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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
8. In the adjoining figure, it is given that ABII CD, LABO=50° and-A50%2CDO = 40°.Find the measure of ZBOD.Hint, Through O draw EOF AB.40° |
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| 2. |
3 \sqrt{98}+8 \sqrt{242}-5 \sqrt{50} |
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| 3. |
8. In the adjoining figure, it is given that ABII CD, LABO=50° and-A<CDO = 40°.Find the measure of ZBOD.Hint. Through O draw EOF II AB.50140 |
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Answer» angle BOD= 50°+ 40°= 90° (alternate pair) |
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| 4. |
(1ma)g .uis 7 — पथ. _l‘&\ ™y8 ue) |
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Answer» (sinθ- 2 sin^3θ)/(2cos^3θ-cosθ) =sinθ(1-2sin^2θ)/cosθ(2cos^2 θ-1) = sin θ [1- 2(1- cos^2 θ)/cos θ(2cos ^2 θ -1) = sin θ [1-2+ 2cos ^2 θ]/ cos θ (2 cos^2θ -1) = sin θ(2cos^2 θ -1)/ cos θ(2 cos ^2θ-1) = sin θ/ cos θ = tanθ |
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| 5. |
( 2v + 3y) ^2 + ( 5v + 2y) ^3 |
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| 6. |
LT gS0D = L urs -~ UIS-hialid SR "६ 1051 |
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Answer» We will use following trigonometric formulas in this solution - •sin^-1(x) - sin^-1(y) = sin^-1{x√(1-y^2) - y√(1-x^2)}---------(1) •sin^-1(x) = cos^-1{√(1 - x^2)}---------(3) sin^-1(3/5) - sin^-1(8/17) = cos^-1(84/85) LHS = sin^-1(3/5) - sin^-1(8/17) LHS = sin^-1{ 3/5 × √(1 - (8/17)^2 - 8/17 × √1 - (3/5)^2 } => LHS = sin^-1{ 3/5 × √(289-64)/17^2 - 8/17 × √8/17 × √(25-9)/25 } => LHS = sin^-1{ (3/5 × √(225/289) - 8/17 × √(16/25) } => LHS = sin-1 { (3/5 × 15/17) - (8/17 × 4/5) } => LHS = sin^-1 (45/85 - 32/85) => LHS = sin^-1(13/85) LHS = cos^-1{√(1 - (13/85)^2)} => LHS = cos^-1{√(7225-169)/7225} => LHS = cos^-1(√7056/7225) => LHS = cos^-1{√(84×84)/(85×85)} => LHS = cos^1(84/85) = RHS |
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| 7. |
6 पष्य » _. 8 809 - 6 S00Z ६७_— 0 UIS z - g uls |
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Answer» (sinθ- 2 sin^3θ)/(2cos^3θ-cosθ) =sinθ(1-2sin^2θ)/cosθ(2cos^2 θ-1) = sin θ [1- 2(1- cos^2 θ)/cos θ(2cos ^2 θ -1) = sin θ [1-2+ 2cos ^2 θ]/ cos θ (2 cos^2θ -1) = sin θ(2cos^2 θ -1)/ cos θ(2 cos ^2θ-1) = sin θ/ cos θ = tanθ please like the solution 👍 ✔️ |
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| 8. |
рео09 8090 рео0рем UIS + рек0рем 809,09 uIs |
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| 9. |
242/1000 |
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Answer» 0.242 is correct answer of following question 242/1000============0.252 242/1000=============.242 242/1000=.242 is the answer .242 is the right answer This question correct answer is 0.242. 0.242 is the correct answer to this question. 0.242 is the correct answer |
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| 10. |
48x+120=242 |
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Answer» add the following integer using number line (-3)+(-8)+(-5) 48×+12048×120=4760 answet 48x+120=24248x=242-12048x=122x=122/48x=61/24 61/24 is the correct answer of the given question 48x + 120 = 24248x = 242 - 12048x = 122x = 122/48× = 2.5416666667x = 2.5 x=61/24 is the best answer. x=61/24 is the best answer. x=2.54166. is the best answer. 48x+120=242 One solution was found : x = 61/24 = 2.542 Rearrange: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : 48*x+120-(242)=0 Step by step solution : Step1: Pulling out like terms : 1.1 Pull out like factors: 48x - 122=2•(24x - 61) Equation at the end of step1: Step2: Equations which are never true: 2.1Solve:2=0 This equation has no solution.A a non-zero constant never equals zero. Solving a Single Variable Equation: 2.2Solve:24x-61 = 0 Add61 to both sides of the equation:24x = 61Divide both sides of the equation by 24:x = 61/24 = 2.542 One solution was found : x = 61/24 = 2.542 Processing ends successfully 61/24 is the correct answer of the given question 61/24 is the answer of the following 48x=242-120x=122/48x=2.541is right and safe answer 48x=242-120x=122/48x=2.541Answer |
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| 11. |
