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sin60° =√3/2

sin60=sin(90-30)=sin90cos30-cos90sin30=1*root(3)/2-0*/2=root(3)/2

thank you

sin 60*cos 30=(3½/2)*(3½/2)=3/4

P = {1,2} P×P = {(1,1),(1,2),(2,1),(2,2)} P×P×P = {(1,1,1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1),(2,2,2)}

thanks mahn

P = {1,2} P×P = {(1,1),(1,2),(2,1),(2,2)} P×P×P = {(1,1,1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1),(2,2,2)}

When we convert 12.5 into fraction we get:

125/10

Co primes should have their HCF as 1

125 and 10 are not co primes as their HCF is 5.

Simplifying 125/10

We get 25/2

25 and 2 are co primes

Therefore, 25/2 is the answer.

Let x= 0.3737...........eq(1)multiply eq (1) by 100100×x=100×0.3737100x= 37.3737.........eq(2)subtract eq 1 from eq299x=37x=37/99

x=23.43.... (1st term)

Multiply 100 on both sides

100x= 2343.4343... (2nd term)

Subtract first term from the second term

100x- x= 23.434343.....-2343.434343.....99x=2302x=2302/99

cos 30=√3/2 sin 60= √3/2 cos 60=1/2 sin 30 =1/2 (√3/2 +√3/2)/ (1+1/2+1/2 )= √3/(2)=√3/2.

cos(a-b)= cosA cosB- SinA sinB= cos30° = cos 2x2x= 30sin(30)= 1/2like the solution 👍 ✔️

sin30cos60 + cos30sin60= (1/2)(1/2)+(V3/2)(V3/2) =1/4+3/4=1+3/4=4/4=1

sin 30=1/2cos 60=1/2cos 30=/3/2cos 60=/3/2sin30cos60+cos30sin60=1/4+3/4 =4/4 =1

thanks bro..

area of a cricle = πr²

πr² = (14+6√5)π

r² = (14+6√5)

r = √(14+6√5)

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Let us simplify step by step sin 80 = sin (90-10) = cos 10, sin 65 sin 35= 1/2( cos (65-35)- cos(65+35)= 1/2( cos 30 - cos 100), sin 80 sin 65 sin 35 = 1/2(cos 10 ( cos 30 - cos 100)) cos 100= cos (90+10) = sin 10. Therefore. = 1/2 cos 10 cos 30 -1/2 cos 10 sin 10= 1/4 (cos 40 + cos 20 ) -1/4 sin 20 This gives 96 sin 80 sin 65 sin 35 = 24 (- sin 20 + cos 20 + cos 40) Let us simplify the denominator. sin 50 = sin (90-40) = - cos 40 sin 110 = sin (90 +20) cos 20. This gives sin 20 + sin 50 + sin 110= sin 20 + cos 20 -cos40. It seems that there is sign mistake. You check it. You will get -24 or +24.

tq sir

LHS = 8sin20° .sin40°.sin80°

=4{ 2sin20.sin80°}sin40°

=4{cos(60) - cos100°} sin40°

=4×1/2sin40° - 4cos100°.sin40°

=2sin40° - 2{2sin40°cos100°}

=2sin40° -2{ sin140° -sin60° }

=2sin40° -2sin40° + 2sin60°

=√3 = RHS

applying COSC- COS D formulaand sinC- sinD formula

cos60=1/2cos30=√3/2sin60=√3/2sin30=1/2hence1/2*√3/2+1/2*√3/2=√3/4+√3/4=√3/2hencecosx=√3/2x=30°

समीकरण 2x-3y= 8and-x-3y= 12 होगे

समझ नहीं आया

((√3/2)+(√3/2)/(1+1/2+1/2)=((2√3)/2)/2=√3/2

I asking to you my questions answer.this extra question.if you dont know please dont answer.

