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Explanation:-Since AB = AC, hence angle opposite to equal sides are also equal.

< ACB = < ABC.

Since, ext < ACB = < QAC + < AQC

Then, < ABC = < QAC + < AQC

Now, because < ABP = < PBC.

Therefore, 2 < PBC = < ABC

Hence, 2 < PBC = < QAC + < AQC

Now, < QAC = < PAC and because ∆PAC and ∆PBC lie on same line segment PC.

So, < PAC = < PBC

Hence < PBC = < QAC

2 < PBC = < PBC + AQC

< PBC = < AQC

< PAC = < AQC

< QAC = < AQC

Side opposite to equal sides are equal.

QC = AC

Hence prove.

Sum of all angles of a triangle = 180°angle a + angle B+ angle C. = 180°

Angle B = angle C ( Base angles of an isosceles triangle are equal).Let angle B an c be x° each.

52° + x° +x °. = 180°2x °. = 180° - 52°x. = 122° /2x. = 61°

Ans.= Angle B is 61°

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180-52=128 not 122 ok please correct it

let,AM=MB=x and AL=LC=y.

now,LHS=4(BL^2+CM^2)

=4[{(2x)^2+(y)^2}+{(2y)^2+(x)^2}] (by pythagoras theorem)

=4[4x^2+y^2+4y^2+x^2]

=4[5x^2+5y^2]

=20x^2+20y^2 …… (1)

and now,RHS=5BC^2

=5[(2x)^2+(2y)^2] (since by pythagoras theorem h^2=b^2+p^2,p=perpndicular, b=base and h=hypotenuse)

=5[4x^2+4y^2]

=20x^2+20y^2 ……(2)

LHS=RHS

hence,proved

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In ΔAEB and ΔADC AE = AD (Given) ∠EAB =∠DAC (common) AB = AC { :- AE = AD and EC = BD }

thus

Δ AEB ≈ΔADC (SAS congurency)

In ΔAEB and ΔADC AE = AD (Given) ∠EAB =∠DAC (common) AB = AC { :- AE = AD and EC = BD }

thus

Δ AEB ≈ΔADC (SAS congurency)

in non leap year of February month at most 4 Sundays come and Mr. Sharma goes 4 days a week so total of 24 days.so total time to reach office is 24(1 h 10 m)= 24h240m=24h+4h=28h

sum of zeros = 0 and one zero = -8 so the other zero is x=0-(-8) = 8 the quadratic equation is (x-8)(x+8) = x²-64

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area of square is (side)²(side)² = 7744side = √7744side = 88mperimeter of sqaure is 4×side4×88m = 352m

thank u

α + β = 42-15 +β =42β= 42+15β= 57αβ= 57*-15=-855p(x) = x² -42x -855

Akbar did not have peace of mind because akbar was a negetive man. akbar has no real love of his life.

im sorrry the second part is incorrect

which one is apsalutly correct pls tell me which one is easy

I didn't understand pls tell me again

6a square =600or, a square=600/100or, a square=100or, a= √100or, a = 10 isliye ghan ka bhuja 10cm hoga

6a square =600 or, a square =600/100 or, a square =100 or, a, a= /100 or, a=is 10 isliye ghan ka bhuja 10cm hoga

10cm is correct answer.

10 cm is the right answer

SA of cube = 6a2600= 6a2a2= 600/6a2=100a = + 10 & -10here the side of cube will be 10 cm because it cannot be negative

how 1 ÷1=1who can say this answer to me

Surface of cube = 6×side2600=6side2600/6=side2100=side2√100 =side10=side so side =10cm

Slope =(y1-y2)/(x1-x2)here, Slope=3-0/√3-0=3/√3=√3

For two lines to be perpendicular , the product of their direction rations should be 0

(a1a2+b1b2+c1c2=0(a1a2+b1b2+c1c2=0

Let the given point beA(1,−1,2)B(3,4,−2)A(1,−1,2)B(3,4,−2)andC(0,3,2)C(0,3,2)andD(3,5,6)D(3,5,6)

The direction ratios of the line joinigAandBis(3,1),(4−(−1)),(−2−2)(3,1),(4−(−1)),(−2−2)

On simplifying we get,

(2,5,−4)(2,5,−4)

The direction ratios of the line joinigCandDis(3−0),(5−3),(6−2)(3−0),(5−3),(6−2)

On simplifying we get,

(3,2,4)(3,2,4)

For the linesABandCDto be perpendicular

a1a2+b1b2+c1c2=0

On substituting fora1,b1,c1a1,b1,c1anda2,b2,c2a2,b2,c2we get,

=(2×3)+(5×2)+(−4×4)=(2×3)+(5×2)+(−4×4)

=6+10−16=6+10−16

=0=0

HenceAB⊥CD

fgh

InΔABC, we have CD/DA = CE/EB as AB || DEinΔCDB, we have CF/FD = CE/EB as EF || BD

hence we get CD/DA = CF/FD=>Reciprocals: DA/CD = FD /CF=> Add 1 on bothsides: AC /CD = DC/CF

=> DC^2 = CF * AC

From the property of interior angle of same side. we get sum of interior angles on same is 180 degrees. Let p and q are the interior angles on same side, such that p+q=180. Now consider thetriangle formed by transversal line, the two angular bisectors of interior angles of p and q. we can find two angles as p/2 and q/2. SInce there are angular bisectors. Let the third angle be 't'. As we know sum of angles in a triangle is 180. then we get p/2+q/2+t=180 ⇒1/2(p+q)+t=180 ⇒1/2(180)+t=180 ⇒90+t=180 ⇒t=90.

Since t is the angle intersected by two angular bisectors of interior angles on same side. They intersect at right angles. Hence proved.

∴length of chord = 2√(OA²-OP²)

= 2√(10²-10²⁄2)

= 2√(100-100⁄2)

= 2√(100-50)

= 2√(50)

= 2×7.07

= 14.14cm Ans