Quantitative Aptitude

Trigonometry

435 Questions
All Height and Distance 268 Triangle Properties and Rules 50 Standard Angle Evaluation 33 Complementary and Odd-Even Angles 30 Value of one ratio from another 26 Trigonometric Identities 14 Maxima and Minima 6 General and Principal Solutions 5 Quadratic Substitution 3

Trigonometry Questions

Multiple choice

Two poles standing on a horizontal ground are of heights $5m$ and $10m$ respectively. The line joining their tops makes an angle of ${15}^{o}$ with ground. Then the distance (in m) between the poles, is:
  1. $\cfrac { 5 }{ 2 } \left( 2+\sqrt { 3 } \right) $
  2. $5\left( \sqrt { 3 } +1 \right) $
  3. $5\left( 2+\sqrt { 3 } \right) $
  4. $10\left( \sqrt { 3 } -1 \right) $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the poles be AB=5 and CD=10. The distance between them is x. The line joining the tops makes an angle of 15 degrees with the ground. This means the angle of elevation of the top of the taller pole from the top of the shorter pole is 15 degrees. tan(15) = (10-5)/x = 5/x. x = 5/tan(15). tan(15) = 2 - sqrt(3). x = 5 / (2 - sqrt(3)) = 5(2 + sqrt(3)).

Multiple choice

If an angle A of a $\Delta$ABC satisfies $5\cos A+3=0$, then the roots of the quadratic equation, $9x^2+27x+20=0$ are.
  1. $\sin A, \sec A$
  2. $\sec A, \tan A$
  3. $\tan A, \cos A$
  4. $\sec A, \cot A$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given 5cos A + 3 = 0, cos A = -3/5. In a triangle, this implies A is in the second quadrant. Thus, sec A = -5/3 and tan A = -4/3. The roots of 9x^2 + 27x + 20 = 0 are found using the quadratic formula: x = (-27 +/- sqrt(729 - 720)) / 18 = (-27 +/- 3) / 18. Roots are -24/18 = -4/3 and -30/18 = -5/3. These match tan A and sec A.

Multiple choice

The angle of elevation of a stationary cloud from a point $2500$ m above a lake is $15^0$ and from the same point the angle of depression of its reflection in the lake is $45^o$. The height (in meters) of the cloud above the lake, given that $\cot\,15^o=2+\sqrt{3}$, is
  1. $2500$
  2. $2500\sqrt{2}$
  3. $2500\sqrt{3}$
  4. $5000$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let h be the height of the cloud. Elevation from 2500m is 15 deg. tan(15) = (h-2500)/d. Depression of reflection is 45 deg. The reflection is at -h. tan(45) = (h+2500)/d. Since tan(45)=1, d = h+2500. Substitute into first: (h-2500)/(h+2500) = tan(15) = 2-sqrt(3). Solve for h: h-2500 = (2-sqrt(3))(h+2500). h-2500 = 2h + 5000 - sqrt(3)h - 2500sqrt(3). h(sqrt(3)-1) = 7500 - 2500sqrt(3) = 2500(3-sqrt(3)) = 2500sqrt(3)(sqrt(3)-1). h = 2500sqrt(3).

Multiple choice

If $A,B,C$ are the angles of $\triangle ABC$ then $\cot { A } .\cot { B } +\cot { B } .\cot { C } +\cot { C } .\cot { A } $ is equal to
  1. $0$
  2. $1$
  3. $2$
  4. $-1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In any triangle ABC, the sum of the angles is A + B + C = 180 degrees. Using the trigonometric identity for the sum of three angles, we have tan A + tan B + tan C = tan A * tan B * tan C. Dividing this entire equation by the product tan A * tan B * tan C yields the identity cot B * cot C + cot A * cot C + cot A * cot B = 1.

Multiple choice

The angle of elevation of the top of a tower from the top of a house is ${ 60 }^{ o }$ and the angle of depression of its base is ${ 30 }^{ o }$. If the horizontal distance between the house and the tower be $12 m$, then the height of the tower is
  1. $48\sqrt { 3 } m$
  2. $16\sqrt { 3 } m$
  3. $24\sqrt { 3 } m$
  4. $\dfrac { 16 }{ \sqrt { 3 } } m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let h be the height of the house. Height of tower = h + x. tan(60) = x / 12, so x = 12 * sqrt(3). tan(30) = h / 12, so h = 12 / sqrt(3) = 4 * sqrt(3). Total height = 12 * sqrt(3) + 4 * sqrt(3) = 16 * sqrt(3).

