Explore topic-wise InterviewSolutions in Current Affairs.

5kg/600g=5000g/600g=50/6g=25/3

25:3

GIVEN: An equilateral triangle ABC, Medians AP, BQ, &CR. Their point of concurrency is O, which is the centroid of the triangle.

TO PROVE THAT: Centroid O is the circumcentre of the triangle ABC.

If we prove that centroid O is the circumcentre of the triangle, then it automatically becomes the centre of the circumcircle.

PROOF: Since AP is median, so P is mid point of BC.

ie, BP = PC.

AB = AC ( as triangle ABC is equilateral)

AP=AP ( common side)

Hence triangle ABP is congruent to ACP( by SSS congruence criterion)

=> angle APB = angle angleAPC ( corresponding parts of congruent triangles)

But their sum = 180°

So, each angle has to be 90°.

That shows that AP is perpendicular bisector of BC.

Similarly, pyove that BQ & CR are perpendicular bisectors of AC & AB respectively.

So now, The point of concurrency ‘O' of these perpendicular bisectors becomes circumcentre of the triangle. ( as circumcentre is the point of concurrency of 3 perpendicular bisectors of the sides of the triangle). And this centre is also the centre of circum circle.

This way centroid O coincides with circumcentre O…

1

2

hence, option (A) is correct.

1). 7cg = 0.07 gram ( 1cg= 0.01gram) 6mg = 0.006 gram( 1mg= 0.001gram)

total = 0.07+0.006 = 0.076gram

2).5mg = 0.005 gram (1mg= 0.001gram)

3). 3078dg = 307.8 gram( 1dg = 0.1gram)

{ please like if you got your answer}

1. 0.076 g2. 0.005 g3. 307.8 g

1. 7 cg 6 mg = 0.076g2. 5 mg = 0.005g3. 3078 dg = 3.078 g

1.0.076g. 2.0.005g3.307.8g

1. 7cg 6mg =0.076g 2. 5mg =0.005g 3. 3078 dg = 3.078 g

a decimal=3078,ek decimal=3078 hota hai

(i) 0.076g(ii) 0.005g(iii) 307.8g

a) 56/100 = 0.56

b) 95/100 = 0.95

c) 62/100 = 0.62

d) 49/100 = 0.49

the correct answer is 2

hypotanius

opposite side of the right angle triangle triangle is hypotenus. This is the best answer for you

in right angle triangle angle opposite side is= opposite sideright angle opposite side =hypotenuseremaining side =adjacent side

side oppsite to right angle in a right angled triangle is known as hypotenuse

sin36° sin72° sin108° sin144°=sin36° sin72°sin(90° + 18°) sin(90° + 54°)=sin36° sin72°cos18° cos 54°= sin236° sin272° {since, cos18° = sin72° and cos54° = sin36°}= { ( √(10 - 2√5 ) / 4 }2[ { √(10 + 2√5) } / 4 ]2{since, sin36° = ( √(10 - 2√5 ) / 4 and sin72° = { √(10 + 2√5) } / 4}

= [ (10 - 2√5 ) / 16 ] [ (10 + 2√5) } / 16 ]= { (10)2- (2√5)2} / ( 16 * 16 )

{since, (A - B)(A + B) = A2- B2}= (100 - 20) / 256= 80 / 256= 5 / 16

sin 30 degree is 0. 5.

1/2 is the value of sin30

Explanation : sin36/cos54 - sin54/cos36 = sin(90-54)/cos54 - sin(90-36)/cos36 = cos54/cos54 - cos36/cos36 = 1 - 1 = 0

solution : 0

If you find this answer helpful then like it

since we know that sides of square are equal then32=5x-832+8=5x40=5x8=xx=8

53*53= 2809Hence a is right answer

53*532809varg ka question my post karo

2809 is the correct answer

2809 is the right answer

53*53=2809square root of 2809 is 53

option a is the right answer

option a is the right answer

53 का वर्ग 2809 is the answer

the correct answer is C

2809 is the right answer

2809 is the right answer

2809 is the right answer

2809 is the right answer

2809 is the right answer

53*53=2809This is the right answer

Sin²0+cos²0=1 is correct answer

sin² theta + cos² theta = 1

1 is right answer....

