Mathematics MCQ on Trigonometry for JEE and Engineering Exam 2022
MCQ on Trigonometry: To be an expert in JEE Mathematics, it is absolutely necessary to practice and be familiar will all the concepts as well as the questions of different types. This is essential to gain mastery over the subject. We have also often heard the common saying, “Practice Makes a Man Perfect”, hence students have to practice, practice and practice till they master the subject.
In this post we are providing you MCQ on Trigonometry, which will be beneficial for you in upcoming JEE and Engineering Exams.
MCQ on Trigonometry
Q1. A vertical pole consists of two parts, the lower part being one-third of the whole. At a point in the horizontal plane through the base of the pole and distance 20 meters from it, the upper part of the pole subtends an angle whose tangent is 1 / 2. The possible heights of the pole are _________. Solution: [H / 3] cotα = d and d = 150 cotφ = 60m or [H / 3d] = tanα and [H / d] = tanβ tan (β − α) = 1 / 2 = {[H / d] − [H / 3d]} / {1 + [H2 / 3d2]} ⇒ 1 + [H2 / 3d2] = 4H / 3d ⇒ H2 − 4dH + 3d2 = 0 ⇒ H2 − 80H + 3 * (400) = 0 ⇒ H = 20 or 60m
a) 20 cm or 45 cm
b) 25 cm or 60 cm
c) 20 cm or 60 cm
d) none
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Q2. tan [(π / 4) + (1 / 2) * cos−1 (a / b)] + tan [(π / 4) − (1 / 2) cos−1 .(a / b)] = _________. Solution: tan [(π / 4) + (1 / 2) * cos−1 (a / b)] + tan [(π / 4) − (1 / 2) cos−1 .(a / b)] Let (1 / 2) * cos−1 (a / b) = θ ⇒ cos 2θ = a / b Thus, tan [{π / 4} + θ] + tan [{π / 4} − θ] = [(1 + tanθ) / (1 − tanθ)]+ ([1 − tanθ] / [1 + tanθ]) = [(1 + tanθ)2 + (1 − tanθ)2] / [(1 − tan2θ)] = [1 + tan2θ + 2tanθ + 1 + tan2θ − 2tanθ] / [(1 + tan2θ)] = 2 (1 + tan2θ) / [(1 + tan2θ)] = 2 sec2θ = 2 cos2θ = 2 / [a / b] = 2b / a
a) 2b/a
b) b/a
c) a/b
d) 2a/b
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Q3. A balloon is observed simultaneously from three points A, B and C on a straight road directly under it. The angular elevation at B is twice and at C is thrice that of A. If the distance between A and B is 200 metres and the distance between B and C is 100 metres, then the height of balloon is given by _________. Solution: x = h cot3α …..(i) (x + 100) = h cot2α ……(ii) (x + 300) = h cotα ……(iii) From (i) and (ii), −100 = h (cot3α − cot2α), From (ii) and (iii), −200 = h (cot2α − cotα), 100 = h ([sinα] / [sin3α * sin2α]) and 200 = h ([sinα] / [sin2α * sinα]) or sin3α / sinα = 200 / 100 ⇒ sin3α / sinα = 2 ⇒ 3 sinα − 4 sin3α − 2 sinα = 0 ⇒ 4 sin3α − sinα = 0 ⇒ sinα = 0 or sin2α = 1 / 4 = sin2 (π / 6) ⇒ α = π / 6 Hence, h = 200 * sin [π / 3] = 200* [√3 / 2]
a) 200* [√3 / 5]
b) 200* [√5 / 2]
c) 200* [√3 / 2]
d) none
