐𝑜𝑠55 . 𝑐𝑜𝑠35 =

(A) 𝑐𝑜𝑠40 (B) 𝑐𝑜𝑠20

(C) 𝑠𝑖𝑛40 (D) 𝑠𝑖𝑛20

47- 𝑐𝑜𝑠 − 𝜃 + 𝑐𝑜𝑠 +𝜃 =

(A) √2 𝑐𝑜𝑠𝜃 (B) 2𝑐𝑜𝑠𝜃

(C) √2 𝑠𝑖𝑛𝜃 (D) 2𝑠𝑖𝑛𝜃

48- tan(−𝜃) × 𝑐𝑜𝑠(−𝜃) =

(A) −𝑠𝑖𝑛𝜃 (B) 𝑠𝑖𝑛𝜃

(C) cos𝜃 (D) 1

49- 𝑡𝑎𝑛30 . 𝑡𝑎𝑛60 . 𝑡𝑎𝑛180 =

(A) 0 (B) 1

(C) √3 (D)

50- 𝑠𝑒𝑐 22 − 𝑐𝑜𝑡 68 =

(A) 0 (B) 2

(C) 1 (D) -1

51- ;fn 𝑐𝑜𝑠20 − 𝑐𝑜𝑠40 = 𝑠𝑖𝑛𝜙 rks 𝜙 dk eku gksxk &

(A) 10 (B) 5

(C) 15 (D) 20

If 𝑐𝑜𝑠20 − 𝑐𝑜𝑠40 = 𝑠𝑖𝑛𝜙 then the value of 𝜙 will be

(A) 10 (B) 5

(C) 15 (D) 20

Page 14 of 35

52- =

(A) (B) 1

(C) √3 (D)
√

53- =

(A) 𝑠𝑖𝑛2𝑥 (B) 𝑐𝑜𝑠2𝑥

(C) 𝑡𝑎𝑛2𝑥 (D) 𝑐𝑜𝑡2𝑥

54- 2𝑠𝑖𝑛 𝑐𝑜𝑠 =

(A) (B) 1
√

(C) (D)

55- ;fn 𝑐𝑜𝑠𝐴 = 0.5 rks 𝑐𝑜𝑠3𝐴 dk eku gksxk &

(A) 1 (B) 0

(C) −1 (D)

If 𝑐𝑜𝑠𝐴 = 0.5 then the value of 𝑐𝑜𝑠3𝐴 will be -

(A) 1 (B) 0

(C) −1 (D)

56- 1 + 𝑐𝑜𝑠30 =

(A) 2𝑐𝑜𝑠 15 (B) 2𝑠𝑖𝑛 15

(C) 𝑐𝑜𝑠 15 (D) 𝑠𝑖𝑛 15

Page 15 of 35

57- 𝑐𝑜𝑠 − 𝑠𝑖𝑛 =

(A) (B)
√

√
(C) 1 (D)

58- =

(A) 𝑠𝑖𝑛𝐴 (B) 𝑐𝑜𝑠𝐴

(C) 𝑠𝑖𝑛𝐴 (D) 𝑐𝑜𝑠𝐴

59- −𝑐𝑜𝑠2𝑥 + 2𝑐𝑜𝑠 𝑥 =

(A) 1 (B) −1

(C) 0 (D)

60- 8𝑐𝑜𝑠 15 − 6𝑐𝑜𝑠15 =

(A) √2 (B) 1

(C) (D) 2
√

61- 1 − 2𝑠𝑖𝑛 =
(A) 𝑠𝑖𝑛𝑥 (B) 𝑐𝑜𝑠𝑥

(C) 𝑐𝑜𝑠2𝑥 (D) 𝑠𝑖𝑛2𝑥

62- =

(A) 𝑐𝑜𝑠𝑒𝑐 (B) 𝑠𝑒𝑐
(C) 𝑡𝑎𝑛 (D) 𝑐𝑜𝑡 +

Page 16 of 35

63- 2 cos 22 −1 =

(A) (B)
√

(C) 0 (D) 1

64- =

𝑥 𝑥
(A) 𝑡𝑎𝑛 (B) 𝑐𝑜𝑠
3 3

(C) 𝑡𝑎𝑛3𝑥 (D) 𝑡𝑎𝑛𝑥

65- ;fn 2𝑥 = 𝑠𝑒𝑐𝜃 rFkk = 𝑡𝑎𝑛𝜃 rks 2 𝑥 − =

(A) 1 (B)

