Here,  We have given, sin (α+β) =  k sin (α-β) or, sin (α+β)/  sin (α-β) = k/1 or,[ sin (α+β) + sin (α-β)] / [ sin (α+β) - .....

We are not here to do all the solution. First we givw you some steps and hints. Then you must try yourself. If you still in confusion, send us your practice paper, then we will provide full solution with correction in your solution. This time I .....

LHS =  cos2π8+A2-⁡cos2π8-A2      = { cos(π/8 +A/2) + cos(π/8 -A/2)}{cos(π/8 +A/2)- cos(π/8 -A/2)} Now use formula of cosC + cosD and cosC - cosD and simplify to get .....

Are the translation t and T .....

Afforestation Conservation of soil 1. It is the act of planting trees in barren land. 2. It is carried out by planting trees. 3. It prevents soil erosion, flood, landslide, etc.      .....

Here, We have given; SinA/3 = 1/2(p + 1/p) We know, SinA = 3SinA/3 - 4Sin3A/3         = 3/2(p + 1/p) - 4/8(p + 1/p)3         = 1/2( p + 1/p)[ 3 -(p + 1/p)2] Solve and find the .....

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Trade diversification is defined as the process of expanding trade or different items to different countries or variation in the products. Trade diversification takes place in two forms: territory diversification and variety diversification. In .....

In geometrical optics, a focus, also called an image point, is the point where light rays originating from a point on the object converge. Although the focus is conceptually a point, physically the focus has a spatial extent, called the blur .....

Here, p Sinθ = q or,Sinθ = q/p  So, h = √[q2 + p2] Now,  Tanθ = q/√[q2 + p2] Sec θ = √[q2 + .....

Plane means flat, in simple language. A plane is a two-dimensional surface that is not curved in the third dimension. A plane mirror is simply a mirror with a flat surface. It’s not got a curved surface like a convex mirror or a concave .....

A plane that includes the incident ray and a line drawn normal to the surface is called the plane of incidence. This plane also contains the reflected and refracted rays. A refracted ray is transmitted into the second medium and travels in a .....

Here, Let  a be the length of sides. Then, TSA = 400cm2 or, LSA + Area of base = 400 or, 3 Area of base + Area of base = 400 or, Area of base = 100 or, a2 = 100cm2.               [ Since, .....

Here, Let  a be the length of sides. Then, TSA = 56cm2 or, 4 ⨯ 10  + Area of base = 56 or, Area of base = 56 - 40 or, Area of base = 16 or, a2 = 16cm2.               [ Since, pyramid .....

Here, Let  a be the length of sides. Then, TSA = 96cm2 or, LSA + Area of base = 96 or, 2al + (6cm)2 = 96 or, 12 l = 60 or, l  = 5cm.              .....

Here, a = 15cm l = 18cm TSA =? We know, TSA =  LSA + Area of base            = 2al + a2            = 2 ⨯ 15cm ⨯ 18cm + (15cm)2 Find .....

Here, a = 25cm LSA = 1500cm2 Slant height = l = ? We know, LSA = 2al or, 1500 = 2⨯ 25 ⨯ l or, 50l = 1500 Find .....

Here, Here, a = 20cm LSA = 1200cm2 Slant height = l = ? We know, LSA = 2al or, 1200 = 2⨯ 20 ⨯ l or, 40l = 1200 Find .....

Here, Let height be h = OP. length of  base = a = 2h Volume = 288cm3 We know, Volume = 1/3 ⨯ A ⨯ h or, 288 = 1/3 ⨯ (2h)2 ⨯ h Calculate h .....

Here, Area of base = A = a2 = (16cm)2 Height of pyramid = h = 6cm, Volume = ? We know, Volume = 1/3 ⨯ A ⨯ h Use values from above and .....

Here, Slant height = l = 13cm Length of side of base = a = ? LSA = 260cm2 2al = 260 or, 2 ⨯ a ⨯ 13 = 260 or, 26a = 260 Find .....

Here, Slant height = l = 10cm Length of side of base = a = 16cm LSA = ? We know, LSA = 2al  = 2⨯ 16cm ⨯ 10cm Calculate .....

Here, LSA = 660cm2 Height = h = 22cm Perimeter of base = P = ? We know, LSA = P⨯h or, 660cm2 = 22cm ⨯ h Find .....

Here, Height of prism = h = ? Area of base = A = 1/2⨯ 6cm ⨯ 8cm = 24cm2 Volume = 960cm3 We know, Volume of prism = A ⨯ h or, 960cm3   = 24cm2 ⨯ h Calculate .....

Here, Height of prism = h = ? Area of base = A = 1/2⨯ 9cm ⨯ 8cm = 36cm2 Volume = 864cm3 We know, Volume of prism = A ⨯ h or, 864cm3   = 36cm2 ⨯ h Calculate .....

Here, Height of prism = h = 15cm Area of base = A = 1/2⨯ 6cm ⨯ 8cm = 24cm2 Volume = ? We know, Volume of prism = A ⨯ h                             = .....

Here, Height of prism = h = ? Area of base = A = 1/2⨯ 3cm ⨯4cm = 6cm2 Volume = 48cm3 We know, Volume of prism = A ⨯ h or, 48cm3   = 6cm2 ⨯ h Calculate .....

