JEE Maths Quiz on Trigonometry
JEE Maths Quiz on Trigonometry : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you maths quiz on Trigonometry.
Maths Quiz on Trigonometry
Q1. 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 sin^{3}α − 2 sinα = 0
⇒ 4 sin^{3}α − sinα = 0
⇒ sinα = 0 or
sin2α = 1 / 4 = sin^{2 }(π / 6)
⇒ α = π / 6
Hence, h = 200 * sin [π / 3]
= 200* [√3 / 2]
Q2. A tower of height b subtends an angle at a point O on the level of the foot of the tower and a distance a from the foot of the tower. If a pole mounted on the tower also subtends an equal angle at O, the height of the pole is ___________.
Solution:
tanα = b / a, tan^{2}α = [2 (b / a)] / [1 − (b / a)^{2}] = [p + b] / a
⇒ 2ba / [a^{2} − b^{2}] = [p + b] / a
⇒ [2ba^{2} − a^{2}b + b^{3}] / [a^{2} − b^{2}] = p
⇒ p = [b * (a^{2} + b^{2})] / [a^{2} − b^{2}]
Q3. A vertical pole consists of two parts, the lower part being onethird 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 _________.
a) 30 cm
b) 60 cm
c) 20 cm
d) 20 or 60m
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 + [H^{2} / 3d^{2}]}
⇒ 1 + [H^{2} / 3d^{2}] = 4H / 3d
⇒ H^{2} − 4dH + 3d2 = 0
⇒ H^{2 }− 80H + 3 * (400) = 0
⇒ H = 20 or 60m
Q4. sec^{2 }(tan^{−1 }2) + cosec^{2 }(cot^{−1} 3) = _________.
a) 15
b) 17
c) 19
d) 21
a) 15
Solution:
Let (tan^{−1 }2) = α
⇒ tan α = 2 and cot^{−1} 3 = β
⇒ cot β = 3 sec^{2 }(tan^{−1 }2) + cosec^{2 }(cot^{−1} 3)
= sec^{2 }α + cosec^{2 }α
= 1 + tan^{2}α + 1 + cot^{2}α
= 2 + (2)^{2} + (3)^{2}
= 15
Q5. The number of real solutions of tan^{−1 }√[x (x + 1)] + sin^{−1 }[√x2 + x + 1] = π / 2 is ___________.
a) 8
b) 6
c) 4
d) 2
d) 2
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.
Q6. tan [(π / 4) + (1 / 2) * cos^{−1 }(a / b)] + tan [(π / 4) − (1 / 2) cos^{−1 .}(a / b)] = _________.
a) 2a / b
b) 2b / a
c) 4b / 2
d) b / 2a
b) 2b / a
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 − tan^{2}θ)]
= [1 + tan^{2}θ + 2tanθ + 1 + tan^{2}θ − 2tanθ] / [(1 + tan^{2}θ)]
= 2 (1 + tan^{2}θ) / [(1 + tan^{2}θ)]
= 2 sec2θ
= 2 cos2θ
= 2 / [a / b]
= 2b / a
Q7. If cos^{−1 }p + cos^{−1} q + cos^{−1} r = π then p^{2} + q^{2} + r^{2} + 2pqr = ________.
a) 2
b) 3
c) 5
d) 1
d) 1
Solution:
According to given condition, we put p = q = r = 1 / 2
Then, p^{2} + q^{2} + r^{2} + 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
Q8. In triangle ABC, if ∠A = 45∘, ∠B = 75∘, then a + c√2 = __________.
a) 2b
b) 4b
c) 6b
d) None of these
a) 2b
Solution:
∠C = 180^{o} − 45^{o} − 75^{o} = 60^{o}
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 / sin75^{o }([1 + √3] / [√2])
= 2b
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 = 60^{o} cosB = [a^{2} + c^{2} −b^{2}] / [2ac]
⇒ 1 / 2 = [100 + a^{2 }− 81] / [20a]
⇒ a^{2} + 19 = 10a
⇒ a^{2 }− 10a + 19 = 0
a = 10 ± (√[100 − 76] / [2])
⇒ a + c√2 = 5 ± √6
Q10. If a, b are different values of x satisfying a cosx + b sinx = c, then tan ([α + β] / 2) = _______.
Solution:
a cosx + b sinx = c ⇒ a {[(1 − tan^{2 }(x / 2)] / [1 + tan^{2 }(x / 2)]} + 2b {[tan (x / 2) / 1 + tan^{2 }(x / 2)} = c
⇒ (a + c) * tan^{2 }[x / 2] − 2b tan [x / 2] + (c − a) = 0
This equation has roots tan [α / 2] and tan [β / 2].
Therefore, tan [α / 2] + tan [β / 2] = 2b / [a + c] and tan [α / 2] * tan [β / 2] = [c − a] / [a + c] Now
tan (α / 2 + β / 2) = {tan [α / 2] + tan [β / 2]} / {1 − tan [α / 2] * tan [β / 2]} = {[2b] / [a + c]} / {1− ([c − a] / [a + c])} = ba
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