3.28=328/100=164/50=82/25

1/ 400/2 x+y is answer

1/400/2 x+y is the correct answer

1/400/2 x + y is correct

1/400/2 X+Y is answer

103x+97y=503----(1)97x+103y=497-----(2)by adding200x+200y=1000200(x+y)=1000x+y=5-----(3)103x+97y=50397x+103y=497by substrate6x-6y=66(x-y)=6x-y=1-----(4)x+y=5----+(3)by adding 2x=6x=6/2=3putting the value of x in eq3x+y=53+y=5y=5-3=2

1/400/2x+y is my answer as right

1000,5,6x, 1,2 is the answer of the given question

Value of Cos(271 degree) = 0.01745241

x=281/22 y = 90/44

answer is 7pie units

option c 14 pie units

Area of circle is 49 pie sq. unitTo find perimeter of circleSolu- pie (r) 2=49 pieBoth pie are cancelNow (r) 2=49r=7Perimeter =2 pie r2 pie 7=14 pie ANS

sin60°= √3/2cosec 30° = 2so2+√3/24+√3/2 Answer

LHS = sin(60°) = √3/2

RHS = 2 sin(60°) = 2×(1/2) = 1

So, LHS is not equal to RHS

thanks

=> sin60[sin20.sin40.sin80]

=>√3/2[sin20.sin(60-20).sin(60+20)]

=>√3/2[sin 3(20)/4]

=>√3/2[sin 60/4]

=>√3/2[√3/2*4]

=>√3/2*√3/8

=3/16

thanks

It will be√3/2*1/2+1/2*√3/2√3/4+√3/4√3/2

Sinθ+2cosθ=1Squaring both sides,sin²θ+2.sinθ.2cosθ+4cos²θ=1or, (1-cos²θ)+4sinθcosθ+4(1-sin²θ)=1 [∵, sin²θ+cos²θ=1]or, 1-cos²θ+4sinθcosθ+4-4sin²θ=1or, -4sin²θ+4sinθcosθ-cos²θ=1-1-4or, -(4sin²θ-4sinθcosθ+cos²θ)=-4or, (2sinθ)²-2.2sinθcosθ+(cosθ)²=4or, (2sinθ-cosθ)²=2²or, 2sinθ-cosθ=2 (Proved)

152x - 378y= -74 ----> [ 1 ]

-378x+152y = -604 ----> [ 2 ]

Adding equation 1 and 2

====> -226x + -226y = -678

x + y = 3 -----> [3]

Now ,Subtracting equation 1 from 2 ,-530x + 530y = -530

530y - 530 x = -530

x - y = 1 ----> [ 4 ]

Adding equation 3 and 4

2x = 4

x = 4/2

x = 2

Lets put this value of x in equation 3

2 + y = 3

y = 3 - 2

y = 1

x+y : x-y = 2+1 : 2-1

= 3:1

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hi

Let x= 0.235bar .........(1)1000x = 235.35 bar------2Subtract 1 from 21000x-x = 235.35 bar - 0.35 bar999x = 235x = 235/999Therefore, 0.235 bar = 235/999

bar is only on 35 .

check the steps now.

In ∆ ABD,tan(30°) = AD/BD

1/(√3) = 7√3/BD

BD = 7 × 3 cm = 21 cm

In ∆ADC,

tan(60°) = AD/DC

√3 = AD/DC

DC = 7√3/√3

DC = 7 cm

So, BC = BD + DC = 21 cm + 7 cm = 28 cm

√1-sina/1+sina√(1-sina)(1+sina)x√(1-sina)/(1-sina)√(1-sina)²/(1-sin²a)√(1-sina)²/cos²a(1-sina)/cos(1-cosa-sina/cosa)seca-tena.

Diameter of cylinder (d) = 2 mRadius of cylinder (r) = 1 mHeight of cylinder (H) = 5 mVolume of cylindrical tank, Vc = πr2H = π×(1)2×5=5π m

Length of the park (l) = 25 mBreadth of park (b) = 20 mthe height of standing water in the park = h Volume of water in the park = lbh = 25×20×h

Now water from the tank is used to irrigate the park. So, Volume of cylindrical tank = Volume of water in the park

⇒5π=25×20×h⇒5π/25×20=h⇒h=π/100 m⇒h=0.0314 m

Through recycling of water, better use of the natural resource occurs without wastage. It helps in reducing and preventing pollution. It thus helps in conserving water. This keeps the greenery alive in urban areas like in parks gardens etc.

The value of sin30° is 1/2

sin(30°) + sin(60°)

1/2 + √3/2

(1+√3)/2