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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. |
What is the value of sec theta |
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Answer» Hypotenuse/base or 1/cos theta 1÷cos theta |
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| 2. |
All chapter formula |
| Answer» Sin90•=1 | |
| 3. |
Prove that (cosecA-sinA)(secA-cosA)sec^2A=tanA |
| Answer» Proved | |
| 4. |
A coin is tossed 20 times. What is the probability of getting at least 1head |
| Answer» | |
| 5. |
sin°18÷cos72° |
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Answer» Cos(90-18)/cos72=cos72/cos72=1 1 1 |
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| 6. |
important question of marhematic coordinates geometryy |
| Answer» | |
| 7. |
abx² +(b²-ac) x-bc = 0 |
| Answer» We have, abx2 + (b2 -ac) x-bc = 0{tex}\\implies{/tex}abx2 + b2 x - acx - bc = 0{tex}\\implies{/tex}bx ( ax+b) - c (ax + b) = 0{tex}\\implies{/tex}(ax + b) (bx - c) = 0Either ax+b = 0 or bx - c = 0{tex}\\implies x = -{b \\over a},\\, {c \\over b}{/tex}Hence, {tex}x = -{b \\over a},\\, {c \\over b}{/tex} are the required solutions. | |
| 8. |
Triangle ABC is right angled at C and CD is perpendicular to AB. Prove that BC2×AD=AC2×BD |
| Answer» | |
| 9. |
Solve triangle |
| Answer» Which triangle | |
| 10. |
Fraction word problem in qudatirc chapter |
| Answer» | |
| 11. |
Find the 10th term of the A.P 2,7,12,.... |
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Answer» Answer is 47....use formula a+(n-1)d Use formula a+(n-1)da=2 d=-52+(9)-52-45-43 |
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| 12. |
5-0 |
| Answer» | |
| 13. |
How to prove root 5 irrational |
| Answer» In ncert book ...nd see ncert solutions in cbse app. | |
| 14. |
is evergreen sample paper of maths is good |
| Answer» I think it depend on question | |
| 15. |
If sin(40+x)=cos60° find value of x |
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Answer» Sin (40+ x) = sin (90-60)40+x = 30X=-10 -10 X=10 X=-10 |
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| 16. |
Plz give me the blue print of maths exam |
| Answer» | |
| 17. |
Any odd no |
| Answer» 1 | |
| 18. |
If the mode of the following data is 10, then find the value of p.6,3,5,7,10,7,8,10,2,1,7,10, p-2. |
| Answer» 12 | |
| 19. |
Ch 7 NCERT book ka formula |
| Answer» | |
| 20. |
Why are use phyathagores theorem in right angle triangle |
| Answer» To find the sides of a rt. Triangle | |
| 21. |
Give important question of sst |
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Answer» Ncert question read it All are important. |
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| 22. |
4-4/2-2 =? Find |
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Answer» 4 Not defined . Not defined 0 not defined.. It may be 0 Infinity |
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| 23. |
What is section formula in coordinate |
| Answer» In chapter 5, 5.2 example | |
| 24. |
Tan 30 - cos 30 |
| Answer» -√3÷4 I think so | |
| 25. |
Basic ofcordinate geometry |
| Answer» | |
| 26. |
26*26*26 |
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Answer» 17576 Sorry by mistake |
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| 27. |
If cos theta +sin theta =root 2 sin( 90- theta).show that cos theta-sin theta=root 2 sin theta |
| Answer» | |
| 28. |
a+b=? |
| Answer» C | |
| 29. |
Find the sum of first 50 terms of an AP whose nth term is 1-4n |
| Answer» a= -3 , d = -4 S50 = 5050 | |
| 30. |
70×5 |
| Answer» 350 | |
| 31. |
Prove the perimetèr of the shaded region is r(tanø+secø+pieø/180-1) |
