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InterviewSolution
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.
| 37101. |
Kx+3y=k- - 3 12x+ky=k. Find the value of k. |
| Answer» The given equations arekx + 3y - (k - 3) = 0 ......... (i)12x + ky - k = 0 ........... (ii)The system of linear equations is in the form of\xa0a1x + b1y + c1\xa0= 0a2x + b2y + c2\xa0= 0Compare (i) and (ii), we geta1= k\xa0,b1= 3, c1\xa0= -(k - 3),a2=12\xa0,b2= k\xa0,c2\xa0= -kFor a unique solution, we must have{tex}\\frac { a _ { 1 } } { a _ { 2 } } \\neq \\frac { b _ { 1 } } { b _ { 2 } }{/tex}{tex}\\frac { k } {1 2 } \\neq \\frac { 3 } { k }{/tex}{tex}\\Rightarrow k ^ { 2 } \\neq 36 {/tex}{tex}\\Rightarrow k \\neq \\pm 6{/tex}Thus, for all real value of k other than {tex}\\pm 6{/tex}, the given system of equations will have a unique solution. | |
| 37102. |
For how many integers n is √9 - (n+2)^2 is a real number? |
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| 37103. |
any1 online here |
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Answer» Dhanyawad...??? Ya wa bhai tere gaane हां भाई।।।। |
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| 37104. |
X-3=4 |
| Answer» 4/3 | |
| 37105. |
Divides |
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Answer» Hlow Kiska |
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| 37106. |
All the formula of all chaptera |
| Answer» You can get them in the revision notes of the chapters :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 37107. |
Proove that sqare of any +ve inteGer is of form 4m+1 and 4m |
| Answer» Let a be the positive integer and b = 4.Then, by Euclid’s algorithm, a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3.{tex}(4q)^3\\;=\\;64q^3\\;=\\;4(16q^3){/tex}= 4m, where m is some integer.{tex}(4q+1)^3\\;=\\;64q^3+48q^2+12q+1=4(\\;16q^3+12q^2+3q)+1{/tex}= 4m + 1, where m is some integer.{tex}(4q+2)^3\\;=\\;64q^3+96q^2+48q+8=4(\\;16q^3+24q^2+12q+2){/tex}= 4m, where m is some integer.{tex}(4q+3)^3\\;=\\;64q^3+144q^2+108q+27{/tex}=4×(16q3+36q2+27q+6)+3= 4m + 3, where m is some integer.Hence, The cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m. | |
| 37108. |
√7-√3=0 √5+√2=0 |
| Answer» Isme krna kya h | |
| 37109. |
Find the value of k for which the system of equations 4x-5y=k, 2x-3y=12 has a unique solution. |
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Answer» Hlow shreya k is any real number |
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| 37110. |
Obain all zeroes of 2x^4+x cube-14x square -19x-6 if two of its zeroes are -2 & -1 |
| Answer» if -2 and -1 are zeros of\xa0f(x) = 2x4 + x3 - 14x2 - 19x - 6x+2 and x+1 are factors of f(x)So (x + 2)(x + 1) = x2 + x + 2x + 2 = x2 + 3x + 2 is a factor of f(x)On long division of f(x) by\xa0x2 + 3x + 2 we getf(x) = 2x4 + x3 - 14x2 - 19x - 6 = (2x2 - 5x - 3)(x2 + 3x + 2)= (2x + 1)(x - 3)(x + 2)(x + 1)Therefore, zeroes of the polynomial = {tex}\\frac{{ - 1}}{2}{/tex}, 3, -2, -1. | |
| 37111. |
Find the value of k, for which one root of the quadratic equation k×2-14×+8=0 is 2 |
| Answer» Given:\xa0Equation: kx²-14x+8= 0Also, One root of this Equation is 2Substitute the Given value in Equation!kx²-14x+8 = 0k(2)² -14(2)+8 = 0k4-28+8 = 04k-20 = 04k= 20•°• k = 20/4=> k = 5\xa0 | |
| 37112. |
Find a quadratic polynomial whose zeros are . |
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| 37113. |
If one zero polynomial is square minus 4x plus k is 2 then what is the value of k |
| Answer» Pls write in eq.form | |
| 37114. |
Please some one answer the question |
| Answer» Yes...... | |
| 37115. |
Find the term of an AP 2,7,12 |
| Answer» which term you required? please specify | |
| 37116. |
36x+27y =0 find the value of x and y |
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Answer» 1)x = 0, y = 02)x = 1, y = -4/3There are infinite solutions as for each value of x there is a value of y. So simple |
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| 37117. |
