Saved Bookmarks
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 16951. |
If the zeroes of x^2-kx+6 are in the ratio 3:2,findx. |
| Answer» | |
| 16952. |
If n is an odd integer then show n²-1is divisible by 8 |
| Answer» Let n = 4q + 1 (an odd integer){tex}\\therefore \\quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \\quad \\text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}{tex}= 16{q^2} + 8q{/tex}{tex}= 8 \\left( 2 q ^ { 2 } + q \\right){/tex}= 8m, which is divisible by 8. | |
| 16953. |
Find the root of 2x-3/x-2+2x-7/x-4=16/3 |
| Answer» {tex}Question:\\space\\large{ {2x-3\\over x-2} +{2x-7\\over x-4} = {16\\over 3}, find \\space x?}{/tex}{tex}\\large {Solution:\\\\}{/tex}\t{tex}\\large {\\underline {Step\\space 1:\\space Fraction\\space solving}}{/tex}{tex}\\large\\implies {{\\big((2x-3)(x-4)\\big)} +{\\big((2x-7)(x-2)\\big)}\\over {(x-2)(x-4)}} = {16\\over 3}{/tex}{tex}\\large\\implies {{\\big((2x^2-8x-3x+12)\\big)} +{\\big((2x^2-4x-7x+14)\\big)}\\over {(x^2-4x-2x+8)}} = {16\\over 3}{/tex}{tex}\\large\\implies {{\\big(2x^2-11x+12+2x^2-11x+14\\big)}\\over {(x^2-6x+8)}} = {16\\over 3}{/tex}{tex}\\large\\implies {{\\big(4x^2-22x+26\\big)}\\over {(x^2-6x+8)}} = {16\\over 3}{/tex}{tex}\\large{ \\underline {Step\\space 2:\\space Cross-multiplication}\\\\\\implies {{3\\times \\big (4x^2-22x+26\\big)} } = {16\\times {(x^2-6x+8)}}}{/tex}\xa0{tex}\\large{ \\underline {Step\\space 3:\\space Simplification:}\\\\\\implies {{\\big (12x^2-66x+78\\big)} } = {{(16x^2-96x+128)}}}{/tex}\xa0{tex}\\large {\\underline {Step\\space 4:\\space Equation-solving:}\\\\\\implies 16x^2-96x+128-12x^2+66x-78=0}{/tex}{tex}\\large {\\underline {Step\\space 5:\\space Arranging\\space like-terms:}\\\\\\implies 16x^2-12x^2-96x+66x+128-78=0}{/tex}{tex}\\large{\\\\\\implies 4x^2-30x+50=0}{/tex}\xa0{tex}\\large {\\underline {Step\\space 6:\\space Taking\\space out\\space common-terms:}\\\\\\implies 2\\times (2x^2-15x+25)=0}{/tex}{tex}\\large {\\\\\\implies 2x^2-15x+25=0}{/tex}{tex}\\large {\\underline {Step\\space 7:\\space Middle-term\\space splitting\\space :}\\\\\\implies 2x^2-10x-5x+25=0}{/tex}{tex}\\large{ \\underline {Step\\space 8:\\space Grouping\\space :}\\\\\\implies 2x(x-5)-5(x-5)=0}{/tex}{tex} \\\\\\implies \\large(2x-5)(x-5)=0{/tex}{tex} \\\\\\implies\\huge \\therefore \\boxed{x=5\\space or \\space {5\\over2}}{/tex}\t\xa0 | |
| 16954. |
Solve : 2/x+3/3y =1/6 and 3/x+2/y =0 |
||
|
Answer» 3(2/x +3/3y = 1/6)2(3/x+2/y = 0)6/x +3/y = 1/26/x +4/y = 0(-) (-)0 + (-y) = 1/2y = -22/x - 3/6 = 1/62/x = 1/6 + 3/62/x = 4/64x = 12x = 3 This is a very simple question of linear equations in two variables chapter, I think you have to practice more on fractions arithmetic.\xa0{tex}\\large Question:\\frac 2x +\\frac 3{3y}=\\frac 16\\space and\\space \\frac 3x +\\frac 2y=0,\\space find\\space x,y?{/tex}Solution: | \t\t\t{tex}\\large\\implies\\boxed {\\frac {2\\times 3y +3\\times x}{x\\times 3y}=\\frac 16 \\space \\dots eq.