(C) (D)

If 2𝑥 = 𝑠𝑒𝑐𝜃 and = 𝑡𝑎𝑛𝜃 then 2 𝑥 − =

(A) 1 (B)

(C) (D)

66- ;fn 𝜙 = 22 rks 𝑡𝑎𝑛2𝜙 − 1 dk eku gksxk &
(A) 1 (B) −1

(C) 0 (D) √3

If 𝜙 = 22 then the value of 𝑡𝑎𝑛2𝜙 − 1 will be -

(A) 1 (B) −1

(C) 0 (D) √3

Page 17 of 35

67- ;fn 𝐴 + 𝐵 + 𝐶 = 𝜋 rks 𝑠𝑖𝑛(𝐴 + 𝐶) =

(A) 𝑠𝑖𝑛𝐵 (B) −𝑠𝑖𝑛𝐵

(C) 𝑐𝑜𝑠𝐵 (D) −𝑐𝑜𝑠𝐵

If 𝐴 + 𝐵 + 𝐶 = 𝜋 then 𝑠𝑖𝑛(𝐴 + 𝐶) =

(A) 𝑠𝑖𝑛𝐵 (B) −𝑠𝑖𝑛𝐵

(C) 𝑐𝑜𝑠𝐵 (D) −𝑐𝑜𝑠𝐵

68- ;fn 𝐴 + 𝐵 + 𝐶 + 𝐷 = 2𝜋 rks 𝑐𝑜𝑡 =

𝐴+𝐶 𝐴+𝐶
(A) tan (B) 𝑐𝑜𝑡
2 2

𝐴+𝐶 𝐴+𝐶
(C) −𝑡𝑎𝑛 (D) −𝑐𝑜𝑡
2 2

If 𝐴 + 𝐵 + 𝐶 + 𝐷 = 2𝜋 then 𝑐𝑜𝑡 =

𝐴+𝐶 𝐴+𝐶
(A) tan (B) 𝑐𝑜𝑡
2 2

𝐴+𝐶 𝐴+𝐶
(C) −𝑡𝑎𝑛 (D) −𝑐𝑜𝑡
2 2

69- ;fn 𝑠𝑒𝑐𝐴 = 𝑐𝑜𝑠𝑒𝑐𝐴 rks 𝐴 dk eku gksxk

(A) 45 (B) 30

(C) 60 (D) 0

If 𝑠𝑒𝑐𝐴 = 𝑐𝑜𝑠𝑒𝑐𝐴 then the value of A will be

(A) 45 (B) 30

(C) 60 (D) 0

Page 18 of 35

70- 𝑐𝑜𝑡36 − 𝑡𝑎𝑛54 =

(A) −1 (B) 1

(C) 0 (D)

71- fuEufyf[kr esa dkSu laHko ugha gS \

(A) 𝑠𝑖𝑛𝐴 = (B) 𝑐𝑜𝑠𝐴 =

(C) 𝑐𝑜𝑠𝑒𝑐𝐴 = (D) 𝑠𝑒𝑐𝐴 = 2

Which of the following is not possible?

(A) 𝑠𝑖𝑛𝐴 = (B) 𝑐𝑜𝑠𝐴 =

(C) 𝑐𝑜𝑠𝑒𝑐𝐴 = (D) 𝑠𝑒𝑐𝐴 = 2

72- ;fn 𝑠𝑖𝑛5𝐴 = 𝑠𝑖𝑛225 rks 𝐴 dk eku gksxk
(A) 0 (B) 30

(C) 45 (D) 60

If 𝑠𝑖𝑛5𝐴 = 𝑠𝑖𝑛225 then the value of A will be

(A) 0 (B) 30

(C) 45 (D) 60

73- ;fn 𝑡𝑎𝑛 ∝= −1(tgk¡ 0 ≤∝≤ 180 ) rks ∝ dk eku gksxk

(A) 45 (B) 60

(C) 120 (D) 135

If 𝑡𝑎𝑛 ∝= −1(where 0 ≤∝≤ 180 ) then the value of ∝ will be

(A) 45 (B) 60

(C) 120 (D) 135

Page 19 of 35

74- ;fn 8𝐴 = 360 rks 1 − 𝑐𝑜𝑠 𝐴 =

(A) (B)
√

(C) 1 (D)

