TS Polycet (Polytechnic) 2019 Previous Question Paper with Answers And Model Papers With Complete Analysis

41)Total surface area of a hemisphere is

A) 2π r2

B) 3π r2

C) 4π r2

D) π r2

View Answer

B) 3π r2
Explanation:Total surface area = 3π r2

42)If the radius of base of a cylinder is doubled and the height remains unchanged, then its curved surface area is

A) double

B) three times

C) four times

D) no change

View Answer

A) double
Explanation:r = 2r, h = h
Curved surface area = 2πrh
= 2π (2r) h
= 2(2πrh)

43)If a right-angled triangle is revolved about its hypotenuse, then it will form a

A) Sphere

B) Cube

C) Cone

D) Cylinder

View Answer

C) Cone
Explanation:Cone

44)tan 85° tan 65° tan 45° tan 25° tan 5° =

A) 1

B) 0

C) -1

D) None of these

View Answer

A) 1
Explanation:tan 85°.tan 65°.tan 45°.tan (90 – 65°).tan (90-85°)
= tan 85°.tan 65°.l.cot 65°.cot 85° = 1

45)If sin 18° = cos x, then x =?

A) 73°

B) 37°

C) 72°

D) 84°

View Answer

C) 72°
Explanation:sin 18° = cos x
sin 18° = sin (90 -x)
90 -x = 18° => 90 – 18 = x = 72°

46)\sqrt{(sec\theta-1)(sec\theta+1)} =

A) sec θ

B) tan θ

C) sin θ

D) cot θ

View Answer

B) tan θ
Explanation:\sqrt{sec^2\theta-1}= \sqrt{tan^2\theta-1}= tanθ

47)If ∠A = 60°, then 3sin3A – 4sinA =

A) \frac{-7\sqrt3}8

B) \frac{\sqrt3}8

C) \frac{-7}8

D) \frac{7\sqrt3}8

View Answer

A) \frac{-7\sqrt3}8
Explanation:3(sin 60°)3 – 4sin 60° = 3\;x\left(\frac{\sqrt3}2\right)3\;-\;4\;x\;\frac{\sqrt3}2
= \frac{3.3.\sqrt3}8\;-\;2\sqrt3 = \frac{9\sqrt3\;-\;16\;\sqrt3}8
= \frac{-7\sqrt3}8

48)If a cos θ + b sin θ = p; a sin θ- b cos θ = q, then which of the following conditions is true?

A) a2 + b2 = p2 + q2

B) a2 + b2 = P2-q2

C) a2-b2 = p2 + q2

D) a2-b2=p2-q2

View Answer

A) a2 + b2 = p2 + q2
Explanation:a2cos2θ + b2sin2θ + 2ab sinθ cosθ = p2
a2sin2θ + b2cos2θ – 2ab sinθ cosθ = q2
a2 + b2 = p2 + q2

49)If tan α = 1/2., tan β = 1/3, then (α+β)=

A) 30°

B) 60°

C) 45°

D) 0°

View Answer

C) 45°
Explanation:tanα = 1/2., tan β = 1/3,
Tan (α+β) = \frac{\tan\alpha\;+\tan\;\beta}{1-\tan\alpha.\tan\;\beta}
= \frac{\frac12+\frac13}{1-\;\frac16}=\frac{\frac{3+2}6}{\frac56} = 1
∴ (α+β) = 45°

50)If the angle of elevation of the sun is 60°, then the ratio of the height of a tree with its shadow is

A) 1 : 1

B) 1 : √3

C) √3 : 1

D) None of these

View Answer

C) √3 : 1
Explanation:tan 60°= h/x
⇒ √3/1 = h/x ⇒ h : x = √3 : 1

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