(i)}{/tex}{tex}\\large{\\implies {\\frac {6y +3x}{3xy}=\\frac 16 \\space }}{/tex}{tex}\\large {\\underline {Step\\space 2:\\space Cross-multiplication:} }{/tex}{tex}\\large{\\implies {6\\times{(6y +3x)}=1\\times {3xy} \\space }}{/tex}{tex}\\large{\\implies {36y +18x}={3xy} \\space }{/tex}{tex}\\large{\\implies {36y +18x}-{3xy}=0 \\space }{/tex}{tex}\\large {\\implies\\boxed {\\frac {3\\times y +2\\times x}{x\\times y}=0 \\space\\dots eq.(ii) }}{/tex}{tex}\\large {\\implies\\frac {3y +2x}{xy}=0 \\space }{/tex}{tex}\\large {\\underline {Step\\space 2:\\space Cross-multiplication:} }{/tex}{tex}\\large {\\implies{3y +2x}=0\\times {xy}=0 \\space }{/tex}{tex}\\large {\\implies\\boxed {{3y}=-2x \\space \\dots eq.(iii)}}{/tex}\xa0\xa0\t\t{tex}\\large {\\underline {Step\\space 3:\\space Putting\\space relation\\space of\\space eq.(iii):} }{/tex}{tex}\\large{\\implies {{\\underline {3y}}(12-x) +18x}=0 \\space }{/tex}{tex}\\large{\\implies {\\underline{(-2x)}(12-x) +18x}=0 \\space }{/tex}{tex}\\large{\\implies {-24x+2x^2 +18x}=0 \\space }{/tex}{tex}\\large{\\implies {2x^2-6x}=0 \\space }{/tex}{tex}\\large{\\implies {2x^2=6x \\space} }{/tex}{tex}\\large{\\implies\\boxed {x=2 \\space} }{/tex}{tex}\\large {\\underline {Step\\space 4:\\space Putting\\space value\\space of\\space \'x\'\\space in\\space eq.(iii):} }{/tex}{tex}\\large {\\implies{3y}=-2x \\space }\\\\\\large {\\implies{3y}=-2\\times 2 \\space }\\\\\\large\\implies\\boxed {{y}={-4\\over3} \\space }\\\\{/tex}\t\xa0 | ||
|---|---|---|---|
| 16955. |
Solve 41x- 17y =19; 17x -41y = 75 |
| Answer» Don\'t you think you have to perform any of the three methods told in the NCERT book,1. Substitution method2. Elimination method3. Cross-multiplication methodLiterally, you don\'t want to do anything by yourself.Ans: x= -31/87 , y=-172/87 | |
| 16956. |
Which is the greatest numberof 5digits exactly divisible by 15,24,36 |
| Answer» Answer= 99720 | |
| 16957. |
What is the H C f of smallest prime number and smallest composite number? |
| Answer» Solution: Smallest prime number = 2 Smallest composite number = 4Thus, HCF ( 2,4 ) = 2\xa0This question is also pathetic, study more and work hard. | |
| 16958. |
root of 588 |
| Answer» Very bad question,{tex}\\implies \\sqrt {588}\\\\\\implies \\sqrt {196\\times3}\\\\\\implies\\boxed{14 \\sqrt {3}}\\\\{/tex} | |
| 16959. |
If p,q,r are jeroes of p(x) = 9x^-3x^-7x+1, then p^-1 +p^-1+ r^-1 is |
| Answer» | |
| 16960. |
If HCF of 408,1032 is written in the form of 1032x - 408x5 then find x |
| Answer» Very low-quality question.Solution: HCF of 408 and 1032 is 24=>\xa01032x - 408x5 = HCF=> 1032x - 2040 = 24=> 1032x = 24+2040=> x= \xa0= 2=> x=2 (Answer) | |
| 16961. |
If two zeroes of polynomial p(x) = x4 - 6x3 - 26x2 + 138x - 35 are 2 + root 3 |
| Answer» Two zeros are {tex}2\\pm\\sqrt3{/tex}Sum of Zeroes\xa0{tex}2 + \\sqrt { 3 } + 2 - \\sqrt { 3 } = 4{/tex}and product of zeroes =\xa0{tex}( 2 + \\sqrt { 3 } ) ( 2 - \\sqrt { 3 } ) = 4 - 3 = 1{/tex}Hence quadratic polynomial formed out of this will be a factor of given polynomial,So, x2 - (sum of zeroes)x + product of zeroes= x2 - 4x + 1 will be a factor of given polynomial,Divide given polynomial with x2 - 4x + 1 to get other zeroes.Now,x2 -2x - 35= x2 - 7x + 5x - 35= x(x - 7) + 5(x - 7)= (x - 5) (x - 7){tex}\\therefore {/tex}\xa0Zeros arex = 7 and x = -5{tex}\\therefore {/tex} Other two zeros are 7 and -5\xa0 | |