254,Evaluate (80(-1) x 5"3)/ (2^(-4)).(a) 245 (b) 240 () 215 (d)50 |
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Answer» Option d is correct. |
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| 12. |
: S ;mt मी al SO0 B *= =g uisi =i 8४ : ही 8 . £ p e |
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| 13. |
कक, नि ee SO0 न. fl__u 502] है. uis L i\3 C CS पट o "kO\'\'\LOशा न “राडी; c8 301नर» |
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Answer» (1/3 * √3/2)/(1/2 * 1/√2) + (√3)/(√3/2) = (1/2√3)/((1/2√2) + (2) = √2/√3 + 2 = (√2+2√3)/√3 Like my answer if you find it useful! I am not under stande |
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| 14. |
=z"19कि v 2gu prw) P p ;Z;ugm ही1. o g 25 3P e11m v nonb सुरनवा गए पारणुर SO0\ gy |T ——— |
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Answer» Like if you find it useful |
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| 15. |
Fig. 10.39CD is a cyclic quadrilateral w hose diagonals intersect at a point E. IfDBC=70°is 30°, find 4 BCD. Further, if AB -BC, find ZECD. |
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| 16. |
9500 + g uIss "o (SO0 + g uIS |
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| 17. |
982-9(p + q)x + (2p" + 5pq +242) = 0 |
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Answer» hit like if you find it useful |
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| 18. |
ha deamaplacus deimal |
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| 19. |
is CE ) v w7 F |
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Answer» (-2)⁴x(3/2)⁴(2)⁴x(3/2)⁴3⁴ |
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| 20. |
Find the LCM of the following nuother(a) 5, 20 (b) 6, 18 (c) 12, 48What do you observe in the results obtained?Il.(d) 9, 45 |
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Answer» Thankyou ☺️☺️ |
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| 21. |
2yif cose ce--cot θV2 cota, then prove that cosecθ + cot θcosec θ. |
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| 22. |
Ulliy Taurial number by a point on the nuEXERCISE 1BRepresent each of the following numbers on the number line:(ii) 17(iv) 2-(V)01(viii) 8 |
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| 23. |
22. In shufflinga pack of cards, four are accidently dropped. Find the chance that themissing cards should be one from each suit |
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| 24. |
The bisectors of ZA and ZBAD and BC.ofaquadrilateralBCDintersectatP.ProvethatPisequidistantfromm |
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| 25. |
Two sides AB, BC and medianAM of one triangle ABC arerespectively equal to sides PQ andQR and median PN ofΔPQR (Seefigure). Show that:i)∆ABM~∆PQNii)∆ABC=∆PQR |
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| 26. |
polygรณn 1(1)0JSquare roots of(i) 1024 (ii)4489iii)nrn are in |
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Answer» square root of !)√1024= 32!!)√4489=87 |
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| 27. |
SB are the resesHR0-19 aß IBG |
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| 28. |
Two numbers are such that the ratio between themis 3:5. If each is increased by 10, the ratio betweenthe new numbers so formed is 5: 7. Find the originalnumbers |
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| 29. |
Find the value of n if C,"C, and C, are in arithmetic progression, |
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| 30. |
www.eenadupratibha.netth term in Arithmetic progression is n + $. Find the sum of first 10 terms13. |
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Answer» If nth term is n+5 hence 10th term will be 10+5=15and first term will be 1+5=6henceSum of first 10 termsSn=n/2(first term +last term)S10=10/2(15+6)S10=5(21)=105 |
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| 31. |
If the sum of first 7 terms of an arithmetic progression is 49 and that of 17 terms is289, find the sum of first n terms |
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| 32. |
In the adjoining figure, ABCD is aparallelogram in which the bisectors of ZAand ZB intersect at a point P. Prove thatZAPB 90 |