sin^2theta + cos^2theta =1

sin²theta+cos²theta=1

iska maan hots hai jiii

si

1 is right answer...

it's answer will be 1

1 is the correct answer

1 is the right answer

sin ^2A +cos^2A = 1 the right answer

sin²theeta+cos²theeta=1

1 is the right answers

1 is a correct answer

sin2θ + cos2θ = 1. is correct answer

Pythagorean identities

a)sin2θ+ cos2θ = 1.

1 is the correct answer of this question

1 is the valueand it is correct answer

sin²0+cos²0=1 is right

1 is the right answer

1 is the correct answer

5

Iska answers 1 hoga 1is right answer

1 is the right answer according to your question

1 is the correct answer

(1) .is right answer

1 is a right answer ,

1 is the answer:))))))))

sin2a +cos2a =1 , 1 is right answer

this is an identity.and answer is one

1. is. right answer

1 is the correct answer h.

sin^2+cos^2=1 is a right ans

1 is the value of these question

sin^2

ans_1

iska answer one hai yeh trignomatric formula hai

(sec∅-1)/(sec∅+1)= (1/cos∅-1)/(1/cos∅+1)= 1-cos∅/1+cos∅= (1-cos∅)(1+cos∅)/(1+cos∅)(1+cos∅)= (1-cos²∅)/(1+cos∅)²= (sin∅)²/(1+cos∅)²

Consider the LHS:(1-sinA+cosA)^2= [(1-sinA) + cosA]^2= (1-sinA)^2+ cos^2A+ 2(1-sinA)cosA= 1 + sin^2A− 2sinA+ cos^2A+ 2(1-sinA)cosA= 1 + (sin^2A+ cos^2A)− 2sinA+ 2(1-sinA)cosA= 1 + 1− 2sinA+ 2(1-sinA)cosA [Since,sin^2A+ cos^2A =1]= 2− 2sinA+ 2(1-sinA)cosA= 2(1− sinA) + 2(1-sinA)cosA= 2(1− sinA)(1 +cosA)= RHS

Like my answer if you find it useful!

sinA = msinB....(1)

tanA= ntanB

(sinA/cosA) = n(sinB/cosB)....(2)

substuting sinB value from equation (1)

cosB = (n/m) cosA......(3)

sin^2A = m^2sin^2B

1-cos^2A = m^2(1-cos^2B)

substituting equation (3)

1-cos^2A = m^2[1 – ((n^2/m^2)cos^2A)]

1 – cos^2A = m^2– n^2cos^2A

n^2cos^2A – cos^2A = m^2– 1

cos^2A = (m^2– 1)/(n^2– 1)

sin^6 x + cos^6 x

= (sin^2 x + cos^2 x) ( sin^4 x - sin^2 cos^2 x + cos^4 x)

= (sin^4 x + cos^4 x) - sin^2 x cos^2 x

= (sin^2 x + cos^2 x)^2 - 2 sin^2 cos^2 x - sin^2 x cos^2 x

= 1 - 3 sin^2 x cos^2 x

please like my answer if you find it useful

( a - b) + ( b - c) + c - a)

a - a + b - b + c - c

0

HCF = 113

LCM = 56952

One no: = 904

Other no: = x

HCF x LCM = Product of two no:s

113 x 56952 = 904x6435576 = 904x

x = 6435576 / 904

x = 7119

Hence ,the other no: is 7119

hit like if you find it useful

CosA-sinA+1/cosA+sinA-1=(cosA-sinA+1)(cosA+sinA+1)/(cosA+sinA-1)(cosA+sinA+1)=(cos²A-cosAsinA+cosA+cosAsinA-sin²A+sinA+cosA-sinA+1)/{(cosA+sinA)²-(1)²}=(cos²A-sin²A+2cosA+1)/(cos²A+2cosAsinA+sin²A-1)={cos²A+2cosA+(1-sin²A)}/(1+2cosAsinA-1) [∵, sin²A+cos²A=1]=(cos²A+2cosA+cos²A)/2cosAsinA=(2cos²A+2cosA)/2cosAsinA=2cosA(cosA+1)/2cosAsinA=(cosA+1)/sinA=cosA/sinA+1/sinA=cotA+cosecA=cosecA+cotA (Proved)

Like my answer if you find it useful!