| 16962. |
Find the quadratic polynomial whose zeroes are 2+√3 and2-√3 |
| Answer» Two zeros are {tex}2\\pm\\sqrt3{/tex}Sum of Zeroes\xa0{tex}2 + \\sqrt { 3 } + 2 - \\sqrt { 3 } = 4{/tex}and product of zeroes =\xa0{tex}( 2 + \\sqrt { 3 } ) ( 2 - \\sqrt { 3 } ) = 4 - 3 = 1{/tex}Hence quadratic polynomial formed out of this will be a factor of given polynomial,So, x2 - (sum of zeroes)x + product of zeroes= x2 - 4x + 1 | |
| 16963. |
X^2-13x-14 find the two facters |
|
Answer» Thanks x=14, -1 |
|
| 16964. |
2x+7x |
|
Answer» 9x 9x 9× X=0 9x |
|
| 16965. |
√18-√8= ? |
|
Answer» Thanks every one for a great help Square root of2 Square root of 2. √18=3√2√8=2√2So,√18-√8=3√2-2√2 = √2 |
|
| 16966. |
Root3xsqaure-2root2x-2root3=3 |
| Answer» | |
| 16967. |
3(36) |
|
Answer» 3(36)3×36=108 108 108 3×36=108 3 x 36 = 108 |
|
| 16968. |
How i start mu study |
|
Answer» Maths Which subject are you doing???? |
|
| 16969. |
what are real numbers? |
|
Answer» All rational and irrational number are called real numbers . The group of rational and irrational nubers The type of number we normally use, such as 1, 15.82, −0.1, 3/4, etc. |
|
| 16970. |
Find the two numbers whose sum is 27and product is 182 |
|
Answer» Find the probality of getting a doublet in a throw of a pair of dice Let the first number be\xa0x\xa0and the second number is 27 -\xa0x.Therefore, their product =\xa0x\xa0(27 -\xa0x)It is given that the product of these numbers is 182.Therefore,\xa0x(27 -\xa0x) = 182⇒\xa0x2\xa0– 27x\xa0+ 182 = 0⇒\xa0x2\xa0– 13x\xa0- 14x\xa0+ 182 = 0⇒\xa0x(x\xa0- 13) -14(x\xa0- 13) = 0⇒ (x\xa0- 13)(x\xa0-14) = 0Either\xa0x\xa0= -13\xa0= 0 or\xa0x\xa0- 14 = 0⇒\xa0x\xa0= 13 or\xa0x\xa0= 14If first number = 13, thenOther number = 27 - 13 = 14If first number = 14, thenOther number = 27 - 14 = 13Therefore, the numbers are 13 and 14. |
|
| 16971. |
7x2-12x+18=0 |
|
Answer» 7x² - 12x + 18\xa0Find the sum of the roots:Sum of the roots = α + βα + β = - b/aα + β = - (-12 ) / 7 = 12/7\xa0Find the product of the roots:Product of the roots = αβαβ = c/aαβ = 18/7Find the ratio:α + β : αβ = 12/7 : 18/7Multiply by 7:α + β : αβ = 12 : 18Divide by 6:α + β : αβ = 2 : 3 14-12x+18 => 32=12x => x=32/12 => x=8/3 |
|
| 16972. |
2x^2 -7 +3=0 |
| Answer» 2x2 -7 x+3=02x2 -6x - 1x +3=02x ( x - 3) - 1( x- 3) = 0(x - 3)(2x - 1) = 0x - 3 = 0 or 2x - 1 = 0x = 3 or x = 1/2 | |
| 16973. |
Sec3A=cosec(A-10) where 3A is an acute angle so find the value of A |
|
Answer» Sec3A=Cosec(A-10)Sec3A=Sec(100-A) 3A=100-A4A=100A=25.............(ans) A is 25degree.sec(3A)=cosec (A-10)sec3A =sec(90-{A-10)}3A=90+10-A3A+A=1004A=100A=100/4A=25 |
|
| 16974. |
If the 3rd and the 9th term of an ap are 4 and -8 respectively, which of which term of this is zero |
|
Answer» a+2d=4. (1)a+8d=-8. (2)Subtract eq 1 and 2a+2d=4a+8d=-8(-)(-). (+)----------------6d=12 d= -2Replace in eq 1a + 2(-2)=4a-4=4a=4+4a=8an=a+(n-1)d 0= 8+(n-1)-2 0-8 =(n-1)-2-8/-2 =n-14=n-14+1 =n5=n--------- ನಂಗ್ ಗೊತ್ತಿಲ್ಲ?????? |
|
| 16975. |
Proved Thales theorem. |
|
