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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.
| 68551. |
Example 9. A farmer connects α pipeof internal diameter 20 em from a canal intoutes.a cylindrical tank in her field, which is 10 min diameter and 2 m deep. If water flowsthrough the pipe at the rate of 3 km /h, in howLmuch time will the tank be filledSol. For cylindrical tank22 n |
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| 68552. |
FEXERCISE 11.41. Given a cylindrical tanka cylindrical tank, in which situation will you find surface area and inwhich situation volume.(a) To find how much it can hold.(h) Number of cement bags required to plaster it.e To find the number of smaller tanks that can be filled with water from it.11:14 |
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| 68553. |
6. A closed cylindrical tank of diameter 14 m and height 5 m is made froma sheet of metal.How much sheet of metal will be required? |
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| 68554. |
A cylindrical tank has a capacity of 6160 cu. m. Find its depth, if ts radinus is14 m. Calculate the cost of painting its curved surface (outer) at the rate of3/m2. |
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Answer» As the unit ofvolume given in the question is too small so considering it 6160 m³Volume of cylindrical tank =π r²×h⇒6160=π×(14)²×h⇒h = 10 mnow, the area of curved surface= 2π r×h =2π×14×10 =880 m²cost of painting that surface = Rs 3×880 = Rs2640 please like the solution 👍 ✔️ |
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| 68555. |
TUD14 cm.0. The diameter and height of a cylindtank is 1.75 m and 3.2 m respectively.Is the capacity of the tank in litresof a cylindricalspectively. How221Solution :The diameter of a cylindrical tank =indrical tank = 1.75 m2.. its radius (r) = 1.75 mits height (h) = 3.2 m142 |
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| 68556. |
Find the shortest distance between the lines7 612 |
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| 68557. |
A rectangular sheet of paper is 33 cm long and 32 cm wide. It isalong its length to make a cylinder of height 32 em. A circular sheetpaper is attached to the bottom of the cylinder formed. Find theof the cylinder formed.TT可啊醃 |
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Answer» Solution :- Given : Length of the rectangular sheet = 33 cm And, Width of the rectangular sheet = 32 cm As we know that, length of the rectangular sheet = circumference of the circle Circumference of the circle = 2πr ⇒ 33 cm = 2*22/7*r ⇒ r = (33*7)/(22*2) ⇒ r = 231/44 ⇒ r = 5.25 cm Capacity or volume of the cylinder =πr²h ⇒ 22/7*5.25*5.25*32 ⇒ 22/7*27.5625*32 ⇒ 19404/7 ⇒ 2772 cm³ Hence, the capacity or the volume of the cylinder is 2772 cm³. |
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| 68558. |
59-I302-61 4-33 | 2 |
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| 68559. |
The area of a square fi5 m/s. In how much time will he return to the starting point?eld is 60025 m2. A man cycles along its boundary at |
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| 68560. |
t0-9-x+-T(1) |
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Answer» Consider the given equation:2x² + x - 6 = 0 Split the middle term: 2x² + 3x - 4x - 6 = 0 Rearrange the terms; 2x² - 4x + 3x - 6 = 0 Take common term out:2x(x - 2) + 3(x - 2) = 0 (x - 2)(2x + 3) = 0 => x - 2 = 0 or 2x + 3 = 0=> x = 2 or x = -3/2 |
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| 68561. |
Find te sum of finst t0 multiples of 6 |
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Answer» 6+12+18+24+30+36+42+48+54+60=330 |
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| 68562. |
frustimThe perimeter ofcm Find its volume.ends of a frustum are 48 cm. and 36 cm. If height of the |
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Answer» Given height of frustum(h)= 11 cm Let the radius, r1 and radius, r2 be the radii of the circular ends of the frustum and h be its height. perimeter = 36 cm (given) 2 π r1=36r1=36/2π r1=18/π perimeter=48 cm 2 π r2=48 r2= 24/π Volume of Frustum (V)= 1/3 π {(r1)²+ (r2)² + (r1r2)} h V= 1/3 x π x h {(18/π)²+(24/π)²+(18/π x 24/π)} V= 1/3 x π x 11 {324/π²+576/π²+432/π²} V= 1/3 x π x 11x 1/π² (324+576+432) V= 1/3 x π x 11x 1/π² (1332) V= 11/3x 1/π (1332) V= 11/3 x 7/22 (1332) V= (11 x 7× 1332)/ (22 ×3) V= 7 x 222 V= 1554 cm³ Hence, the volume of frustum is 1554 cm³ |
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| 68563. |
13. The base of an isosceles triangle is 16 cm. If its perimeter is 36 cm,find its area |