If 8𝐴 = 360 then 1 − 𝑐𝑜𝑠 𝐴 =

(A) (B)
√

(C) 1 (D)

75- ;fn √3𝑐𝑜𝑠𝑒𝑐2𝑥 − 2 = 0 rks 𝑥 =

(A) 15 (B) 30

(C) 40 (D) 90 a

If √3𝑐𝑜𝑠𝑒𝑐2𝑥 − 2 = 0 then 𝑥 =

(A) 15 (B) 30

(C) 40 (D) 90

76- ;fn 𝑠𝑖𝑛𝜙 = − (tgk¡ 0 ≤ 𝜙 ≤ 360 ) rks 𝜙 =

(A) 210 , 330 (B) 60 , 240

(C) 30 , 150 (D) 150 , 210

If 𝑠𝑖𝑛𝜙 = − (where 0 ≤ 𝜙 ≤ 360 ) then 𝜙 =

(A) 210 , 330 (B) 60 , 240

(C) 30 , 150 (D) 150 , 210

Page 20 of 35

∝
77- ;fn = 40 rks 𝑠𝑒𝑐 ∝=

(A) 2 (B) 3

(C) 4 (D)

∝
If = 40 then 𝑠𝑒𝑐 ∝=

(A) 2 (B) 3

(C) 4 (D)

( )( )
78- fdlh ∆𝐴𝐵𝐶 esa =
( )

(A) tan (B) tan

(C) tan (D) buesa ls dksbZ ugha

( )( )
In any ∆𝐴𝐵𝐶 =
( )

(A) tan (B) tan

(C) tan (D) None of these

79- ;fn ∆𝐴𝐵𝐶 esa Hkqtk 𝑎 = 2 lsehŒ] 𝑏 = 3lsehŒ vkSj 𝑠𝑖𝑛𝐴 = rks ∠𝐵 =

(A) 120 (B) 90

(C) 30 (D) 45

In a ∆𝐴𝐵𝐶 if the side 𝑎 = 2𝑐𝑚. , 𝑏 = 3𝑐𝑚. and 𝑠𝑖𝑛𝐴 = 𝑡ℎ𝑒𝑛 ∠𝐵 =

(A) 120 (B) 90

(C) 30 (D) 45

Page 21 of 35

80- fdlh ∆𝐴𝐵𝐶 esa Hkqtk 𝑎 = 2 lsehŒ] ∠𝐶 = 105 rFkk ∠𝐵 = 45 rks Hkqtk 𝑏 =

(A) √2 lsehŒ (B) 2√2 lsehŒ

(C) 4 lsehŒ (D) 2 lsehŒ

In any ∆𝐴𝐵𝐶 the side a=2cm., ∠𝐶 = 105 𝑎𝑛𝑑 ∠𝐵 = 45 then the side b =

(A) √2cm. (B) 2√2cm.

(C) 4cm. (D) 2cm.

81- ;fn ∆𝐴𝐵𝐶 esa Hkqtk 𝑎 = 6 lsehŒ] 𝑐 = 4lsehŒ rFkk ∠𝐵 = 30 rks ∆𝐴𝐵𝐶 dk {ks=Qy
gksxk &
(A) 4lsehŒ2 (B) 2lsehŒ2

(C) 5lsehŒ2 (D) 6lsehŒ2

In ∆𝐴𝐵𝐶 if the side 𝑎 = 6𝑐𝑚. , 𝑐 = 4𝑐𝑚. and ∠𝐵 = 30 then the area of ∆𝐴𝐵𝐶
will be -
(A) 4𝑐𝑚. (B) 2𝑐𝑚.

(C) 5𝑐𝑚. (D) 6𝑐𝑚.