Answer» Let ABC be the triangle.The line l parallel to BC intersect AB at D and AC at E.To prove AD\u200b/DB =AE\u200b/ECJoin BE,CDDraw EF⊥AB, DG⊥CASince EF⊥AB,EF is the height of triangles ADE and DBEArea of △ADE=1/2 × base × height=1/2\u200bAD×EFArea of △DBE= 1/2 \u200b×DB×EFareaofΔDBEareaofΔADE\u200b=1/2\u200b×AD×EF/ 1/2\u200b×DB×EF\u200b=AD/DB\u200b ........(1)Similarly,areaofΔDCEareaofΔADE\u200b= 1/2 ×AE×DG/ 1/2 ×EC×DG \u200b=AE/EC\u200b ......(2)But ΔDBE and ΔDCE are the same base DE and between the same parallel straight line BC and DE.Area of ΔDBE= area of ΔDCE ....(3)From (1), (2) and (3), we haveAD\u200b/DB =AE\u200b/ECHence proved. Basic Proportionality Theorem states that, if a line is parallel to a side of a triangle which intersects the other sides into two distinct points,then the line divides those sides of the triangle in proportion.Let\xa0ABC\xa0be the triangle.The line\xa0l\xa0parallel to\xa0BC\xa0intersect\xa0AB\xa0at\xa0D\xa0and\xa0AC\xa0at\xa0E.To prove\xa0AD\u200b/DB =AE\u200b/ECJoin\xa0BE,CDDraw\xa0EF⊥AB,\xa0DG⊥CASince\xa0EF⊥AB,EF\xa0is the height of triangles\xa0ADE\xa0and\xa0DBEArea of\xa0△ADE=1/2 ×\xa0base\xa0×\xa0height=1/2\u200bAD×EFArea of\xa0△DBE= 1/2 \u200b×DB×EFareaofΔDBEareaofΔADE\u200b=1/2\u200b×AD×EF/ 1/2\u200b×DB×EF\u200b=AD/DB\u200b ........(1)Similarly,areaofΔDCEareaofΔADE\u200b= 1/2 ×AE×DG/ 1/2 ×EC×DG \u200b=AE/EC\u200b ......(2)But\xa0ΔDBE\xa0and\xa0ΔDCE\xa0are the same base\xa0DE\xa0and between the same parallel straight line\xa0BC\xa0and\xa0DE.Area of\xa0ΔDBE=\xa0area of\xa0ΔDCE ....(3)From (1), (2) and (3), we haveAD\u200b/DB =AE\u200b/ECHence proved. |
|
| 16976. |
(1296)1_4*(1296)1_2 |
| Answer» | |
| 16977. |
If x=-1/2 is a root of the quadratic equation 3x^2+2kx+3=0, find the value of k. |
|
Answer» K=-15\\4 X ke place pe -1/2 daal do aur equate kro aa jayega Putting x = (-1/2) in\xa03x2\xa0+ 2kx - 3 =\xa00,orGiven :\xa0If\xa0\xa0is a solution of the quadratic equation\xa0.To find :\xa0The value of k ?Solution :To get the value of k substitute the value of x in the equation,Equation\xa0Put\xa0,Therefore, The value of k is\xa0 or |
|
| 16978. |
Can anyone be my girlfriend |
|
Answer» Padhne aaya hai is app mein ya ladkiyaan patane ?????????????? \tDon\'t post personal information, mobile numbers and other details.\tDon\'t use this platform for chatting, social networking and making friends. This platform is meant only for asking subject specific and study related questions.\tBe nice and polite and avoid rude and abusive language. Avoid inappropriate language and attention, vulgar terms and anything sexually suggestive. Avoid harassment and bullying.\tAsk specific question which are clear and concise.If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section. Suspended users will receive error while adding question or answer. Question comments will be disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html |
|
| 16979. |
State and prove two theorems of triangle chapter |
| Answer» Free may kar de | |
| 16980. |
How will learn every subject in just four month |
|
Answer» nhi krenge By working hard |
|
| 16981. |
Please tell me that in this app their is case study based questions are their |
|
Answer» But where Jaldi batao bhi kaha hai case study Arjun please tell me where is it yes there r case study based questions in this app |