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| 33. |
AB=AC.21. In AABC, AB = AC and the bisectors of ZBand ZC meet at a point O. Prove thatBO CO and the ray AO is the bisector of ZA. |
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| 34. |
In IABC, AB-AC and the bisectors of LB and LC meet at a point O. Prove that BO*CO and the ray Ao is the bisector8 |
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Answer» since AB=BCso <B=<C (angles opposite to equal sides)so <OBC=<OCBso BO=OC (sides opposite to equal angles)since BO=OC so <BAO=<OACso AO is bisector of <A |
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| 35. |
18. In AABC, ABAC and the bisectors of LB and<C meet at a point O. Prove that BO = CO andthe ray AO is the bisector of LA. |
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Answer» thanks |
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| 36. |
18. In AABC, AB - AC and the bisectors of 4B andZC meet at a point O. Prove that BO CO andthe ray AO is the bisector of ZA |
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| 37. |
21. In AABC, AB=AC and the bisectors of ZBand ZC meet at a point 0. Prove thatBO = CO and the tay AO is the bisector of ZA. |
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| 38. |
18. In AABC, AB AC and the bisectors of LB andCO andZC meet at a point O. Prove that BOthe ray AO is the bisector of ZA. |
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| 39. |
(c) 20 cm48. If figure sides AB and AC of a AABC areextended to points P and Q respectively. AlsoZPBC <20CB. The correct relation isO |
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| 40. |
1.List five rational numbers(i) -1 and0 (ii) |
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| 41. |
List five rational numbers between(1)2 and 0and |
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| 42. |
Listfive rational numbers between:1) -1 and0 i) -2 and -1 |
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Answer» give more details |
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| 43. |
LEVEL-11. In Fig. 15.120, O is the centre of the circle. If ZAPB 50°, find ZAOB and ZOAB50°80110Fig. 15.120Fig. 15.121 |
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| 44. |
1. List five rational numbers between0) -1 and 0 ()-2 and-1 |
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| 45. |
List five rational numbers between:(G) -1 and 0(ii)-2 and-1 |
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| 46. |
In the adjoining figure, if PA and PB are tangents to the circlewith centre O, such that ZAPB 50 Then, find COA |
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Answer» Since OA is perpendicular to PA and also, OB is perpendicular to PB ∠APB + ∠AOB = 180° 50°+ ∠AOB = 180° ∠AOB = 180° – 50° = 130° In △AOB, OA = OB = radii of same circle ∠OAB = ∠OBA = x ( say ) Again, ∠OAB + ∠OBA + ∠AOB = 180° x +x + 130° = 180° 2x = 180° – 130° = 50° X = 25° Hence, ∠OAB =25° |
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| 47. |
In the adjoining figure, if PA and PB are tangents to the circlewith centre O, such that ZAPB 50 Then, find COAR |
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Answer» the required answer is 25degree |
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| 48. |
3. In fig. O is the centre of the circle and PA, PB are tangents to the circle. IfZAPB-60°, then find ZACB6ă) 20 |
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| 49. |
L IF PE and FB are tangents to the circle with centre O such that ZAPB- 50, find the 204BA.P-15090 |
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Answer» Since OA is perpendicular to PA and also, OB is perpendicular to PB ∠APB + ∠AOB = 180° 50°+ ∠AOB = 180° ∠AOB = 180° – 50° = 130° In △AOB, OA = OB = radii of same circle ∠OAB = ∠OBA = x ( say ) Again, ∠OAB + ∠OBA + ∠AOB = 180° x +x + 130° = 180° 2x = 180° – 130° = 50° X = 25° Hence, ∠OAB =25° |
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| 50. |
. In the given figure, O is the centre of thecircle with AC 24 cm, AB7cm andBOD 90°. Find the area of shaded region.Use n 3.14.[][CBSE 2012] |
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Answer» so , thanks |
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