Prove that cube of any positive integer can be written as 3m,3m+1 or 3m+2 |
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Answer» Let, \'n\' be a positive integer.By using Euclid\'s Division Lemma,\'n\' can be written in the form of—3q+r, where 0 <= r <3.Therefore, r = {0,1,2}Case 1:r = 0n = 3q=>n³ = 27q³ = 3m (where m = 9q³)Case 2:r = 1n = 3q+1=>n³ = 27q³+27q²+9q+1 = 3m +1(where m = 9q³+9q²+3q)Case 3:r = 2n = 3q+2=>n³ = 27q³+54q²+36q+8 = 3m+2(where m = 9q³+18q²+12q+2)Hence, it is proved that the cube of any positive integer can be written in the form of 3m or 3m+1 or 3m+2. ORWe know that the cube of any positive integer is a positive integer.On applying Euclid\'s division lemma,n³ = 3m+r, where 0 ≤ r < 3Therefore, r = {0,1,2}Therefore, n³ = 3m or 3m+1 or 3m+2 Ex: agar tum 3m le rahe ho vaha pe eg taur par 3q lena aur use square kar \'q\' se aage jo bhi tumhe milega use \'m\' consider karo aur doubt katam |
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| 37118. |
Prime factors of 213478 |
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| 37119. |
Find the ratio between root 2 and root 7 |
| Answer» Since √2=1.414 and √7=2.646 as1.414<2<2.646Thus,there is only one integer (2) between them.but there are many irrational number between them such as √3 ,√4,√5 and √6. | |
| 37120. |
1/x - 1/x - 3 =3 : slove the polynomial by middle team splitting method ??? |
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| 37121. |
13/3125 Question 1.4 1 |
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| 37122. |
Hii 10th kesi ja rhi h sbki?? |
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Answer» ? Meri bhi mst ek dm? Badia chal rhi hai khushi??? Mst Now we are in 11th |
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| 37123. |
How to know where we have to find hcf and where lcm? |
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| 37124. |
Find the quadratic polynomial whose sum and product of zeroes are a and 1/a |
| Answer» ax^2-ax+1 | |
| 37125. |
Hhvhnym |
| Answer» ?? | |
| 37126. |
prove that( 1÷cosecA- cotA ) - (1÷sinA ) = (1÷sinA) -(1÷cosecA + cotA) |
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| 37127. |
Verify the relationship between the zeroes and the coefficients. 1.2x^2-3+5x2.2y^2+6y3.x^2+1/6x -2. |
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| 37128. |
tan square theta minus sin square theta |
| Answer» tan2θ– sin2θ{tex} = {\\tan ^2}\\theta - \\frac{{{{\\sin }^2}\\theta }}{{{{\\cos }^2}\\theta }} \\cdot {\\cos ^2}\\theta \\left[ {\\because {{\\tan }^2}\\theta = \\frac{{{{\\sin }^2}\\theta }}{{{{\\cos }^2}\\theta }}} \\right]{/tex}= tan2θ– tan2θcos2θ= tan2θ(1 – cos2θ) {tex}\\left[ \\because \\sin ^ { 2 } \\theta = 1 - \\cos ^ { 2 } \\theta \\right]{/tex} | |
| 37129. |
X+y=14, x+y=4 |
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Answer» Not possible What I have to do here,should I solve it for (x) and (y)??? Or what?? |
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| 37130. |
X+y=14, x-y=4 |
| Answer» x + y\xa0= 14 ....(1)x - y = 4 ....(2)x\xa0= 4 + y from equation (2)Putting this in equation (1), we get4+y +y =14⇒ 2y =10⇒ y = 5Putting value of y in equation (1), we getx + 5 = 14⇒ x = 14 - 5 = 9Therefore, x =9 and y =5 | |
| 37131. |
If alpha and beta are the zeros of the polynomial x2-5x+m such that alpha -beta = 1, find m |
| Answer» Since\xa0{tex}\\alpha , \\beta{/tex} are the zeros of the polynomial f(x) = x2\xa0- 5x + m.Compare f(x) = x2\xa0- 5x + m\xa0with ax2 + bx + c.So, a = 1 , b = -5 and c = m{tex}\\alpha + \\beta = - \\frac { ( - 5 ) } { 1 }{/tex}\xa0= 5{tex}\\alpha \\beta = \\frac { mk } { 1 } = m{/tex}Given,\xa0{tex}\\alpha - \\beta{/tex}\xa0= 1Now,\xa0{tex}( \\alpha + \\beta ) ^ { 2 } = ( \\alpha - \\beta ) ^ { 2 } + 4 \\alpha \\beta{/tex}{tex}\\Rightarrow{/tex}\xa0(5)2\xa0= (1)2\xa0+ 4m{tex}\\Rightarrow{/tex}\xa025 = 1 + 4m{tex}\\Rightarrow{/tex}\xa04m\xa0= 24{tex}\\Rightarrow{/tex}\xa0m\xa0= 6Hence the value of m\xa0is 6. | |
