Problem 1 :
Simplify the following
Solution :
L.H.S
= (1 + tan2A) / (1+cot2A)
sec2A - tan2A = 1
sec2A = 1 + tan2A
cosec2A - cot2A = 1
cosec2A = 1 + cot2A
= sec2A / cosec2A
= (1/cos2A) / (1/sin2A)
= sin2A/cos2A
= tan2A
Problem 2 :
From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (take √3 = 1.732)
Solution :
In triangle BCD,
tanθ = Opposite side / Adjacent side
tan 30 = BC/CD
1/√3 = 10/CD
CD = 10√3 ------(1)
In triangle ACD,
tan 45 = AC/CD
1 = (10 + x) /CD
CD = 10 + x ------(2)
(1) = (2)
10√3 = 10 + x
x = 10(√3 - 1)
x = 10(10.732 - 1)
x = 10(0.732)
x = 7.32
The height of the flag staff is 7.32 m.
Problem 3 :
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower is 30°. Find the height of the tower.
Solution :
Height of the tower = AB, Distance between the foot of the tower and point of observation = BC.
tanθ = Opposite side / Adjacent side
tan θ = AB/BC
tan 30 = AB/30
1/√3 = AB/30
AB = 30/√3
AB = (30/√3) ⋅ (√3/√3)
AB = 30√3/3
AB = 10√3
Problem 4 :
The coefficient of variation of a series is 69% and its standard deviation is 15. Find the mean.
Solution :
Coefficient of variation = (σ/x̅) ⋅ 100%
69/100 = (15/x̅) ⋅ 100%
0.69 = 15/x̅
x̅ = 15/0.69
x̅ = 21.73
So, the required mean is 21.73.
Problem 5 :
If the two dice are thrown. what is the probability of getting the same number in both the dice.
Solution :
When two dice are thrown, the sample space
n(S) = 36
Let A be the event of getting doublet.
A = { (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6) }
n(A) = 6
p(A) = n(A)/n(S)
p(A) = 6/36
Problem 6 :
2 cubes each of volume 64 cm3 are joined end to end. Determine the surface area of the resulting cuboid.
Solution :
Volume of one cuboid = 64 cm3
a3 = 64
Side length of cube is 4 cm.
Side length of cuboid = 4 + 4 = 8 cm
breadth = 4 cm and height = 4 cm
Surface area of cuboid = 2(lb + bh + hl)
= 2(8⋅4 + 4⋅4 + 4⋅8)
= 2(32 + 16 + 32)
= 2(80)
= 160 cm2
Problem 7 :
A metallic sphere of radius 4.2 cm is melted and recast into a shape of a cylinder of radius 6 cm. Determine the height of the cylinder.
Solution :
Radius of sphere = 4.2 cm
Radius of cylinder = 6
Volume of sphere = Volume of cylinder
(4/3) πr3 = πr2 h
(4/3) 4.23 = 62 h
h = 98.784/36
h = 2.744
So, the height of the cylinder is 2.744 cm.
Problem 8 :
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Determine the curved surface area of the frustum.
Solution :
Slant height (l) = 4 cm
Circumference of larger circle = 18
2πR = 18 R = 18/2π R = 9/π |
2πr = 6 r = 6/2π r = 3/π |
Surface area of the frustum cone = πl (R + r)
= π ⋅ 6(9/π + 3/π)
= 6(9+3)
= 12(6)
= 72
Problem 9 :
Find the HCF of 96 and 404 by the prime factorization method. Find their LCM.
Solution :
= 2 ⋅ 2
= 4
Problem 10 :
The angle in a semi circle is
Solution :
Angle in a semi circle is right angle.
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