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Q4. The general solution of sin x − 3 sin2x + sin3x = cos x − 3 cos2x + cos3x is _________. Solution: sinx − 3 sin2x + sin3x = cosx − 3 cos2x + cos3x ⇒ 2 sin2x cosx − 3 sin2x − 2 cos2x cosx + 3 cos2x = 0 ⇒ sin2x (2cosx − 3) − cos2x (2 cosx − 3) = 0 ⇒ (sin2x − cos2x) (2 cosx − 3) = 0 ⇒ sin2x = cos2x ⇒ 2x = 2nπ ± (π / 2 − 2x) i.e., x = nπ / 2 + π / 8
a) x = nπ / 2 + π / 4
b) x = nπ / 2 + π / 6
c) x = nπ / 2 + π / 8
d) x = nπ / 4 + π / 8
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Q5. In triangle ABC, if ∠A = 45∘, ∠B = 75∘, then a + c√2 = __________. Solution: ∠C = 180o − 45o − 75o = 60o Therefore, a + c√2 = k (sin A + √2 sinC) = k (1 / √2 + [√3 / 2] * √2) = k ([1 + √3] / [√2]) and k = b / [sin B] ⇒ a + c√2= b / sin75o ([1 + √3] / [√2]) = 2b
a) 4b
b) 2b
c) 6b
d) 11b
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Q6. The number of real solutions of tan−1 √[x (x + 1)] + sin−1 [√x2 + x + 1] = π / 2 is ___________. Solution: tan−1 √[x (x + 1)] + sin−1 [√x2 + x + 1] = π / 2 tan−1 √[x (x + 1)] is defined when x (x + 1) ≥ 0 ..(i) sin−1 [√x2 + x + 1] is defined when 0 ≤ x (x + 1) + 1 ≤ 1 or 0 ≤ x (x + 1) ≤ 0 ..(ii) From (i) and (ii), x (x + 1) = 0 or x = 0 and -1. Hence, the number of solution is 2.
a) -2
b) 2
c) 4
d) none
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Q7. If sec 4θ − sec 2θ = 2, then the general value of θ is __________. Solution: sec 4θ − sec 2θ = 2 ⇒ cos 2θ − cos 4θ = 2 cos 4θ cos 2θ ⇒ −cos 4θ = cos 6θ ⇒ 2 cos 5θ cosθ = 0 ⇒ H = [h cot 15o] / [cot 15o − 1] or nπ/5 + π/10
a) nπ/10+ π/10
b) nπ/5 + π/10
c) nπ/5 + π/5
d) nπ/5 + π/2
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Q8. sec2 (tan−1 2) + cosec2 (cot−1 3) = _________. Solution: Let (tan−1 2) = α ⇒ tan α = 2 and cot−1 3 = β ⇒ cot β = 3 sec2 (tan−1 2) + cosec2 (cot−1 3) = sec2 α + cosec2 α = 1 + tan2α + 1 + cot2α = 2 + (2)2 + (3)2 = 15
a) 10
b) 12
c) 15
d) 14
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Q9. In a triangle, the length of the two larger sides are 10 cm and 9 cm, respectively. If the angles of the triangle are in arithmetic progression, then the length of the third side in cm can be _________. Solution: We know that in a triangle larger the side, larger the angle. Since angles ∠A, ∠B and ∠C are in AP. Hence, ∠B = 60o cosB = [a2 + c2 −b2] / [2ac] ⇒ 1 / 2 = [100 + a2 − 81] / [20a] ⇒ a2 + 19 = 10a ⇒ a2 − 10a + 19 = 0 a = 10 ± (√[100 − 76] / [2]) ⇒ a + c√2 = 5 ± √6
a) ⇒ a + c√2 = 5 ± √8
b) ⇒ a + c√2 = 5 ± √6
c) ⇒ a + c√2 = 5 ± √3
d) ⇒ a + c√2 = 8 ± √6
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Q10. If cos−1 p + cos−1 q + cos−1 r = π then p2 + q2 + r2 + 2pqr = ________. Solution: According to given condition, we put p = q = r = 1 / 2 Then, p2 + q2 + r2 + 2pqr = (1 / 2)2 + (1 / 2)2 + (1 / 2)2 + 2 * [1 / 2] * [1 / 2] * [1 / 2] = [1 / 4] + [1 / 4] + [1 / 4] + [2 / 8] =1
a) 0
b) 1
c) -1
d) none
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