| Answer» Given, Radius\xa0= r{tex} \\triangle{/tex}AOB is a right triangle.In the right {tex} \\triangle{/tex}OAB,{tex} \\tan \\theta = \\frac { \\mathrm { AB } } { \\mathrm { OA } }{/tex}{tex} \\Rightarrow A B = O A \\times \\tan \\theta{/tex}{tex} = r \\tan \\theta{/tex}Now, area of {tex} \\triangle{/tex}OAB ={tex}\\frac12 b \\times h{/tex}={tex} \\frac { 1 } { 2 } \\mathrm { OA } \\times \\mathrm { AB }{/tex}{tex} = \\frac { 1 } { 2 } \\times r \\times r \\tan \\theta{/tex}{tex} = \\frac { 1 } { 2 } r ^ { 2 } \\tan \\theta{/tex}and area of sector OAC ={tex} \\pi r ^ { 2 } \\times \\frac { \\theta } { 360 ^ { \\circ } }{/tex}\xa0length of arc AC =\xa0{tex} 2 \\pi r \\times \\frac { \\theta } { 360 ^ { \\circ } }{/tex}{tex} = \\frac { 2 \\pi r \\theta } { 360 ^ { \\circ } } = \\frac { \\pi r \\theta } { 180 ^ { \\circ } }{/tex}{tex} \\therefore{/tex}Perimeter of the shaded region= Arc AC + AB + BC{tex} = \\frac { \\pi r\\theta } { 180 } + r \\tan \\theta + ( \\mathrm { OB } - \\mathrm { OC } ){/tex}{tex} = \\frac { \\pi r\\theta } { 180 } + r \\tan \\theta + ( r \\sec \\theta - r ){/tex}{tex} = r \\left( \\frac { \\pi \\theta } { 180 } + \\tan \\theta + \\sec \\theta - 1 \\right){/tex}{tex} = r \\left( \\tan \\theta + \\sec \\theta + \\frac { \\pi \\theta } { 180 } - 1 \\right){/tex} | |
| 32. |
12×4 |
| Answer» 48.. | |
| 33. |
Define theorm 6.1(Thales theorm)...? |
| Answer» bhai ncert me dekh le.....yha define nhi ho payega.... | |
| 34. |
Find the area of a sector of a circle radius, |
| Answer» Area of sector of radius R = (theta/360)* πR² | |
| 35. |
Smallest composite no |
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Answer» 4 9 |
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| 36. |
Find HCF and LCM of 56 and 112 by prime factorisation method |
| Answer» 56 = 2 {tex}\\times{/tex} 2 {tex}\\times{/tex} 2 {tex}\\times{/tex} 7 and 112 = 2 {tex}\\times{/tex} 2 {tex}\\times{/tex} 2 {tex}\\times{/tex} 2 {tex}\\times{/tex} 7Hence HCF is 2 {tex}\\times{/tex} 2 {tex}\\times{/tex} 2 {tex}\\times{/tex} 7 = 56 andLCM is 2 {tex}\\times{/tex} 2 {tex}\\times{/tex} 2 {tex}\\times{/tex} 2 {tex}\\times{/tex} 7 = 112 | |
| 37. |
How find in class intervel in the 0 & above type question. |
| Answer» Very easy | |
| 38. |
Use Euclid division algorithm to find whether the pair of number 847 and 2160 are compaire or not |
| Answer» No | |
| 39. |
sample questions paper |
| Answer» Which subject | |
| 40. |
Hey any one give preboard questions |
| Answer» Sorry mere pre boards 16th January se start honge | |
| 41. |
Datesheet?? |
| Answer» Cbse.in pe search karlo waise to 6 march se 28 march tak h | |
| 42. |
Prove that : sin20°.sin40°.sin80°=√3/8 |
| Answer» prove that sin 30° and 45° sin 80 degree is equal to under root 3 by 8 | |
| 43. |
Square |
| Answer» | |
| 44. |
If sinø + cosø =√3 then find the value of tanø + cotø =1 |
| Answer» | |
| 45. |
Sin4/a+cos4=1/a+b then prove sin8/a3+cos8/b3=1/(a+b)3 |
| Answer» | |
| 46. |
Find the minimum rating of fuse to run two. Gysers of 1.1KW simultaneously |
| Answer» | |
| 47. |
What is the marking scheme of board 2018 for class 10 |
| Answer» You can get this on various sites | |
| 48. |
Show that exacty one of the numbers n,n+2 or n+4 is divisible by 3. |
| Answer» On dividing n by 3,let q be the quotient and r be the remainder Then n=3q+r where r is greater or equal to 0 and Lessthan 3n = 3q+r where r=0,1,2n =3q or 3q+1or 3q+2Case1 if n=3q then n is divisible by 3 Case2 if n=3q+1 then(n+2) =3q+3=3(q+1) which is divisible by3Case3 if n=3q+2 then (n+4) =3q+6=3(q+2) which is divisible by 3 Hence in all cases one any one out of n,n+2,n+4 is divisible by 3 | |
| 49. |
Sin0 -cos0+1÷sin0+cos0-1=1÷sec0-tan0 |
| Answer» | |
| 50. |
Exam ka kaisa taiyari chal RHA hai |
| Answer» | |