82- ∆𝐴𝐵𝐶 esa =

( )
(A) (B)
( )

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

In ∆𝐴𝐵𝐶 =

( )
(A) (B)
( )

( )
(C) (D) None of these
( )

Page 22 of 35

83- ;fn ∆𝐴𝐵𝐶 esa Hkqtk 𝑎 = 16 lsehŒ] 𝑏 = 24lsehŒ rFkk 𝑐 = 20lsehŒ rks 𝐶𝑜𝑠 =

(A) (B)

(C) (D)

In ∆𝐴𝐵𝐶 if the side 𝑎 = 16𝑐𝑚. , 𝑏 = 24𝑐𝑚. and 𝑐 = 20𝑐𝑚. then 𝑐𝑜𝑠 =

(A) (B)

(C) (D)

84- fdlh ∆𝐴𝐵𝐶 esa 𝑎 + 𝑏 − 2𝑎𝑏. 𝑐𝑜𝑠𝐶 =

(A) 0 (B) 𝐶

(C) abc (D) 1

In any ∆𝐴𝐵𝐶 𝑎 + 𝑏 − 2𝑎𝑏. 𝑐𝑜𝑠𝐶 =

(A) 0 (B) 𝐶

(C) abc (D) 1

85- ∆𝐴𝐵𝐶 esa tan tan =

(A) (B)

(C) (D)

In ∆𝐴𝐵𝐶, tan tan =

(A) (B)

(C) (D)

Page 23 of 35

86- ;fn ,d 6 ehVj Å¡ps [kaHks dh Nk;k i`Foh ij 2√3 ehVj yEch gS] rks lw;Z dk mUu;u

dks.k gS &

(A) 60 (B) 45

(C) 30 (D) 15

If the shadow of a 6 metre high pole is 2√3 metre long on the earth then the angle of elevation
of the sun is -

(A) 60 (B) 45

(C) 30 (D) 15

87- fcUnqvksa (𝑎 𝑠𝑖𝑛15 , 0) rFkk (0, 𝑎 𝑐𝑜𝑠15 ) ds chp dh nwjh gS &

(A) a bdkbZ (B) 2a bdkbZ

(C) √2 𝑎 bdkbZ (D) buesa ls dksbZ ugha

The distance between the points (𝑎 𝑠𝑖𝑛15 , 0) and (0, 𝑎 𝑐𝑜𝑠15 ) is

(A) a units (B) 2a units

(C) √2 a units (D) None of these

88- fcUnq 𝑃(4𝑐𝑜𝑠𝜙, 4𝑠𝑖𝑛𝜙) dh ewy fcUnq ls nwjh gS

(A) 8 bdkbZ (B) 4 bdkbZ

(C) 2 bdkbZ (D) 16 bdkbZ

The distance between the point 𝑃(4𝑐𝑜𝑠𝜙, 4𝑠𝑖𝑛𝜙) from the origin is.

(A) 8 units (B) 4 units

(C) 2 units (D) 16 units

Page 24 of 35

89- fcUnq 𝐴(0, 8) fLFkr gS

(A) 𝑥 − v{k ij (B) 𝑦 − v{k ij

(C) izFke ikn esa (D) prqFkZ ikn esa

The point 𝐴(0, 8) lies

(A) on 𝑥 − 𝑎𝑥𝑖𝑠 (B) on 𝑦 − 𝑎𝑥𝑖𝑠

(C) in first quadrant (D) in fourth quadrant

90- ;fn fcUnq A rFkk B laikrh gksa rks AB dh yackbZ gS

(A) 1 (B) 0

(C) 2 (D) buesa ls dksbZ ugha

If points A and B coincide then the length of AB is

(A) 1 (B) 0
(C) 2 (D) None of these

91- js[kk 𝑦 = dk vkys[k fuEu esa ls fdl fcUnq ls gksdj xqtjsxh \

(A) 3, (B) ,

(C) , 6 (D) ,

The graph of the line 𝑦 = will pass through which of the following points?

(A) 3, (B) ,

(C) , 6 (D) ,

Page 25 of 35

92- fdlh o`Ùk ds O;kl ds Nksjkas ds funsZ'kkad 𝐴 (2, 6) vkSj 𝐵 (8, 12) gSaA o`Ùk ds dsUnz ds

funsZ'kkad gksaxs &
(A) (10, 18) (B) (5, 9)

(C) (6, 6) (D) (3, 3)a

The Co-ordinates of the ends of diameter of a circle are A(2, 6) and (8, 12). The Co-ordinates of
the centre of the circle will be -

(A) (10, 18) (B) (5, 9)

(C) (6, 6) (D) (3, 3)