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Answer» who was the Girl who fell |
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| 68564. |
elosed cylindrical tank of height 1.4 m. and radius of the base is 56 cm.is made up of a thick metal sheet. How much metal sheet is required (Expressin square meters) |
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| 68565. |
1. A closed cylindrical tank ofheight 1.4 m. and radius of the base is 56 cm. ismade up of a thick metal sheet. How much metal sheet is required (Express insquare meters) |
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Answer» tnq |
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| 68566. |
A closed cylindrical tank of height 14 m. and radius of the base is 56 cm. ismade up of a thick metal sheet. How much metal sheet is required (Express insquare meters) |
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| 68567. |
How much metal sheet is required to make an open cylindrical tank of radius 4.2 m and height 15 m? |
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| 68568. |
10. How much metal sheet is required to make an open cylindrical tank of radius 4.2 m and height 15 m. |
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Answer» Sheet required=Curved surface area of cylinder+ area code f base=2πrh+πr²=2(22/7)(4.2)(15)+(22/7)(4.2)²=395.84+55.41=451.25m² |
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| 68569. |
\operatorname { tan } 3 A = \frac { 3 \operatorname { tan } A - \operatorname { tan } ^ { 3 } A } { 1 - 3 \operatorname { tan } ^ { 2 } A } |
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Answer» tan 3A= tan (A + 2A)= (tan A + tan 2A) / (1 - tan A tan 2A)= [ tan A + 2 tanA / (1 - tan²A) ] / [ 1 - 2tan²A / (1 - tan²A) ]= [ (tanA - tan³A + 2tan A) ] / [ 1 - tan²A - 2tan²A ]= (3tanA - tan³A) / (1 - 3tan²A) |
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| 68570. |
\operatorname { tan } 3 \theta - \operatorname { tan } 2 \theta - \operatorname { tan } \theta = \operatorname { tan } 3 \theta \cdot \operatorname { tan } 2 \theta \cdot \operatorname { tan } |
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Answer» Let theta be x |
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| 68571. |
= 1 + _ + _ + _ +e + e=-n=0(2n)!2!4!612(2n-2)!n=1 |
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Answer» thanks for the answer |
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| 68572. |
From a rectangular sheet of paper 30 cm by 21cm. Leena wants to cut out equal square pieces of paper.What is the maximum length of the side of these squares? How many square pieces will Leena getfrom the sheet of paper? |
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Answer» AREA of rectangular sheet = 30*21 = 630 Sq.cm ER. RAVI KUMAR ROY we have to take side of square = 3 cm so that the area of square I.e 9 Sq.cm devides equally the area of sheet..number of squares = 630/9 = 70. ANS... 70 is the correct answer of the given question |
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| 68573. |
A rectangular sheet of paper is rolled along its length to make a cylinder, Thesheet is 33 cm long and 32 cm wide. A circular sheet of paper is attached to thebottom of the cylinder formed. Find the capacity of cylinder so formed. |
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Answer» Given : Length of the rectangular sheet = 33 cm And, Width of the rectangular sheet = 32 cm As we know that, length of the rectangular sheet = circumference of the circle Circumference of the circle = 2πr ⇒ 33 cm = 2*22/7*r ⇒ r = (33*7)/(22*2) ⇒ r = 231/44 ⇒ r = 5.25 cm Capacity or volume of the cylinder =πr²h ⇒ 22/7*5.25*5.25*32 ⇒ 22/7*27.5625*32 ⇒ 19404/7 ⇒ 2772 cm³ Hence, the capacity or the volume of the cylinder is 2772 cm³. |
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| 68574. |
. 2 cubes each of volume 64 cm are joined eld t0 tlu. Tmting cuboid |
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Answer» Given,Volume of cube = 64(side)^3 = 64side = 4 cm When two cubes of side 4 cm joined end to end then dimensions cuboid formed areLength l = 8 cm, Breadth b = 4 cm, Height h = 4 cm Surface area of cuboid= 2(l*b + b*h + l*h)= 2(8*4 + 4*4 + 8*4)= 2(32 + 16 + 32)= 2(80) = 160 cm^3 |
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| 68575. |
A rectangular sheet of paper is rolled along its length to make a cylinder. Thesheet is 33 cm long and 32 cm wide. A circular sheet of paper is attached to thebottom of the cylinder formed. Find the capacity of cylinder so formed. |
