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Ask QuestionPosted by Aadya Singh 5 years, 8 months ago
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Posted by Jerlin Christy 5 years, 9 months ago
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Sia ? 5 years, 9 months ago
Sin 18/cos72 cos(90-72)/cos72
Cos72/sin72=1
Tan26/tan(90-74)=1
Cos48-sin42=sin(90-42)-sin42=0
Cosec31-Cosec(90-69) Cose31-Cosec31=0
Tanisha Yadav 5 years, 9 months ago
Posted by Suniksha Bhatia 5 years, 9 months ago
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Sia ? 5 years, 9 months ago
Consider a triangle ABC in which ∠A=θ and ∠B=90o
Let AB = 12k and AC = 13k
Then, using Pythagoras theorem,
BC=√(AC)2−(AB)2=√(13k)2−(12k)2
=√169k2−144k2=√25k2=5k
∴ sinθ=BCAC=5k13k=513
cosθ=ABAC=12k13k=1213tanθ=BCAB=5k12k=512
cotθ=ABBC=12k5k=125cosecθ=ACBC=13k5k=135
Tanisha Yadav 5 years, 9 months ago
Anshuman Singh Chouhan 5 years, 9 months ago
Posted by Anna Shivani 5 years, 9 months ago
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Sia ? 5 years, 9 months ago
Let seven years ago arya's age be x years.
As per given condition
Seven years ago sparsh's age was five times the square of arya's age.
Then, seven years ago sparsh's age was 5x2 years.
Therefore, arya's present age = (x + 7) years
And sparsh's present age = (5x2 + 7) years
Three years hence arya's age = (x + 7 + 3) years = (x + 10) years
And three years hence sparsh's age = (5x2 + 7 + 3) years = (5x2 + 10) years
As per given condition
Three years hence arya's age will be 25 of sparsh's age.
Therefore, x + 10 = 25(5x2 + 10)
⇒ x + 10 = 2(x2 + 2)
⇒ x + 10 = 2x2 + 4
⇒ 2x2 - x - 6 = 0
⇒ 2x2 - 4x + 3x - 6 = 0
⇒ 2 x ( x - 2 ) + 3 ( x - 2 ) = 0
⇒ (2x + 3) (x - 2) = 0
⇒ x - 2 = 0 [∵ 2x + 3 ≠ 0 as x > 0]
⇒ x = 2
Hence, arya's present age = (2 + 7) years = 9 years
sparsh's present age = (5 ×22 + 7) years = 27 years
Posted by Md Shadil 5 years, 9 months ago
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Posted by Shikshita Naryal 5 years, 9 months ago
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Sia ? 5 years, 9 months ago
Here the remainder =6x+2
Therefore, when 6x+2 is subtracted from 8x4 + 14x3 + x2 + 7x + 8, then it will be divisible by 4x2 - 3x + 2.
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Ram Kushwah 5 years, 9 months ago
The less than type CF table:
<th scope="col">Class range</th> <th scope="col">cf</th>Less than 10 | 14 |
Less than 20 | 14+17=31 |
Less than 30 | 31+22=53 |
Less than 40 | 53+26=79 |
Less than 50 | 79+23=102 |
Less than 60 | 102+16=118 |
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Sia ? 5 years, 8 months ago
Let ABCD be a square and B (x, y) be the unknown vertex.

AB = BC
⇒ AB2 = BC2
⇒ (x + 1)2 + (y - 2)2 = (x - 3)2 + (y - 2)2
⇒ x2 + 1 + 2x + y2 + 4 - 4x = x2 - 6x + 9 + y2 + 4 - 4x
⇒ 2x + 1 = - 6x + 9
⇒ 8x = 8
⇒ x = 1 ........ (i)
In △ABC, AB2 + BC2 = AC2
⇒ (x + 1)2 + (y - 2)2 + (x - 3)2 + (y - 2)2 = (3 + 1)2 + (2 - 2)2
⇒x2 + 1 + 2x + y2 - 4x + 4 + x2 + 9 - 6x + y2 + 4 - 2y = 16 + 0
⇒ 2x2 + 2y2 + 2x - 4y - 6x - 4y + 1 + 4 + 9 + 4 = 16
⇒ 2x2 + 2y2 - 4x - 8y + 2 = 0
⇒ x2 + y2 - 2x - 4y + 1 = 0 ..... (ii)
Putting the value of x in eq. (ii),
1 + y2 - 2 - 4y + 1 = 0
⇒ y2 - 4y = 0
⇒ y(y - 4) = 0
⇒ y = 0 or 4
Hence the other vertices are (1, 0) and (1, 4).
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