Quantitative Aptitude
Trigonometry
435 Questions
Trigonometry Questions
-
$\displaystyle \frac{-2}{5}$
-
$\displaystyle \frac{2}{5}$
-
$\displaystyle \frac{2}{3+2\sqrt{3}}$
-
$-2$
B
Correct answer
Explanation
If sin A = sqrt(3)/2, A = 60 degrees. tan 60 = sqrt(3), cot 60 = 1/sqrt(3), cosec 60 = 2/sqrt(3). The expression is (sqrt(3) - 1/sqrt(3)) / (sqrt(3) + 2/sqrt(3)). Numerator: (3-1)/sqrt(3) = 2/sqrt(3). Denominator: (3+2)/sqrt(3) = 5/sqrt(3). Result: (2/sqrt(3)) / (5/sqrt(3)) = 2/5.
-
$\sqrt { 3 } $
-
$\dfrac { 1 }{ \sqrt { 3 } } $
-
$\dfrac { 2 }{ \sqrt { 3 } } $
-
$\dfrac{1}{3}$
B
Correct answer
Explanation
In a right triangle ABD, if angle B is 60 degrees and angle A is 30 degrees, then angle D must be 90 degrees. The value of cot(60 degrees) is 1/sqrt(3).
-
$1$
-
$\sqrt{2}$
-
$\dfrac{1}{\sqrt{2}}$
-
None of above
B
Correct answer
Explanation
cosec(45) = 1 / sin(45) = 1 / (1/sqrt(2)) = sqrt(2).
-
$\dfrac{1}{10}$
-
$\dfrac{2}{10}$
-
$\dfrac{3}{10}$
-
$\dfrac{4}{10}$
C
Correct answer
Explanation
Given tan theta = 1/sqrt(2). Then cot theta = sqrt(2), cosec^2 theta = 1 + cot^2 theta = 3, sec^2 theta = 1 + tan^2 theta = 1 + 1/2 = 3/2. Expression = (3 - 3/2) / (3 + 2) = (1.5) / 5 = 0.3 = 3/10.
-
$\tan { \theta } =3\tan { B } $
-
$\tan { \theta } =3\tan { C } $
-
$\tan { A } =\cfrac { 6\tan { \theta } }{ \tan ^{ 2 }{ \theta } } $
-
$9\cot ^{ 2 }{ \cfrac { A }{ 2 } } =\tan ^{ 2 }{ \theta }$
A
Correct answer
Explanation
Given DB=DE=EC, let each segment be x. In triangle ADE, using the sine rule or trigonometric ratios, one can derive the relationship between the angles. The correct identity derived from the geometry is tan(theta) = 3 tan(B).
-
$10 (\sqrt3 + 1)$m
-
$10 \sqrt3$m
-
$10 (\sqrt3 - 1)$m
-
$\displaystyle \frac{10}{ \sqrt3}$m
-
$\displaystyle\frac{1}{\sqrt{2}}$
-
0
-
1
-
$\displaystyle\frac{1}{2}$
D
Correct answer
Explanation
cot(40) = tan(50), so the first term is 1. cos(35) = sin(55), so the second term is 1/2 * (1) = 1/2. 1 - 1/2 = 1/2.
-
$A = 32^{\small\circ}$
-
$A = 22^{\small\circ}$
-
$A = 41^{\small\circ}$
-
$A = 16^{\small\circ}$
B
Correct answer
Explanation
Using the identity sec(x) = cosec(90 - x), we set 90 - 4A = A - 20. Solving for A gives 5A = 110, so A = 22 degrees.
-
$\displaystyle 30^{\circ}$
-
$\displaystyle 60^{\circ}$
-
$\displaystyle 50^{\circ}$
-
$\displaystyle 24^{\circ}$
C
Correct answer
Explanation
tan(2A) = cot(A-60) = tan(90 - (A-60)) = tan(150 - A). 2A = 150 - A => 3A = 150 => A = 50 degrees.
-
$\displaystyle 58^{\circ}$
-
$\displaystyle 122^{\circ}$
-
$\displaystyle 32^{\circ}$
-
$\displaystyle 158^{\circ}$
A
Correct answer
Explanation
tan(32) * cot(90 - theta) = 1. Since cot(90 - theta) = tan(theta), we have tan(32) * tan(theta) = 1. This implies tan(theta) = 1/tan(32) = cot(32) = tan(90 - 32) = tan(58). Thus, theta = 58 degrees.
B
Correct answer
Explanation
cot(50) = tan(90 - 50) = tan(40). Therefore, tan(40) / tan(40) = 1.
A
Correct answer
Explanation
sin 18 = (sqrt(5)-1)/4. cos 54 = sin 36. cos 36 = sin 54. The expression simplifies: sin 18 * (1/sin 54) * (1/sin 36) * (1/cos 72) * cos 54 * cos 36. Since cos 72 = sin 18, the terms cancel out to 1.
-
$\sin \theta$
-
$\tan \theta$
-
$\cot \theta$
-
None of above
A
Correct answer
Explanation
In a right triangle, the sum of angles is 180 degrees. Since angle B is 90, angle A + angle C = 90. Therefore, angle C = 90 - A = 90 - theta. By definition, cos(C) = sin(A) = sin(theta).
-
$\sec 21^{\circ} + \tan 21^{\circ}$
-
$\sin 21^{\circ} + \cot 21^{\circ}$
-
$\sin 21^{\circ} + \cos 21^{\circ}$
-
$\sec 21^{\circ} + \cot 21^{\circ}$
A
Correct answer
Explanation
Using co-function identities, cosec(69) = sec(90-69) = sec(21) and cot(69) = tan(90-69) = tan(21). Thus, cosec(69) + cot(69) = sec(21) + tan(21).
-
$\cot 22^{\circ} +\text{cosec } 22^{\circ}$
-
$\sin 22^{\circ} + \cos 22^{\circ}$
-
$\cos 22^{\circ} + \tan 22^{\circ}$
-
None of these