93- fcUnq (0, 4) vkSj (6, 0) dks feykus okyh js[kk[k.M ds e/;&fcUnq ds fu;ked gSa &

(A) (6, 4) (B) (4, 6)

(C) (2, 3) (D) (3, 2)

The Co-ordinates of the mid-point of the line segment joining the points (0, 4) and (6, 0) are -

(A) (6, 4) (B) (4, 6)

(C) (2, 3) (D) (3, 2)

94- 𝑥 − v{k ij dkSu lk fcUnq] fcUnqvksa P(7, 6) vkSj Q(-3, 4) ls leku nwjh ij gS \

(A) (3, 0) (B) (0, 4)

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

Which point on the x-axis is equidistant from the points P(7, 6) and Q(-3, 4)?

(A) (3, 0) (B) (0, 4)

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

95- 'kh"kZ A(0, 6), B(0, 0) rFkk C(8, 0) okys Δ𝐴𝐵𝐶 dk ifjeki gS &

(A) 14 bdkbZ (B) 24 bdkbZ

(C) 12 bdkbZ (D) 40 bdkbZ

Page 26 of 35

The perimeter of a Δ𝐴𝐵𝐶 with vertices A(0, 6), B(0, 0) and C(8, 0) is

(A) 14 units (B) 24 units

(C) 12 units (D) 40 units

96- ;fn fcUnq,¡ (p, o), (o, q) vkSj (1, 1) lajs[kh gSa rks p + q =

(A) 𝑝𝑞 (B) −𝑝𝑞

1
(C) (D) −
𝑝𝑞

If points (p, o), (o, q) and (1, 1) are collinear then p + q =

(A) 𝑝𝑞 (B) −𝑝𝑞

1
(C) (D) −
𝑝𝑞

97- ;fn fcUnqvksa (a, b), (b, c) rFkk (c, a) }kjk cus f=Hkqt dk dsUnzd ewy fcUnq gS rks
𝑎 +𝑏 +𝑐 =

(A) 3abc (B) a + b + c

(C) abc (D) 0

If the centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is at the origin, then
𝑎 +𝑏 +𝑐 =

(A) 3abc (B) a + b + c

(C) abc (D) 0

98- fcUnqvksa P(3, 3), Q(3, 5) rFkk R(2, 4) ls cus f=Hkqt dk {ks=Qy gksxk &

(A) 4 oxZ bdkbZ (B) 3 oxZ bdkbZ

(C) 2 oxZ bdkbZ (D) 1 oxZ bdkbZ

Page 27 of 35

The area of the triangle formed by the points P(3, 3), Q(3, 5) and R(2, 4) will be -

(A) 4 Square units (B) 3 Square units
(C) 2 Square units (D) 1 Square unit

99- ewy fcUnq dk fu;ked gS &

(A) (1, 1) (B) (0, 1)

(C) (0, 0) (D) (-1, 0)

The Co-ordinates of the origin is

(A) (1, 1) (B) (0, 1)

(C) (0, 0) (D) (-1, 0)

100- fcUnqvksa A(3, 6) rFkk B(4, 8) ds dksfV ds ekuksa dk ;ksx gksxk

(A) 7 (B) 11

(C) 10 (D) 14

The sum of the ordinate values of the points A(3, 6) and B(4, 8) will be

(A) 7 (B) 11

(C) 10 (D) 14

Section-B

Short Answer Type Questions
iz'u la[;k 1 ls 30 rd y?kq mÙkjh; gSaA bueas ls fdUgha 15 iz'uksa ds mÙkj nsaA izR;sd iz'u
ds fy, 2 vad fu/kkZfjr gSaA 15 x 2 = 30
Question Nos 1 to 30 are short Answer Type. Answer any 15 questions. Each question carries 2
marks. 3x2=6

1- 28 38′11" dks jsfM;u esa cnysaA 2
Convert 28 38′11" in radian.

Page 28 of 35

2- fdlh f=Hkqt ds dks.k 3 % 4 % 5 gSA NksVs dks.k dk eku fMxzh esa rFkk cM+s dks.k dk eku
jsfM;u esa Kkr djsaA 2
The angles of a triangle are in the ratio 3 : 4 :5. Find the smallest angle in degree and the greatest
angle in radian.