| 37132. |
Prove that 1 by root 7 is irrational number |
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Answer» We can prove 1by root 7 irrational by contradiction .Let us suppose that 1by root7 is rational it means we have some co-prime integers a and b (b is not equal to 0)such that 1by root 7=a/bRoot7=b/a .....1RHS of 1 is irrational but we know that root7 is irrational.It is not possible which means our supposition is wrong.Therefore,1by root7 cannot be rational..Hence it is irrational.. for which value of k will the following pair of linear equation has no solution 3 X + Y is equal to 12 ke minus y x x + K minus y + Y 2 |
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| 37133. |
For any positive integers n,prove that n^3-n is divisible by 6 |
| Answer» n3\xa0- n = n (n2\xa0- 1) = n (n - 1) (n + 1)\xa0Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.\xa0Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q, then n is divisible by 2.If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.⇒ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.∴ n (n-1) (n+1) = n3\xa0- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6) | |
| 37134. |
How can we make an effective time table |
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Answer» Could you please suggest me a proper layout Know your schedule and plan your days accordingly |
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| 37135. |
The numder of zaroes lying between -1 and 1 of the polynomial p(x) , |
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| 37136. |
(a+b)=(b+a)... Which property has been use in this equation |
| Answer» Commutative property | |
| 37137. |
How to preparation mathematics compartments paper 2019 |
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Answer» Larn mane things two know we larnwd in ex Study |
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| 37138. |
what is irrrational |
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Answer» No. That cannot be written in the form p/q are irrational. They are non-repeating and endless. For example the value of π The numbers that cannot be written as a ratio of two integers ,they are called Irrational Numbers.Example: π (Pi) is a famous irrational number.π = 3.1415926535897932384626433832795... (and more)We cannot write down a simple fraction that equals Pi.The popular approximation of 22/7 = 3.1428571428571... is close but not accurate. |
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| 37139. |
Check weather 6n can end with the digit 0 |
| Answer» 2×3=6 and as we know 2n×5m if factor are than the number is terminating and 3 is one of the factor of 6. So it means that 6n can not end with the digit 0 | |
| 37140. |
Heyy guys..... anyone on?? |
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Answer» Hello Hiii hii..... or btao kya hal chal Helo |
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| 37141. |
2x+3y=03x+2y=15 |
| Answer» X=9,y=-6 | |
| 37142. |
Chlo ...tata ...byy byyy kl milte h ..... ....... G |
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Answer» Phir ruk jao... Bbye gd nyt Bbye..gud nyt Nhi jana kl scool Bbye good night sweet dreams take care...apko me rok bhi nii skta kl school jana hoga isiliye??? |
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| 37143. |
Hlow frndz |
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Answer» Hlo hlo Hii Rohit... Hlow bro Hi |
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| 37144. |
For any positive integer n prove that n³-n is divisible by 6 |
| Answer» n3\xa0- n = n (n2\xa0- 1) = n (n - 1) (n + 1)\xa0Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.\xa0Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q, then n is divisible by 2.If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.⇒ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.∴ n (n-1) (n+1) = n3\xa0- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)\xa0\xa0 | |
