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sin⁴ (π / 8) + sin⁴ ( 3π/8) + sin⁴( 5π/8 ) + sin⁴( 7π/8 )

= [sin²(π/8)]² + [sin²( 3π/8 )]² + [sin²( 5π/8 )]² + [sin²( 7π/8 )]²

= [{1 - cos(π/4)}/2]² + [{1 - cos(3π/4)}/2]² + [{1 - cos(5π/4)}/2]² + [{1 - cos(7π/4)}/2]²

= [{1 - √2/2 }/2]² + [{1 + √2/2 }/2]² + [{1 + √2/2}/2]² + [{1 - √2/2}/2]²

= 2[{1 - √2/2 }/2]² + 2[{1 + √2/2 }/2]²

= 2[{2 - √2}/4]² + 2[{2 + √2}/4]²

= (1/8)(2 - √2)² + (1/8)(2 + √2)²

= (1/8)[(4 - 4√2 + 2) + (4 + 4√2 + 2)]

= 12/8 = 3/2

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Let the radius of the circle is =r

Then the area of the circle = πr2

Now double radius = 2r

Then the new area = π (2r)2

= π × 4 ×r2

= 4 (πr2)

If the radius of the sphere is double then the area will become four times.

4x + 5 - x - (x-5)/4 = 3

3x + 5 + ( 5 - x)/4 = 3

3x + 2 + ( 5 - x)/4 = 0

12x + 8 + 5 - x = 0

11x = - 13

x = -13/11

हमें ज्ञात है

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Consider the longer of the two sides PQ and PR. Let's say that it is PQ (though it doesn't matter, and you can pick either if they are equal).

PS must be shorter than or equal to PQ. (It's only equal if S and Q are the same point.)

It is a rule that the sum of any two sides of a triangle must be greater than the third side. So we also know that PR + QR > PQ

Since PR + QR > PQWe know PQ + PR + QR > 2PQ(just added PQ to both sides)

And since PQ >= PS,We know 2PQ >= 2PS(multipled both sides by 2)

From those two (PQ + PR + QR > 2PQ, 2PQ >= 2PS), we know:PQ + PR + QR > 2PS

Question 10 karna hai

aap jo question kiy hai vah 11 Ka hai

Given: A △ABC in which AB = AC. D is a point on AC such that BC^2= AC × CD.To prove : BD = BCProof : Since BC^2= AC × CDTherefore BC × BC = AC × CDAC/BC = BC/CD .......(i)Also ∠ACB = ∠BCDSince △ABC ~ △BDC [By SAS Axiom of similar triangles]AB/AC = BD/BC ........(ii)But AB = AC (Given) .........(iii)From (i),(ii) and (iii) we getBD = BC.

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Ans :- Option (a) is correct

Given - AB = AC

To prove -- BD > CD

Proof -- Since AB = AC∠ABC = ∠ACB (By Isosceles Triangle property) ----(i)

Here clearly,∠ABC > ∠CBD ∠ACB > ∠CBD ---from (i)∠DCB > ∠CBDBD > CD (Angle opposite to greater side is greater in a triangle)

Hence proved!

3x3 = 9(multiplication)

Thank

the answer is very simple 3×3 = 9

thanks

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log105+log10(5x+1)=log10(x+5)+1 ⇒log105+log10(5x+1)=log10(x+5)+1 ⇒log10[5(5x+1)]=log10[10(x+5) ] ⇒5(5x+1)=10(x+5) ⇒5x+1=2x+10 ⇒3x=9⇒x =3

log105+log10(5x+1)=log10(x+5)+1

⇒log105+log10(5x+1)=log10(x+5)+1

⇒log10[5(5x+1)]=log10[10(x+5) ]

⇒5(5x+1)=10(x+5)

⇒5x+1=2x+10

⇒3x=9⇒x =3

a circle with Centre P is inscribed in a triangle ABC side a b side BC and side AC touch the circle at points L,m and n respectively the radius of the circle is r

prove that area of triangle ABC =1/2(AB+BC+AC)×r​

Let say center point = O

if we draw line from points A , B & C at point O

we can Divide ΔABC into three triangle

ΔAOB , ΔBOC & ΔCOA

Area of ΔAOB = (1/2) * AB * OL ( Base * Perpendicular)

OL = Radius = r

Area of ΔAOB = (1/2) * AB * r

SImilarly

Area of ΔBOC = (1/2) * BC * r

Area of ΔBOC = (1/2) * BC * r

Area of ΔCOA = (1/2) * AC * r

Area of ΔABC = Area of ΔAOB + Area of ΔBOC + Area of ΔCOA

=> Area of ΔABC = (1/2) * AB * r + (1/2) * BC * r + (1/2) * AC * r

=> Area of ΔABC = (1/2) * (AB + BC + AC) * r

QED

Apply remainder theorem =>7 + 3x =0 => 3x = - 7 => x = - 7/3Replace x by - 7/3 we get =>3x3 + 7x=>3(-7/3)3 + 7(-7/3)=>3(-343/27) – 49/3=> (-343/9) – 49/3This value is not equal to 0 we have checked that 7 + 3x is not a factor of expression 3x3 + 7x

√2 sin135 cos210 tan240 cot300 sec330 sin135 = sin(180-45) = sin45 = 1/√2 cos210 = cos(180+30) = -cos30 = -√3/2 tan240 = tan(180+60) = tan60 = √3 cot300 = cot(360-60) = -cot60 = -1/√3 sec330 = sec(360-30) = sec30 = 2/√3

√2 × 1/√2 × (-√3/2) × √3 × (-1/√3) × 2/√3 = 1

(ii)