3- ;fn fdlh lecgqHkqt dk ,d cfg"dks.k jsfM;u gS rks cgqHkqt dh Hkqtkvksa dh la[;k Kkr
djsaA 2
If one of the exterior angle of a regular polygon is radian then find the number of sides of

polygon.

4- ml o`r dh f=T;k Kkr djs]a ftlesa 38-5 lsehΠdk pki dsUnz ij 600 dk dks.k cukrk
gSA 2
Find the radius of the Circle in which an arc of 38.5cm. makes an angle of 60 0 at the centre.

5- ;fn 𝑠𝑖𝑛𝜙 = − vkSj 𝜙 prqFkZ ikn esa fLFkr gks rks 𝑐𝑜𝑡𝜙 dk eku Kkr djsaA 2

If 𝑠𝑖𝑛𝜙 = − and 𝜙 lies in fourth quadrant then find the value of 𝑐𝑜𝑡𝜙 .

6- fl) djsa fd 2
𝑡𝑎𝑛 𝐴 − 𝑠𝑖𝑛 𝐴 = tan 𝐴. 𝑠𝑖𝑛 𝐴
Prove that
𝑡𝑎𝑛 𝐴 − 𝑠𝑖𝑛 𝐴 = tan 𝐴. 𝑠𝑖𝑛 𝐴.

7- ;fn 3𝑐𝑜𝑡𝐴 = 4 rks dk eku Kkr djsaA 2

If 3𝑐𝑜𝑡𝐴 = 4 then find the value of

8- 2𝑠𝑒𝑐 30 + 3𝑠𝑖𝑛 45 − 𝑐𝑜𝑠0 dk eku Kkr djsAa 2

Find the value of 2𝑠𝑒𝑐 30 + 3𝑠𝑖𝑛 45 − 𝑐𝑜𝑠0 .

Page 29 of 35

9- fl) djsa fd 2
𝑐𝑜𝑠18 + 𝑐𝑜𝑠38 + 𝑐𝑜𝑠162 + 𝑐𝑜𝑠142 = 0
Prove that

𝑐𝑜𝑠18 + 𝑐𝑜𝑠38 + 𝑐𝑜𝑠162 + 𝑐𝑜𝑠142 = 0

10- fl) djsa fd 2
𝑐𝑜𝑠 + 𝑠𝑖𝑛 − 2𝑐𝑜𝑡 =
Prove that

𝑐𝑜𝑠 + 𝑠𝑖𝑛 − 2𝑐𝑜𝑡 =

11- 𝑡𝑎𝑛15 dk eku Kkr djsaA 2
Find the Value of 𝑡𝑎𝑛15 .

12- ;fn A vkSj B U;wudks.k gksa rFkk 𝑐𝑜𝑡𝐴 = , 𝑐𝑜𝑡𝐵 = rks] fl) djsa fd
𝐴+𝐵 = . 2

If A and B are acute angle and 𝑐𝑜𝑡𝐴 = , 𝑐𝑜𝑡𝐵 = then prove that 𝐴 + 𝐵 = .

13- fl) djsa fd

𝑐𝑜𝑠 − 𝐴 𝑐𝑜𝑠 − 𝐵 − 𝑠𝑖𝑛 − 𝐴 . 𝑠𝑖𝑛 − 𝐵 = 𝑠𝑖𝑛(𝐴 + 𝐵) 2
Prove that

𝑐𝑜𝑠 − 𝐴 𝑐𝑜𝑠 − 𝐵 − 𝑠𝑖𝑛 − 𝐴 . 𝑠𝑖𝑛 − 𝐵 = 𝑠𝑖𝑛(𝐴 + 𝐵)

14- ;fn 𝜙 = 110 rks (𝑠𝑖𝑛𝜙 + 𝑐𝑜𝑠𝜙) dk fpà Kkr djsaA 2
If 𝜙 = 110 then find the sign of (𝑠𝑖𝑛𝜙 + 𝑐𝑜𝑠𝜙).

15- fl) djsa fd 2
= 𝑡𝑎𝑛2𝐴

Page 30 of 35

Prove that

= 𝑡𝑎𝑛2𝐴

16- fl) djsa fd 2
𝑐𝑜𝑠𝜃. 𝑐𝑜𝑠(60 + 𝜃). 𝑐𝑜𝑠(60 − 𝜃) = 𝑐𝑜𝑠3𝜃.
Prove that

𝑐𝑜𝑠𝜃. 𝑐𝑜𝑠(60 + 𝜃). 𝑐𝑜𝑠(60 − 𝜃) = 𝑐𝑜𝑠3𝜃..