Multiple choice

A tower $AB$ leans towards west marking an angle $\alpha$ with the vertical. The angular elevation of $B$, the top most point of the tower is $\beta$ as observed from a point $C$ due east of $A$ at a distance $'d'$ from $A$. If the angular elevation of $B$ from a point $D$ due east of $C$ at a distance $2d$ from $C$ is $r$, then $2\tan \alpha$ can be given as
  1. $3\cot \beta - 2\cot \gamma$
  2. $3\cot \gamma - 2\cot \beta$
  3. $3\cot \beta - \cot \gamma$
  4. $\cot \beta - 3\cot \gamma$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the tower height be h and its westward displacement be h tan(alpha). From the two observation points, cot(beta) = d/h + tan(alpha) and cot(gamma) = 3d/h + tan(alpha). Eliminating d/h gives 2 tan(alpha) = 3 cot(beta) - cot(gamma), matching Option C.

Multiple choice

From a light house the angles of depression of two ships on opposite sides of the light house are observed to be ${ 30 }^{ \circ }$ and ${ 45 }^{ \circ }$. If the height of the light house is $h$ metre, the distance between the ships is
  1. $-\dfrac{{\left( {\sqrt 3  + 1} \right)h}}{{\sqrt 3 }}$ meter
  2. $\left( \sqrt { 3 } -1 \right) h$ meters
  3. $\sqrt { 3 } h$ meters
  4. $1+\left( 1+\dfrac { 1 }{ \sqrt { 3 } } \right) h$ meters
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

A railway line is banked with an angle of 0.01 radians. The height of the outer rail over inner rail, if the distance between the two rail of 1.5 m, will be

  1. 0.025 m

  2. 0.035 m

  3. 0.015 m

  4. 0.045 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The height of the outer rail over the inner rail is given by h = d * sin(theta), where d is the distance between rails and theta is the angle. For small angles, sin(theta) is approximately theta. h = 1.5 * 0.01 = 0.015 m.

Multiple choice

The bakery of a small Indiana town has just made a huge strawberry shortcake for the opening of the summer festival. The cake has a radius of 0.5 m. A ladybug lands on the cake and walks around the edge for a distance of 80 cm. What is the angular displacement of the ladybug?

  1. 1.6 radians

  2. 16 degrees

  3. 0.16 radians

  4. 1.6 degrees
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Angular displacement theta = arc length s / radius r. s = 80 cm = 0.8 m. r = 0.5 m. theta = 0.8 / 0.5 = 1.6 radians.

Multiple choice

The value of $\displaystyle \sin 0^{\circ}+\cos 30^{\circ}-\tan 45^{\circ}+$ $cosec 60^{\circ}+\cot 90^{\circ}$ is equal to
  1. $\displaystyle \dfrac{5\sqrt{3}-6}{6}$
  2. $\displaystyle \dfrac{-6+7\sqrt{3}}{6}$
  3. $0$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

sin 0 = 0, cos 30 = sqrt(3)/2, tan 45 = 1, cosec 60 = 2/sqrt(3), cot 90 = 0. Sum = 0 + sqrt(3)/2 - 1 + 2/sqrt(3) + 0 = (3/2sqrt(3) + 4/2sqrt(3)) - 1 = 7/2sqrt(3) - 1 = (7sqrt(3)/6) - 1 = (7sqrt(3)-6)/6.

Multiple choice

Find the value of $\dfrac{sin (-660^o) tan (1050^o) sec (-420^o)}{cos (225^o ) cosec (315^o) cos(510^o)}$
  1. $\dfrac{\sqrt 3}{4}$
  2. $\dfrac{\sqrt 3}{2}$
  3. $\dfrac{2}{\sqrt 3}$
  4. $\dfrac{4}{\sqrt 3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Simplify each term: sin(-660) = sin(60) = sqrt(3)/2. tan(1050) = tan(330) = -1/sqrt(3). sec(-420) = sec(60) = 2. cos(225) = -1/sqrt(2). cosec(315) = -sqrt(2). cos(510) = cos(150) = -sqrt(3)/2. Numerator: (sqrt(3)/2) * (-1/sqrt(3)) * 2 = -1. Denominator: (-1/sqrt(2)) * (-sqrt(2)) * (-sqrt(3)/2) = -sqrt(3)/2. Result: -1 / (-sqrt(3)/2) = 2/sqrt(3).

Multiple choice

Find the angle $\theta $ whose cosine is equal to its tangent
  1. $\theta =\sin^{-1}\left ( 2\sin 18^{0} \right )$
  2. $\theta =\sin^{-1}\left ( 2\sin 36^{0} \right )$
  3. $\theta =\sin^{-1}\left ( 3\sin 18^{0} \right )$
  4. $\theta =\sin^{-1}\left ( 2\sin 54^{0} \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We need cos(theta) = tan(theta) = sin(theta)/cos(theta), so cos^2(theta) = sin(theta). This is 1 - sin^2(theta) = sin(theta), or sin^2(theta) + sin(theta) - 1 = 0. Using the quadratic formula, sin(theta) = (-1 + sqrt(5))/2. This value is equal to 2 * sin(18 degrees).