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Let tan(15°) = tan(45°-30°)We know that tan(A - B) = (tanA - tanB) /(1 + tan A tan B)⇒tan(45°-30°) = (tan45°- tan30°)/(1+tan45°tan30°)= {1- (1/√3)} / {1+(1/√3)}∴ tan15° = (√3 - 1) / (√3 + 1)

thnks alot

Sum of all angles of triangle is 18070 + 40 + angleACB = 180angleACB = 180 - 110 = 70

angleACB + angleACD = 180 (linear pair)angleACD = 180 - 70 = 110

angle BAC+angle ABC=angle ACD(exterior angle )70+40=angle ACD=110°

let angle ACD = x 70+40=x x=110

x + y + z = 180...... (1)

Using triangle angle sum property

z + 55 + 60 = 180 ( linear pair) z = 180 - 115z = 65

x = 60 (alternate angles)

From eq(1)60 + y + 65 = 180y = 180 - 125y = 55

Therefore, x = 60, y = 55, z = 65

couldn't understand

x^2+9x+18=0 Splitting the middle term,x^2+6x+3x+18=0x(x+6)+3(x+6)=0(x+3)(x+6)=0x=-3x=-6

Thank you.. integrate please

no need cause it will be zero. even it do it na. take x² = t and proceed further but of course answer will be zero again.

oh yes I did it

Given: 17p-2=49To find: pSolution:17p-2=4917p=49-217p=47p=47/17p=2.76Answer: p= 2.76

ACB + 120 = 180 ACB = 180 - 120 ACB = 60as AC = AB so ABC isosceles triangle therefore angle ABC = ACB = 60so 60+60 + A = 180 A = 180 - 120 = 60

thanks

l was checking .it works or not . l have done. Thanks

joint equation of line is

L1.L2 = 0

=> (3x+2y-1)(x+3y-2) = 0=> 3x²+9xy-6x+2xy+6y²-4y-x-3y+2 = 0=> 3x²+6y²+11xy-7x-7y+2 = 0

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question is good one…

Here, we have to take two condition-

After removing modulus we can write equation as follow-

+(x^2-4x)<5 …….. (1)

And

-(x^2-4x)<5 ……… (2)

Case-1_

Solving inequation ..(1)

+(x^2-4x)<5

Take 5 to the LHS -

(x^2-4x)-5 < 0

After solving LHS[which is quadratic]-

We get_

(x+1)(x-5)<0

=> x belongs to (-1 to 5)……. (3)

Case 2_

Solving inequation….(2)

-(x^2-4x)<5

Take LHS term to RHS -

We get -

0 < (x^2-4x)+5

But equation in RHS gives imaginary roots .That's why we have to exclude the second case .

Therefore from …(3) , we get -

Answer-> -1 < x < 5

f(x) = x^2 - 4x - 5 < 0

First, find the 2 real roots: f(x) = 0.Since a -b + c = 0, one real root is (-1) and the other is (-c/a = 5)

------------------|-1===|0===========|5-----------------

Test point method. Plot (-1) and (5) on the number line. Use the origin O as test point --> x = 0 --> f (x) = -5 < 0. True. Then the origin O is located on the segment (-1, 5) that is the solution set.Answer: Open interval (-1, 5).

Algebraic Method. Between the 2 real roots (-1) and (5), f(x) is negative, having the opposite sign to a (> 0).Answer: Open interval (-1, 5)

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as it is an isoceles trianglehenceangle B=angle Chence in a triangle sum of angles is 180henceangle A+B+C=180hence40+2C=1802C=140C=70°

In mathematics alinear inequalityis aninequalitywhich involves alinearfunction.

Red line is y = 3,Blue line is 3x + y = 3

let angle A be 2x B be 3x C be xby angle sum property of triangleA+B+C=1802x+3x+x=1806x=180x=180/6x=30angle C=x=30 B=3x=90 A=2x=60

rong answer

angle cannot be in fraction

mine answer is correct

b= 70 ( vertically opposite anglesc+ 100 = 180 c= 80in triangle we see that70+ 80+ a = 180 a = 180 - 150a = 30

Actually in a equlateral triangle all the sides and angles are equal. One side of an equlateral is 60 degree so when 6 sides are are multiplied by 60 we will get 360 degree . This figure is a compination of 2 triangles because of angle sum property one triangle is equal to 180 degree. So 2 triangle is equals to 360 degrees.

In △ACE, By angle Sum property of triangle∠A+∠C+∠E=180° ...(1)Similarly, in △BDF∠B+∠D+∠F=180° ...(2)Adding (1) and (2), we get∠A+∠B+∠C+∠D+∠E+∠F=360°

because fractions are always represented in the form of p/q .

Thnk u

(3x+4)(3x-5) = 3x*3x-5*3x+4*3x-4*5 = 9x*x-3x-20

3(3x-5)+4(3x-5) 9x-15+12x-203x-35

lim x/sin2xMultiplying by 2 in Numerator and denominatorhencelim 2x/sin2x*(1/2)Now lim x to 0 sinx/x=1 by sandwich theoremHence answer will be 1/2