|
| 16982. |
find the co-ordinates of the point which divides (-1,7) and (4,-3) in the ratio 2:3 |
|
Answer» 1,3 is the correct answer Welcome We know that, the coordinates of the point dividing the line segment joining the points x 1\u200b ,y 1\u200b and x 2\u200b ,y 2\u200b in the ratio m:n is given by ( m 1\u200b +m 2\u200b 1m 1\u200b x 2\u200b +m 2\u200b x 1\u200b \u200b , m 1\u200b +m 2\u200b m 1\u200b y 2\u200b +m 2\u200b y 1\u200b \u200b )Given (x 1\u200b ,y 1\u200b )=(−1,7); (x 2\u200b ,y 2\u200b )=(4,−3) (m 1\u200b ,m 2\u200b )=2:3Coordinates of point of intersection of line segment is( 2+3(2)(4)+(3)(−1)\u200b , 2+3(2)(−3)+(3)(7)\u200b )=(5/5,15/5)=(1,3) |
|
| 16983. |
What should be subtracted with the number of CRPF soldiers and |
| Answer» And what?? | |
| 16984. |
Prove that tan theta/1-cot theta |
| Answer» Q :\xa0Prove that tan theta/1-cot theta +cot theta/1-tan theta=1+tan theta+cot thetaA n s w e r : | |
| 16985. |
What is uclids division algorithm |
|
Answer» Bro no need to study this , it\'s deleted form the portion. I think you should focus on the portion which will come first :) a=bq+r where 0 < r < b According to Euclid’s Division Lemma if we have two positive integers a and b, then there exist unique integers\xa0q\xa0and\xa0r\xa0which satisfies the condition\xa0a = bq + r\xa0where 0\xa0≤ r < b.The basis of the Euclidean division algorithm is Euclid’s division lemma. To calculate the Highest Common Factor (HCF) of two positive integers\xa0a\xa0and\xa0b\xa0we use Euclid’s division algorithm. HCF is the largest number which exactly divides two or more positive integers. That means, on dividing both the integers\xa0a\xa0and\xa0b\xa0the remainder is zero. |
|
| 16986. |
previous year question papers |
| Answer» | |
| 16987. |
a^2x+b^2y=c^2b^2x+a^2y=d^2 |
|
Answer» Your question is not clearPlz tell the question clearly I dont know |
|
| 16988. |
If Sec A + Tan A = X, find the value of Sec A- Tan A. |
| Answer» SecA + TanA = X.Sec^2A - Tan^2A = 1 .(SecA + TanA)(SecA - TanA) =1.X(SecA - TanA )= 1.SecA- TanA=1/X. | |
| 16989. |
Prove that 2√5 is irrational , give √5 irrational |
| Answer» The product of rational and irrational is always irrational that\'s why 2√5 is irrational | |
| 16990. |
6.1 |
|
Answer» Tell the which questions answer do you want Please ask full question bro |
|
| 16991. |
Express 156 as product of primes |
|
Answer» 156=2x2x3x13 =2²x3x13 156=2×2×3×13 =2²×3×13 156 =2 x 2 x 3 x 13 =2² x 3 x 13 |
|
| 16992. |
Solve the following system of equations 1/2x - 1/y = -1 and 1/(x ) + 1/2y = 8 |
| Answer» You must let 1/x be p and 1/y be q then you try to solve then you must solve this question | |
| 16993. |
Write the empirical relationship among mean, median and mode. |
|
Answer» Thnku 3 meadian = 2 Mode +Mean Mode\xa0= 2\xa0Mean\xa0– 3\xa0Median. d. 3\xa0Median\xa0= 2\xa0Mode\xa0+\xa0Mean. |
|
| 16994. |
Given that HCF(306,657)=9, find LCM (306,657) |
|
Answer» Given that HCF = 9 and the numbers are 306 and 657 .LCM = ?We know that LCM\xa0× HCF = product of two numbers\xa0⇒LCM\xa0× 9 = 306\xa0× 657⇒LCM = 201042/9 = 22338\xa0 We don\'t need the given part in this question, we can directly find the LCM(306,657) by prime factorisation method.306=2×3×3×17657=3×3×73. So, the LCM of 306 and 657 is2×3×3×17×73=22338 |