Given:A,B,CA,B,Care angles of a triangle, we have thatA+B+C=π

Wehave that

sin(2A)+sin(2B)

=2sin(A+B)cos(A−B)

=2sin(π−C)cos(A−B)

=2sin(C)cos(A−B)sin⁡(2A)+sin⁡(2B)=2sin⁡(A+B)cos⁡(A−B)=2sin⁡(π−C)cos⁡(A−B)=2sin⁡(C)cos⁡(A−B)

Hence,

sin(2A)+sin(2B)+sin(2C)=2sin(C)cos(A−B)+2sin(C)cos(C)

=2sin(C)(cos(A−B)+cos(C))

=2sin(C)(cos(A−B)+cos(π−(A+B)))=2sin(C)(cos(A−B)−cos(A+B))=2sin(C)×2sin(A)sin(B)=4sin(A)sin(B)sin(C)Hence proved

1+sin^2 theta=3 sintheta cos theta (we know that sin^2 theta + cos^2 theta =1)

= ( sin^2 theta + cos^2 theta ) + sin ^2 theta = 3 sin theta cos theta

= sin^2 theta + cos^2 theta + sin ^2 theta = 3 sin theta cos theta

= cos^2 theta + 2 sin^2 theta = 3 sin theta cos theta

On dividing by cos^2 theta, we get

= 1 + 2 tan^2 theta = 3 tan theta

Let tan theta = b

2b^2 - 3b + 1 = 0

= (2b-1)(b-1) = 0

b = 1 or 1/2

So, tan theta = 1 or 1/2.

x:4 = 9:6x/4 = 9/6x = (4×9)/6x = 36/6 = 6x = 6

sin^6ө+cos^6ө =1-3sin^2өcos^2ө(sin^2ө)^3+(cos^2ө)^3= (sin^2ө+cos^2ө)^3− 3 (sin^2ө cos^2ө)(sin^2ө+cos^2ө) [since a + b = (a+b)^3− 3ab(a+b)] = 1 − 3sin^2өcos^2ө [Since sin^2ө+cos^2ө = 1]

let a be any positive integerb=4we know that 0<r<4r=0,1,2,3a can be 4q,4q+1,4q+2,4q+3where q is quotientsince a is odd and a cannot be 4q or 4q+2because they are divisible by 2 therefore any odd integer is of the form 4q,4q+1 if you like my solution so plz give like

incomplete question please post a complete question for answer

rishav rajjjjjjjjjjjjjjjjjjjjjjjjj

lolllllllllllllllllllllllll

The given integers are137and−354137and−354.

Add the given integers.

137+(−354)=137−354=−217137+(−354)=137−354=−217

Thus, the sum of137and−354137and−354is−217−217.

bb

The given integers are−52and52−52and52.

Add the given integers.

−52+52=0−52+52=0

Thus, the sum of−52and52−52and52is−217−217.

cc

The given integers are−312,39and192−312,39and192.

Add the given integers.

−312+39+192=−312+231=−81−312+39+192=−312+231=−81

Thus, the sum of−312,39and192−312,39and192is−81−81.

dd

The given integers are−50,−200and300−50,−200and300.

Add the given integers.

−50+(−200)+300=−50−200+300=−250+300=50−50+(−200)+300=−50−200+300=−250+300=50

Thus, the sum of−50,−200and300−50,−200and300is5050.

ankush answered question in right way

1] 5 -2 = 3 , 2] 5 + 3 = 8 and mark the value c = 3 and h =8