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angle bca equal two *angle bac

can you use AC =bc square +an square

I don't know answer sorry

thank u

(a-b)x+(a+b)y=a²-2ab-b².....(1)

(a+b)(x+y)=a²+b²a(x+y)+b(x+y)=a²+b²(a+b)x+(a+b)y=a²+b²......(2)

Subtract equation 1 from 2

(a+b-a+b)x=a²+b²-a²+2ab+b²2bx=2b²+2abx=b+a

(a+b)x+(a+b)y=a²+b²(a+b)(a+b)+(a+b)y=a²+b²(a+b)y=a²+b²-(a+b)(a+b)(a+b)y=a²+b²-a²-b²-2ab(a+b)y=-2aby=-2ab/(a+b)

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Anisosceles triangleis one with two equal sides. Therefore,every equilateral triangleisisosceles.

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Here ABCD is a square, AEB is an equilateral triangle described on the side of the square and DBF is an equilateral triangle described on diagonal BD of the square.

To Prove: Ar(ΔDBF) / Ar(ΔAEB) = 2 / 1

Proof: If two equilateral triangles are similar then all angles are = 60 degrees.

Therefore, by AAA similarity criterion,△DBF~△AEB

Ar(ΔDBF) / Ar(ΔAEB) = DB2/ AB2 --------------------(i)

We know that the ratio of the areas of two similar triangles is equal tothe square of the ratio of their corresponding sides i .e.But, we haveDB=√2AB

{But diagonal of square is√2times of its side}-----(ii).

Substitute equation (ii) in equation (i), we get

Ar(ΔDBF) / Ar(ΔAEB) = (√2AB)2/ AB2 = 2 AB2/ AB2= 2

∴ Area of equilateral triangle described on one side os square is equal to half the area of the equilateral triangle described on one of its diagonals.

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Now by the property of triangles, sum of any two sides > the third side.Therefore, AB + BC > AC Same way CD +DA > ACSimilar way we can get DA + AB > BD and BC + CD > BDNow adding all these inequalities, 2(AB + BC +CD +DA) > 2(AC + BD)Dividing both sides by 2 we get, (AB + BC +CD +DA) > (AC + BD)Thus, the given first identity is proved.For the second identity,2AC > AB + BC Similar way, 2AC > CD + DAAlso, 2BD > DA + AB and 2BD > BC + CDAdding the inequalities, we get 4(AC + BD) > 2(AB + BC +CD +DA)Dividing by 2 on both sides,we arrive at 2(AC + BD) > (AB + BC +CD +DA)or (AB + BC +CD +DA) < 2(AC + BD) Hence second identity is proved.

Hii

h r u

x=2.13796, x = 2.02685

The algaebric equation is -

3m - 6 = 12

YeS It Is ThE CoRrEcT AnSwEr

Explanation : lim x -> 0 (cosecx - cotx) = lim x->0 (1/sinx - cosx/sinx) = lim x->0 (1-cosx)/sinx If we put x = 0 we will get 0/0 form so using L'hospital rule = lim x-> 9 (sinx)/(cosx) = sin0/cos0 = 0/1 = 0

Solution : 0

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this is not my question

1/x nahi hai mere question me

Given both triangles are equilateral then each angle of both triangles is 60

Sum of angles around point is 360As shown in fig.169 + 60 + 60 + b = 360289 + b = 360b = 360-289b = 71Angle b = 71 degree

bhag

area of a circle =πr^2

in a regular hexagon each angle is 120°

So the amount of circle inside the hexagon is 1/3rd of total area of circle.

so the area of one shaded part = 1/3(πr^2)

there are 6 shaded parts.

so the answer will be 6× 1/3 (πr^2) = 2πr^2

It is given that ΔABC ~ ΔPQRWe know that the corresponding sides of similar triangles are in proportion.∴ AB/PQ = AC/PR = BC/QR ...(i)Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R …(ii)Since AD and PM are medians, they will divide their opposite sides.∴ BD = BC/2 and QM = QR/2 ...(iii)From equations(i)and(iii), we getAB/PQ = BD/QM ...(iv)In ΔABD and ΔPQM,∠B = ∠Q [Using equation(ii)]AB/PQ = BD/QM [Using equation(iv)]∴ ΔABD ~ ΔPQM (By SAS similarity criterion)⇒ AB/PQ = BD/QM = AD/PM

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When 3x² - k√3 + 4 = 0

have real root then discrimanat must be greater or equal to 0

D = (-k√3)² - 4 × 3 × 4

3k² - 48 > = 0

k² > = 16

k >= √16

k > = +/-4

x=1y=2

(-√3,2√3)__+&&____💐

As AB || FE, FA becomes the tranversal line cutting parallel lines. So angle A + angle F = 180°.

sum of other four angles = 720° - 180°= 540°.So 6 x + 4 x + 2 x + 3 x = 15 x = 540°.x = 36°.

Angles B C D E are: 216°, 144°, 72°, 108°

Ans :- join BF , BE and BD4 triangles are formed ABF, FBE,EBD and BDCthe sum of all the angles of a triangle is 180°∴the sum of all the angles in 4 triangles will be =180× 4 = 720°∴the sum of all the angles in a hexagon is 720°∴the sum of all the interior angles of hexagon ABCDEF is 720°

Do you study in DAV? anyways, Thanks a lot