17- ;fn 𝑐𝑜𝑠 ∝= rks 𝑠𝑖𝑛4 ∝ dk eku Kkr djsaA

If 𝑐𝑜𝑠 ∝= then find the value of 𝑠𝑖𝑛4 ∝.

18- fl) djsa fd 2
𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠 𝑥 = 1 − 𝑠𝑖𝑛 2𝑥.
Prove that

𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠 𝑥 = 1 − 𝑠𝑖𝑛 2𝑥.

19- fl) djsa fd 2
𝑠𝑖𝑛2𝐴 =
Prove that

𝑠𝑖𝑛2𝐴 = .

20- ;fn 𝑡𝑎𝑛𝑥 = vkSj < 𝑥 < 𝜋 rks 𝑐𝑜𝑠 dk eku Kkr djsaA

If 𝑡𝑎𝑛𝑥 = and < 𝑥 < 𝜋 then find the value of 𝑐𝑜𝑠 .

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21- fl) djsa fd 2
𝑐𝑜𝑡 − tan = 2𝑐𝑜𝑡𝐴.
Prove that

𝑐𝑜𝑡 − tan = 2𝑐𝑜𝑡𝐴.

22- fdlh ∆𝐴𝐵𝐶 esa 𝐴 + 𝐵 + 𝐶 = 𝜋 rFkk 𝑡𝑎𝑛𝐴 = 2, 𝑡𝑎𝑛𝐵 = 3

rks fl) djsa fd 𝐶 = 2
In any ∆𝐴𝐵𝐶, 𝐴 + 𝐵 + 𝐶 = 𝜋 and 𝑡𝑎𝑛𝐴 = 2, 𝑡𝑎𝑛𝐵 = 3 then prove that 𝐶 = .

23- ;fn A, B, C rFkk D fdlh prqHkqZt ds dks.k gksa rks fl) djsa fd 2
𝑐𝑜𝑡 = − 𝑐𝑜𝑡
If A, B, C and D are the angles of a quadrilateral then prove that

𝑐𝑜𝑡 = − 𝑐𝑜𝑡 .

24- gy djsa % 3𝑐𝑜𝑡 𝑥 − 1 = 0 tgk¡ 0 ≤ 𝑥 ≤ 360 2
Solve: 3𝑐𝑜𝑡 𝑥 − 1 = 0 where 00 ≤ 𝑥 ≤ 3600 .

25- ;fn f=Hkqt dh Hkqtk 8lsehŒ] 10lsehŒ vkSj 12lehŒ gS rks lcls NksVs dks.k dk eku Kkr
djsaA 2
If sides of the triangle are 8cm., 10cm., and 12cm. then find the value of smallest angle.

26- fdlh ∆𝐴𝐵𝐶 esa fl) djsa fd & 2
+ + = .

In any ∆𝐴𝐵𝐶 prove that -

+ + = .

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27- fcUnq 𝐴(−6, −8) vkSj 𝐵(4, 2) dks feykus okys js[kk[k.M dks fcUnq 𝑃(12, 11) fdl
vuqikr esa ckár% foHkkftr djrk gS \
In what ratio does the point 𝑃(12, 11) divide line segment joining the points 𝐴(−6, −8) and

𝐵(4, 2) externally?

28- ;fn nks fcUnqvksa 𝐴(𝑥, −1) rFkk 𝐵(3, 2) ds chp dh nwjh 5 bdkbZ gks rks 𝑥 dk eku Kkr

djsaA
If the distance between two points 𝐴(𝑥, −1) and 𝐵(3, 2) is 5 units then find the value of 𝑥.

29- ml f=Hkqt ds dsUnz dk funsZ'kkad Kkr djsa ftlds 'kh"kZ (𝑎 + 𝑏, 𝑏 − 𝑐)(−𝑎, −𝑏) vkSj
(𝑏, 𝑐) gSaA
Find the Co-ordinates of Centroid of a triangle whose vertices are
(𝑎 + 𝑏, 𝑏 − 𝑐) (−𝑎, −𝑏) and (𝑏, 𝑐).