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Answer» Given : Length of the rectangular sheet = 33 cm And, Width of the rectangular sheet = 32 cm As we know that, length of the rectangular sheet = circumference of the circle Circumference of the circle = 2πr ⇒ 33 cm = 2*22/7*r ⇒ r = (33*7)/(22*2) ⇒ r = 231/44 ⇒ r = 5.25 cm Capacity or volume of the cylinder =πr²h ⇒ 22/7*5.25*5.25*32 ⇒ 22/7*27.5625*32 ⇒ 19404/7 ⇒ 2772 cm³ Hence, the capacity or the volume of the cylinder is 2772 cm³. |
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| 68576. |
11. Find the area of an isosceles triangle with base 10 cm and perimeter 36 cm.Explain linear equation |
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| 68577. |
\left. \begin{array} { l } { 48 ^ { 2 } } \\ { 998 ^ { 2 } } \\ { 28 \times 32 } \\ { 12.1 ^ { 2 } - 7.9 ^ { 2 } } \end{array} \right. |
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| 68578. |
4. Verify that 3,-1,-ar the zeroes of the cubic polynomial p) -3x-5x-1lx-3, and then verify the relationship among the zeroes and the coefficientsrayana Group of Schools |
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| 68579. |
Hence, required difference1.Exercise 2.3Verify property (a - b) + (-a):(a) 42, 56 C(b) 19,17(d) 39, 112(c) 453, 762Verify property a-b-c) (a - b)-c:2. |
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Answer» (45-56)=//(56-45) -9 not equal to 9 1a)(42+56=98)≠(56-42=14) B )(19+17=36)≠(17-19=-2) C )(39+112=151)≠(112-39=73) D )(453+762=1215)≠(762-453=309) |
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| 68580. |
\left. \begin{array} { l } { \text { 28. } 3 \operatorname { cos } ^ { 2 } 60 ^ { \circ } + 2 \operatorname { cot } ^ { 2 } 30 ^ { \circ } - 5 \operatorname { sin } ^ { 2 } 45 ^ { \circ } = } \\ { \text { (A) } \frac { 13 } { 6 } } \end{array} \right. |
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| 68581. |
A closed cylindrical tank, made of thin iron-sheet, has diameter 8.4 m and height 5.4 mHow much metal sheet, to the nearest m2, isused in making this tank, if 1/15 of the sheetactually used was wasted in making the tank? |
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| 68582. |
\left. \begin{array} { l } { \operatorname { tan } 3 A \operatorname { tan } 2 A \operatorname { tan } A = \operatorname { tan } 3 A - \operatorname { tan } 2 A - \operatorname { tan } A } \\ { E N / A S / B N ) } \end{array} \right. |
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| 68583. |
\left. \begin{array} { l l } { ( A ) \sqrt { 3 } \operatorname { tan } ( \frac { \alpha } { 2 } ) - \operatorname { tan } ( \frac { \beta } { 2 } ) = 0 } & { ( B ) \operatorname { tan } ( \frac { \alpha } { 2 } ) - \sqrt { 3 } \operatorname { tan } ( \frac { \beta } { 2 } ) = 0 } \\ { ( C ) \operatorname { tan } ( \frac { \alpha } { 2 } ) + \sqrt { 3 } \operatorname { tan } ( \frac { \beta } { 2 } ) = 0 } & { ( D ) \sqrt { 3 } \operatorname { tan } ( \frac { \alpha } { 2 } ) + \operatorname { tan } ( \frac { \beta } { 2 } ) = 0 } \end{array} \right. |
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| 68584. |
7.The area of a rhombus is equal to the area of atriangle having base 10 cm and height 14 cm. Ifthe length of one of the diagonals length is 14 cm,find the length of the other diagonal. |
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Answer» area of rhombus = area of triangle1/2×d1×d2 = 1/2×base×heighton both sides 1/2 is cancellednow substitute the given values14×d2 =10×14on both sides 14 is cancelledso d2 = 10c.m.length of other diagonal is 10 c.m |
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| 68585. |
\operatorname { sec } A = \frac { 5 } { 4 } , \text { verify that } \frac { 3 \operatorname { sin } A - 4 \operatorname { sin } ^ { 3 } A } { 4 \operatorname { cos } ^ { 3 } A - 3 \operatorname { cos } A } = \frac { 3 \operatorname { tan } A - \operatorname { tan } ^ { 3 } A } { 1 - 3 \operatorname { tan } ^ { 2 } A } |
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| 68586. |
\left. \begin array l 28 ^ \circ - \frac 2 3 \operatorname cot 58 ^ \circ \operatorname tan 32 ^ \circ \\ - \frac 5 3 \operatorname tan 13 ^ \circ \operatorname tan 47 ^ \circ \operatorname tan 45 ^ \circ \operatorname tan 53 ^ \circ \operatorname tan 77 ^ \circ = - 1 \end array \right. |
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Answer» Like if you find it useful! |
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| 68587. |
\operatorname { tan } x + \operatorname { tan } 2 x = \operatorname { tan } 3 x |
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| 68588. |