| 37145. |
Show that n^2 -- 1 is divisible by 8 if n is an odd positive integer. |
| 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. | |
| 37146. |
Solve x square -3x-10=0 by completing square method |
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Answer» Briliant x2\xa0-3x - 10x2\xa0- 3x -10 + 9/4 -9/4 (we square the term obtained by dividing the coefficient of x by 2, after that we add and subtract the same in to the equation)now, (x2\xa0-3x + 9/4) -10-9/4(x-3/2)2\xa0- 49/4 = 0(x-3/2)2\xa0- (7/2)2\xa0= 0(x-3/2-7/2) (x-3/2 +7/2) = 0(x-10/2) (x + 4/2)x=5, -2 is the solution |
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| 37147. |
Verify that 1/2, 3 are zeroes of cubic polynomial 2x cube -17x square + 38x -17x -15 |
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| 37148. |
Draw a graph of equation1. x+3y=6 2x-3y=12 |
| Answer» Graph of the equation {tex}x + 3y = 6{/tex}:We have, {tex}x + 3y = 6{/tex}{tex} \\Rightarrow {/tex}\xa0{tex}x = 6 - 3y{/tex}When y = 1, we have x = 6 - 3 =3When y = 2, we have x = 6 - 6 = 0Thus, we have the following table:\tx30y12\tPlotting the points {tex}A(3,1)\\ and\\ B(0,2){/tex} and drawing a line joining them, we get the graph of the equation x + 3y = 6 as shown in Fig.Graph of the equation {tex}2x - 3y = 12{/tex} :We have,\xa0{tex} 2 x - 3 y = 12 \\Rightarrow y = \\frac { 2 x - 12 } { 3 }{/tex}When x=3, we have\xa0{tex}y = \\frac { 2 \\times 3 - 12 } { 3 } = - 2{/tex}When x=0, we have\xa0{tex}y = \\frac { 0 - 12 } { 3 } = - 4{/tex}\tx30y-2-4\tPlotting the points {tex}C(3,-2)\\ and\\ D(0, - 4){/tex} on the same graph paper and drawing a line joining them, we obtain the graph of the equation {tex}2x - 3y = 12{/tex} as shown in Fig.Clearly, two lines intersect at P(6, 0).Hence, {tex}x = 6, y = 0{/tex} is the solution of the given system of equations.Putting x = 6, y = 0 in {tex}a = 4x + 3y{/tex}, we geta = (4 {tex}\\times{/tex}\xa06) + (3 {tex}\\times{/tex}\xa00) = 24 | |
| 37149. |
Cbse 2019 paper |
| Answer» What cbse 2019 paper | |
| 37150. |
If son theta =a²-b²/a²+b², find the values of all T-ratio of theta ? |
| Answer» We have,{tex}\\sin \\theta = \\frac { \\text { Perpendicular } } { \\text { Hypotenuse } } = \\frac { a ^ { 2 } - b ^ { 2 } } { a ^ { 2 } + b ^ { 2 } }{/tex}So, Let us draw a right triangle ABC in which\xa0{tex}\\angle B{/tex}\xa0is right angle, we havePerpendicular = BC = a2 - b2 Hypotenuse = AC = a2 + b2 and,\xa0{tex}\\angle B A C = \\theta{/tex}By Pythagoras theorem, we haveAC2 = AB2 + BC2{tex}\\Rightarrow \\quad \\left( a ^ { 2 } + b ^ { 2 } \\right) ^ { 2 } = A B ^ { 2 } + \\left( a ^ { 2 } - b ^ { 2 } \\right) ^ { 2 }{/tex}{tex}\\Rightarrow \\quad A B ^ { 2 } = \\left( a ^ { 2 } + b ^ { 2 } \\right) ^ { 2 } - \\left( a ^ { 2 } - b ^ { 2 } \\right) ^ { 2 }{/tex}{tex}\\Rightarrow \\quad A B ^ { 2 } = \\left( a ^ { 4 } + b ^ { 4 } + 2 a ^ { 2 } b ^ { 2 } \\right) - \\left( a ^ { 4 } + b ^ { 4 } - 2 a ^ { 2 } b ^ { 2 } \\right){/tex}{tex}\\Rightarrow \\quad A B ^ { 2 } = 4 a ^ { 2 } b ^ { 2 } = ( 2 a b ) ^ { 2 }{/tex}{tex}\\Rightarrow \\quad A B = 2 a b{/tex}\xa0Now, Let{tex}\\angle B A C = \\theta{/tex}\xa0We haveBase = AB = 2ab, Perpendicular {tex}= \\mathrm { BC } = a ^ { 2 } - b ^ { 2 }{/tex}, Hypotenuse\xa0= AC = a2 + b2Therefore,\xa0{tex}\\quad \\cos \\theta = \\frac { \\text { Base } } { \\text { Hypotenuse } } = \\frac { 2 a b } { a ^ { 2 } + b ^ { 2 } }{/tex},\xa0{tex}\\tan \\theta = \\frac { \\text { Perpendicular } } { \\text { Base } } = \\frac { a ^ { 2 } - b ^ { 2 } } { 2 a b }{/tex}{tex}\\quad cosec \\;\\theta = \\frac { \\text { Hypotenuse } } { \\text { Perpendicular } } = \\frac { a ^ { 2 } + b ^ { 2 } } { a ^ { 2 } - b ^ { 2 } } , \\quad \\sec \\theta = \\frac { \\text { Hypotenuse } } { \\text { Base } } = \\frac { a ^ { 2 } + b ^ { 2 } } { 2 a b }{/tex}and,\xa0{tex}\\cot \\theta = \\frac { \\text { Base } } { \\text { Perpendicular } } = \\frac { 2 a b } { a ^ { 2 } - b ^ { 2 } }{/tex} | |