30- ml o`Ùk dk {ks=Qy Kkr djsa ftldk dsUnz O(−1, −2) gS rFkk o`Ùk ij fLFkr ,d fcUnq

𝑃(3, 4)gSA
Find the area of the circle whose centre is O(−1, −2) and 𝑃(3, 4) is a point lies on the circle.

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Long Answer Type Questions

iz'u la[;k 31 ls 38 nh?kZ mÙkjh; gaSA buesa ls fdUgha 4 iz'uksa ds mÙkj nsaA izR;sd iz'u ds
fy, 5 vad fu/kkZfjr gaSA 4 x 5 = 20
Question Nos 31 to 38 are Long Answer Type. Answer any 4 questions. Each question carries 5
marks. 4 x 5 = 20

31- T;kferh; fof/k ls fl) djsa fd 5
𝑐𝑜𝑠2𝐴 = 𝑐𝑜𝑠 𝐴 − 𝑠𝑖𝑛 𝐴
Prove that geometrically.
𝑐𝑜𝑠2𝐴 = 𝑐𝑜𝑠 𝐴 − 𝑠𝑖𝑛 𝐴

32- fl) djsa fd 5
𝑠𝑖𝑛20 . 𝑠𝑖𝑛40 . 𝑠𝑖𝑛60 . 𝑠𝑖𝑛80 = .
Prove that

𝑠𝑖𝑛20 . 𝑠𝑖𝑛40 . 𝑠𝑖𝑛60 . 𝑠𝑖𝑛80 = .

33- fl) djsa fd 5
= tan + .
Prove that

= tan + .

34- ;fn 𝐴 + 𝐵 + 𝐶 = 𝜋 rks fl) djsa fd 5
𝑠𝑖𝑛 + 𝑠𝑖𝑛 + 𝑠𝑖𝑛 = 1 + 4𝑠𝑖𝑛 𝑠𝑖𝑛 𝑠𝑖𝑛

If 𝐴 + 𝐵 + 𝐶 = 𝜋 then prove that.

𝑠𝑖𝑛 + 𝑠𝑖𝑛 + 𝑠𝑖𝑛 = 1 + 4𝑠𝑖𝑛 𝑠𝑖𝑛 𝑠𝑖𝑛

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35- gy djsaA 5
𝑠𝑒𝑐𝑥. 𝑡𝑎𝑛𝑥 = √2 (tc 0 ≤ 𝑥 ≤ 360 )
Solve.

𝑠𝑒𝑐𝑥. 𝑡𝑎𝑛𝑥 = √2 (when 0 ≤ 𝑥 ≤ 360 )

36- fdlh ∆𝐴𝐵𝐶 esa fl) djsa fd %& 5
𝑠𝑖𝑛2𝐴 + 𝑠𝑖𝑛2𝐵 + 𝑠𝑖𝑛2𝐶 = 0

In any ∆𝐴𝐵𝐶 prove that

𝑠𝑖𝑛2𝐴 + 𝑠𝑖𝑛2𝐵 + 𝑠𝑖𝑛2𝐶 = 0

37- ;fn prqHkqZt 𝐴𝐵𝐶𝐷 ds 'kh"kksaZ ds fu;ked 𝐴(−4, −2), 𝐵(−3, −5), 𝐶(3, −2) rFkk
𝐷(2, 3) gSa rks prqHkqZt dk {ks=Qy Kkr djsaA 5

If vertices of a quadrilateral ABCD are 𝐴(−4, −2), 𝐵(−3, −5), 𝐶(3, −2) and 𝐷(2, 3) then find
the area of quadrilateral.

38- 30ehVj pkSM+h lM+d ds nksuksa vksj leku Å¡pkbZ ds nks [kEHks [kMs+a gSaA lM+d ds fdlh fcUnq
ij] tks nksuksa [kEHkksa ds chp esa gS] [kEHkksa ds 'kh"kZ dk mUu;u dks.k 600 vkSj 300 gSA [kEHkksa
dh Å¡pkbZ Kkr djsAa 5
Two pillars of equal height stand on either side of a road way which is 30m wide. At any point
on the road between the pillars, the elevation of the tops of the pillars are 600 and 300. Find the
height of the pillars.

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