5. A sports company was ordered to prepare 100 paper cylinders for shuttle cocks. TheFindrequired dimensions of the cylinder are 35 cm length/ height and its radius is 7 cm.2the required area of thin paper sheet needed to make 100 cylinders? |
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Answer» ⛧Radius of the cylinder = 7 cm ⛧its given that dimensions are given 35 cm length/height ⛧So height =35 cm ⛧Required total surface of the cylinder 2*22/7*7*35=1540cm^2 ⛧The required paper sheets for 100 such cylinders =>(1540×100)=>154000cm^2 |
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| 68589. |
) cot 12° cot 38° cot 52° cot 60° cot 78。 |
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| 68590. |
\operatorname { cot } 4 x ( \operatorname { sin } 5 x + \operatorname { sin } 3 x ) = \operatorname { cot } x ( \operatorname { sin } 5 x - \operatorname { sin } 3 x ) |
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| 68591. |
\frac{5 \cot \theta+\csc \theta}{5 \cot \theta-\csc \theta}=\frac{7}{3} |
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\left. \begin{array} { c } { \operatorname { cot } 4 \theta ( \operatorname { sin } 5 \theta + \operatorname { sin } 3 \theta ) } \\ { = \operatorname { cot } \theta ( \operatorname { sin } 5 \theta - \operatorname { sin } 3 \theta ) } \end{array} \right. |
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Answer» Lhs = Rhs not prove |
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| 68593. |
13) The curved surface area of a right circular cylinder of height 14 cm is 88 cm2 Find theliameter of the base of the cylinder |
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18. Find the sum to n terms of the sequence, 8, 88, 888, 8888... |
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| 68595. |
8888 + 848 + 88 -? = 7337+ 737 |
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Answer» by basic addition subtraction ? = 7337+737-8888-848-88 = -1750 |
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| 68596. |
Sum of two integers is 88. If the greater is divided bythe smaller, the quotient is 5 and the remainder is 10.The greater integer is:(1) 13(3) 65(2) 75(4) 23 |
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Answer» second option is the right answer. ..75/(88-75=13)=5 and remainder 10 if the larger of two integers whose sum is 88 is divided by the smaller the quotient is 5 and the remainder is 10. what are the 2 nos.?------------Let the two integers be x and 88-x--------------EQUATION: (88-x)/x = 5 + 10/(xMultiply thru by x to get:88-x = 5x + 106x = 78x= 13 (one of the numbers)88-x= 75 (the other number)==========Cheers,Stan H. Answer byankordixie-net.com(21703)(Show Source): You canput this solution on YOUR website!if the larger of two integers whose sum is 88 is divided by the smaller the quotient is 5 and the remainder is 10. what are the 2 nos.?:x + y = 88y = (88 - x):= 5:Multiply equation by yx-10 = 5y:Substitute (88-x) for y:x - 10 = 5(88-x):x - 10 = 440 - 5x:x + 5x = 440 + 10:6x = 450:x = 75:y = 88 - 75 = 13::Check: Divide 75 by 13 you get 5 with a remainder of 10 (2) 75 is the right answer (2) 75 is right answer |
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| 68597. |
\frac 3 \operatorname sin 3 A %2B 2 \operatorname cos ( 5 A %2B 10 ^ \circ ) \sqrt 3 \operatorname tan 3 A - \operatorname cosec ( 5 A - 20 ^ \circ ) , \text when A = 10 |
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Answer» When A= 10then3sin30+2cos(60)/(√3tan30-cosec(30) 3/2+2*1/2= 3/2+1/(1-2) 5/2/(-1)-5/2Answer |
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| 68598. |
8. Barry is wrapping a book with the givendimensions as a present. What is the leastamount of wrapping paper needed to wrapthe book?1.5 in.8 in.5 in.A surface Area |
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Answer» Dimension of bookLength l = 8 inchWidth b = 5 inchHeight h = 1.5 inch Then,Amount of wrapping paper required is equal to surface area of book Surface area of book= 2(l*b + b*h + l*h)= 2(8*5 + 5*1.5 + 8*1.5)= 2(40 + 7.5 + 10.5)= 2(58)= 116 inch Therefore,Wrapping paper required = 116 inch |
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| 68599. |
2. if 6 cot θ t 2 cosec θ = cot θ + 5 cosec θ, then find cos θ. |
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Answer» 6cotA+2cosecA=cotA+5cosecAor, 6cotA-cotA=5cosecA-2cosecAor, 5cotA=3cosecAor, 5(cosA/sinA)=3(1/sinA)or,5cosA=3or, cosA=3/5 Hit like if you find it useful |
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| 68600. |
RF & RR55133.If sin θ=then write the value of cosec θ. |
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