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aoc + bod = 266°

so aoc = bod = 133°

angle aod = cob = 180-133 = 47°

y+25°=75°(alternate angle)y=75°-25°=50°x+25°=180°(adjacent pair)x=180°-25°=155°

4. positive5.negative

both signs are of negative integer

(i) +(ii) -(iii) +(iv) +

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please give me like Sir

Option (a) is correct.

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1 and 4, 2 and 1, 3 and 3 , 4 and 2

1 is 4, 2 is 1,3 is 3,4 is 2

1 and 4,2,and1,3 and 3,4 and 2

A. bar graphB. widthC. two,horizontal and vertical.

Draw the graph of equation y=2x+1

Velocity of the stone at timet,

v= 9.8t

Whent= 4 sec,

v= 9.8 × 4 m/sec = 39.2 m/sec

Thus, velocity of the stone after 4 second is 39.2 m/sec.

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1

let angle A = xangle B = 3x -2angle C = 3x -2 +9 = 3x + 7

Now, angle A + angle B + angle C = 180°

subsitue value of A, B and C:

x + 3x -2 + 3x + 7 = 1807x + 5 = 1807x = 175x = 175/7x = 25

Now ;angle A = 25°angle B = 3 × 25 -2 = 73°angle C = 73 + 9 = 82°

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The tree breaks at point D

∴ seg AD is the broken part of tree which then takes the position of DC

∴ AD = DC

m∠ DCB = 30º

BC = 30 m

In right angled ∆DBC,

tan 30º = side opposite to angle 30º/adjacent side of 30º

∴ tan 30º = BD/BC

∴ 1/√3 = BD/30

∴ BD = 30/√3

∴ BD = (30/√3)×(√3/√3)

∴ BD = 10√3 m

cos 300 = adjacent side of angle 300/Hypotenuse

∴ cos 300 = BC/DC

∴ cos 300 = 30/DC

∴ √3/2 = 30/DC

∴ DC = (30×2)/√3

∴ DC = (60/√3)×(√3/√3)

∴ DC = 20√3 m

AD = DC = 20 √3 m

AB = AD + DB [∵ A - D - B]

∴ AB = 20 3 + 10 3

∴ AB = 30 3 m

∴ AB = 30 × 1.73

∴ AB = 51.9 m

∴ The height of tree is 51.9 m.

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given : distance between the foot of the tree and the broken part of the tree that touches the ground = 30 m

let ,the height ofstanding part of the tree be = h

theheight of fallen part (forms hypotenuse) be = x

then the total height of the tree will be = h + x

now,tan 30 = h/30 m 1/√3 = h/30 m 30/√3 = h ⇒ h= 30/√3 m ..... 1

similarly,cos30 = 30 m/ x √3/2 = 30/ x√3x = (30)2√3x = 60 m⇒x = 60 /√3 m ......2

(we now have both value of h and x )

on adding equation1 & 2 :⇒h + x = 30 /√3 +60 /√3 =90 /√3 m = 60√3 m

so , the total height of the tree is 60√3 m .

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As AB || DCthus 125°+x=180° (property of parallel lines when cut by a transversal)x=55°

As AD||BCthus x+y+56=180°55°+y°+56°=180°y=69°

And z+y=180° (transversal cutting the two parallel lines and angles on the same side of the transversal sum up to 180 degrees)z+69°=180°z=111°

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-8-7-6-5-4-3-2-1 0,1

angle A =angle Chence56+y=125y=125-56=69°now sum of angle C and angle D is 180henceX+69=180X=111°nowAEDC. is a quadrilateralhence 125+X+y+z=360125+111+69+z=360z=55

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Let x and y be the two angles of a parallelogram.It is given that one angle istwo-third of its adjacent angle. So, we assume that angle"x" is two-third of angle "y" .which is written as,x = (2/3). y ........ (1)We also know that the adjacent sides of a parallelogram are supplementary. It means that the sum of adjacent angles is equal to 180.Hence, x + y = 180 ........ (2)Put the value of x from equation (1) in equation (2):(2/3) y + y = 180(2/3 +1) y = 180(5/3) y = 180y = 180. (3/5)y =108Now put this value in equation (2) to get the value of x:x + 108 = 180x = 180 - 108x = 72Hence, the adjacent angles of a parallelogram are 72° and 108°

what is your question

option B is the correct answer

option b is correct answer

but why are u dividing the get by 2

(i) GST payable = 18% of ₹10,000

= 18/100×10000

= ₹1800

CGST and SGST will be divided for the equal parts:

CGST amount = 1800/2

= ₹900

and SGST amount = 1800/2

= ₹900

So the final answer for this question would be ₹900

on expanding,=A^3 - i^3 - 3A^2 + 3A + A^3 + I^3 + 3A^2 + 3A -7A=2A^3 +6A-7A=2A^3-Aas we know that A^2=I A^3=^2 *A=IA=Athe simplified form = 2A-A=Athus A is the required answer..

Option - C

it can further simplfied depending on type of question

it will be positive

Consider 1/secA-tanA - 1/cosA

Multiplying by secA + tanA in the numerator and denominator of

first term,

we get secA + tanA/(secA +tanA)(secA - tanA) - 1/cosA

= secA + tanA - secA (Since sec²A - tan²A = 1)

= tanA

Adding and subtracting secA , we get

secA + tanA - secA

= 1/cosA - (secA - tanA)

Now multiplying and dividing (secA - tanA) by (secA + tanA), we

get 1/cosA - (sec²A - tan²A)/(secA + tanA)

= 1/ cosA - 1/secA + tanA

= R.H.S

Hence, Proved.

(x+3)(2x)-(-2)(-3x)=82x^2+6x-6x=82x^2=8so x^2=4 and x belongs to Nso x=2