|
| 16995. |
If the zero of polynomial ax square+bc+c are in ratio 4:5 prove that 20 b square=81ac |
|
Answer» Let roots 4k and 5k4k+5k =-b/a9k =-b/aK=-b/9a..... (1)4k*5k=c/a20k2=c/aSub ... (1)20 (-b/9a)2=c/a20 (b2/81a2)=c/a20 b2\xa0= 81 ac Let the zeroes be 4p, 5p.We know that the sum of the zeroes=-b/a. Then, 4p+5p=9p=-b/a.p=-b/9a, then, p^2=b^2/81a^2...(1)And the product of zeroes is c/a.Then 4p×5p=c/a.20p^2=c/a then, p^2=c/20a...(2)Using (1) and (2).b^2/81a^2=c/20a.Then, 20b^2=81ac. |
|
| 16996. |
2 prime no |
|
Answer» The numbers that are only divisible by 1 are called prime numbers.Ex-2,3,5.....etc 2, 3 |
|
| 16997. |
X+2=6 |
|
Answer» X=4 x+2=6Send the 2after 6 with negative sign.now it will be...x=6-2=4The answer is 4. x=4 X+2=6x = 6 - 2 = 4 X= 4 ?? |
|
| 16998. |
Find the 10th term of the AP:2,7,12, ... |
|
Answer» 10th term of the Given AP is 47 Kaise kaise log hai 47 As a=2,d=a2-a1= 7-2 = 5an= a+(n-1)×d a10= 2+(10-1)×5a10= 2+(9)×5a10= 2+45a10= 47 ? 10th term of the AP is 47 |
|
| 16999. |
What is computer??? |
| Answer» A computer is a machine that can be instructed to carry out sequences of arithmetic or logical operations automatically via computer programming. Modern computers have the ability to follow generalized sets of operations, called programs. These programs enable computers to perform an extremely wide range of tasks.\xa0A\xa0computer\xa0is an electronic device that manipulates information or data. It has the ability to store, retrieve, and process data. A\xa0computer\xa0is used to type documents, send an email, play games, and browse the Web. | |
| 17000. |
Plzzzz someone tell me the value of all trigonometric ratios. |
|
Answer» The\xa0Trigonometry Ratios\xa0of the angle θ in the triangle APM are defined as follows.Opposite over Hypotenuse – Sin,\xa0Adjacent Over Hypotenuse – Cos, Opposite over Adjacent – Tan, Hypotenuse over Opposite – Cosec, Hypotenuse Over Adjacent – Sec and Adjacent over Opposite – Cotangent,The ratios defined above are abbreviated as sin θ, cos θ, tan θ, cosec θ, sec θ and cot θ respectively. Note that the ratios\xa0cosec θ, sec θ and cot θ\xa0are respectively, the reciprocals of the ratios sin θ, cos θ and tan θ. So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.Opposite of Sin: CosecantOpposite of Cos:\xa0SecantOpposite of Tan: CotangentOpposite of Cosecant:\xa0SinOpposite of Cotangent: TanOpposite of Secant: CosecantTrig Mnemonics –\xa0\xa0Some\xa0People\xa0Have,\xa0Curly\xa0Black\xa0Hair\xa0Through\xa0Proper\xa0Brushing.Here,\xa0Some\xa0People\xa0Have is for\tSin θ=\xa0Perpendicular/\xa0Hypotenuse.Curly\xa0Black\xa0Hair is for\tCos θ=\xa0Base/\xa0Hypotenuse.Through\xa0Proper\xa0Brushing is for\tTan θ=\xa0Perpendicular/BaseTrigonometric Ratios of Some Specific AnglesWe already know about isosceles right angle triangle and right angle triangle with angles 30º, 60º and 90º.Can we find sin 30º or tan 60º or cos 45º etc. with the help of these triangles?Does sin 0º or cos 0º exist? |
|