Here, BC  = √[102 - 62] = 8cm Height = 30cm. TSA = ? We know,  TSA of prism =  2 Area of base + LSA                        = 2⨯1/2⨯ 6cm⨯8cm + .....

Here, Base of triangular base = √[102 - 82] = 6cm Area of base = A = 1/2 ⨯ 8cm ⨯ 6cn = 24cm2. We know, Volume of prism = Area of base ⨯ height or, h = Volume / A = 480/24 =................Cm. Find .....

Here, Height = h = 45cm BC =  √[202 - 162] = 12cm Area of base = A = 1/2 ⨯ 12cm ⨯ 16cn = 96cm2. We know, TSA of prism = LSA + 2 area of base                   .....

Here, Height = h = 30cm EF =  √[102 - 62] = 8cm Area of base = A = 1/2 ⨯ 6cm ⨯8cn = 24cm2. We know, TSA of prism = LSA + 2 area of base                     .....

Here,  Area of base = a2√3/4 = (6cm)2 √3/4 = 9√3cm2    Height = h = 6√3cm Volume = Area of base ⨯ height               = 9√3cm2 ⨯ .....

Given by the question, T.S.A = 300 π cm2 Diameter (d) = 24 cm so, radius(r) = 12 cm By using formula  T.S.A =  π r (r + l) 300  π =  π ⨯ 12( 12 + l)  300/12 = 12 + l l = 13 .....

Let r be the radius and l be the slant height of cone. Then, r + l = 37cm TSA = 814cm2 h = ? We know, TSA = πr(r + l) or, 22/7 ⨯r ⨯ 37cm = 814cm2. Find r . Then find l from r + l = 37cm. Finally, Height = h .....

Here, Slant height = l = 10m Radius = 7cm Rate of cost = Rs.300/m2 Total cost= ? We know, CSA = πrl Find it. Then,  Total cost = CSA ⨯ Rate of cost Find .....

Ler r be the radius of cone. Then,  C= 2πr = 88cm or, πr = 44cm Slant height = l = 30cm CSA = ? We know, CSA = πrl = 44cm ⨯ 30cm =.....................Find .....

Here, Radius = r = 12cm Height = h = 13cm Volume = ? We know, Volume  = 1/3 ⨯πr2h Now, use values of r & h and .....

Here, Volume of sphere = volume of cylinder = 45πcm3   Height of cylinder = 5cm Radius of cylinder  = ? We know Volume = πr2h or, 45π= π r2 ⨯5 or, r2 = 9 Find .....

Let r be the radius of hemisphere, Then, Volume  = 486π cm3 or, 2/3 ⨯π r3 = 486π  Solve & find r. Finally,  Circumference = 2πr = ....................Find .....

Let r be the radius of hemi-sphere, Then Circumference = 2πr or, 44cm = 2πr Find r. Finally, Volume  = 2/3 ⨯πr3. Use value of r and find .....

Here, Let r be the radius of sphere. Then, Volume  = 4/3⨯πr3 or, 1372π/3= 4/3πr3 or, r3 = 1372/4 Find r. At last, TSA = 4πr2 Use value of r and find .....

Let . r be the radius of sphere, Then,  TSA = 36πcm2   or, 4πr2 = 36πcm2                 [ TSA of sphere =  4πr2 .] or,  .....

Please post question again. Your question is .....

Here, Let r be the radius of base of cylinder. Then, TSA = 2πr(r + h) or, 1320/7 = 2 ⨯22/7 ⨯r( r + 7) or, 30 = r2 + 7r or, r2 + 7r - 30 = 0 Factorize it and find r. Ignore the -ve value of .....

Here, Circumference = C =2πr =  88cm Sum of radius & height = r + h = 24cm TSA = ? We know, TSA = 2πr(r + h) Use value of them from above and find .....

Here, height= radius = r     CSA= 2πrh or, 2772cm2= 2 ⨯22/7⨯ r ⨯ r or,  r2= 2772⨯7/44 Find r. At last, Circumference of base = C=2πr =...................find .....

Here, let the no. of students who passed in Maths only be x. then according to the given condition, nₒ(M) = x nₒ(S) = x +10 nₒ(N) = x + 20 nₒ(M∩S) = x +20 nₒ(M∩N) = x +10 nₒ(S∩N) = x +10 n(S) = 150 n(M∩S∩N) = 50 n(M) .....

n(U) = 54 n(M) = 32 n(M∩W) = 20 n(M∩R) = 23 n(W∩R) = 24 nₒ(W) = 6 nₒ(M) = 4 nₒ(R) = y (say) Let n(M∩W∩R) = x Now, by using Venn-diagram, Now, form above Venn-diagram, From set M 20 - x + x +23 - x + 4 = 32 or,  - x .....

There is conventional rain throughout the year on either side of the equator. There is heavy rainfall in the Amazon basin which is at the equator. Central Chile recieves rainfall in winter, so the climate is mediterranean. Summer is hot and dry .....

Here,  let a, b and c be the number  who like only short and tshirt, pants  and tshirt, shirt and pants respectively According to question, 13 bought only two items so,  a + b + c = 13.......(a) From Venn-diagram, a + b .....