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

The horizontal distance between two stations P and Q is 100 m. The vertical angles from P and Q to the top of a vertical tower at T are 30 and 50 above horizontal, respectively. The vertical angles from P and Q to the base of the tower are 0.10 and 0.50 below horizontal, respectively. Station P, Q and the tower are in the same vertical plane with P and Q being on the same side of T. Neglecting earth's curvature and atmospheric refraction, the height (in m) of the tower is

  1. 6.972

  2. 12.387

  3. 12.540

  4. 128.745
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let h be the height of the tower. Using trigonometry, the height of the tower top above the horizontal plane of P and Q is H_top = d1 * tan(3) = d2 * tan(5). The base is H_base = d1 * tan(0.1) = d2 * tan(0.5). With d2 = d1 + 100, solving these simultaneous equations for the height difference gives the tower height.

Multiple choice

Two poles of equal heights are standing opposite to each other on either side of a 100 m wide road. From a point between them on the road, the angles of elevation of the top of the poles are 60o and 30o, respectively. Find the height of the poles and the distance of the point from the nearer of the two poles.

  1. 25 m and 43.3 m, respectively

  2. 43.3 m and 25 m, respectively

  3. 25 m and 25 m, respectively

  4. 43.3 m and 43.3 m, respectively
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let AC be the road and poles AB and CD are standing opposite to each other.

AB = CD  = x m Let    PC  =   y m   AP  = (100 - y) m,                  Angle  CPD  = 60° and angle APB = 30° In $\triangle$DCP, DC/PC = tan 60° x/y = tan 60° = $\sqrt 3$ x = $\sqrt 3$y.........(1) In $\triangle$ABP, AB/AP = tan 30° x/(100-y) = 1/$\sqrt 3$ $\sqrt 3$x = 100-y ...... (2) Substituting the value of x from (1) into (2), we get$\sqrt 3$ x $\sqrt 3$y = 100 - y 3y = 100 - y y =25. Putting the value of y in (1), we get x = $\sqrt 3$ x 25 = 25 (1.73) = 43.3 m The height of pole is 43.3 m and distance of point P from the nearer of the 2 poles is 25 m.  

Multiple choice maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

One angle of a triangle is $\displaystyle \frac{2x}{3}$ grades another is $\displaystyle \frac{3x}{2}$ degrees, whilst the third is $\displaystyle \frac{2\pi x}{75}$ radians ; express them all in degrees.

  1. ${ 55 }^{ o },\quad { 28 }^{ o }\quad \& \quad { 97 }^{ o }\\$
  2. ${ 65 }^{ o },\quad { 22 }^{ o }\quad \& \quad { 93 }^{ o }\\$
  3. $\\{ 60 }^{ o },\quad { 24 }^{ o }\quad \& \quad { 96 }^{ o }\\$
  4. ${ 70 }^{ o },\quad { 15 }^{ o }\quad \& \quad { 95 }^{ o }\\$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Convert all angles to degrees: (2x/3) grades = (2x/3) * (9/10) = 0.6x degrees. (3x/2) degrees = 1.5x degrees. (2*pi*x/75) radians = (2*pi*x/75) * (180/pi) = 4.8x degrees. Sum = 0.6x + 1.5x + 4.8x = 6.9x = 180. x = 180/6.9 = 26.08. Checking the options, C gives 60, 24, 96 which sum to 180.

Multiple choice maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

The angles of elevation of the top of the tower from two points at distances '$a$' and '$b$' from the base and in the same straight line with it complementary The height of the tower is

  1. $a + b$
  2. $\displaystyle \sqrt{ab}$
  3. $\displaystyle a\times b$
  4. $\displaystyle a\sqrt{b}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

AP=a,AQ=b
$\displaystyle \tan \theta =\frac{h}{a}$....(i)
$\displaystyle \tan \left ( 90-\theta  \right )=\frac{h}{b}$....(ii)
$\displaystyle \Rightarrow \cot \theta =\frac{h}{b}$
$\displaystyle \Rightarrow \tan \theta \times \cot \theta = \frac{h}{a}\times \frac{h}{b}=1$
$\displaystyle \Rightarrow h^{2}=ab$
or $\displaystyle h=\sqrt{ab}$

Multiple choice maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

Find the exact value of cosec $\theta$ trigonometric functions for the angle formed when the terminal side passes through $(3, 4).$

  1. $\dfrac{4}{5}$
  2. $\dfrac{3}{5}$
  3. $\dfrac{5}{3}$
  4. $\dfrac{5}{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

By using Pythagoras theorem, we will find the value of hypotenuse.
$3^{2}+4^{2}=c^{2}$
$9 + 16 = c^{2}$
$c = 5$
So, $\csc \theta$ = $\dfrac{hypotenuse}{opposite \space\ side}$

$\csc \theta$ = $\dfrac{5}{3}$

So, option C is correct.

Multiple choice maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

Find cot $\theta$, if $\theta$ be an angle in standard position with (x, y) a point on the terminal side of $\theta$ and r = $\sqrt{x^{2}+y^{2}}\neq 0$.

  1. $\dfrac{y}{r}$
  2. $\dfrac{x}{y}$
  3. $\dfrac{x}{r}$
  4. $\dfrac{r}{x}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By using Pythagoras theorem, we will find the value of hypotenuse.
$x^{2}+y^{2}=r^{2}$
So, cot $\theta$ = $\dfrac{adjacent\space\ side}{opposite\space\ side}$= $\dfrac{x}{y}$


cot $\theta$ = $\dfrac{x}{y}$

So, option B is correct.

Multiple choice maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

Find cosec $\theta$, if $\theta$ be an angle in standard position with (x, y) a point on the terminal side of $\theta$ and r = $\sqrt{x^{2}+y^{2}}\neq 0$.

  1. $\dfrac{y}{r}$
  2. $\dfrac{r}{y}$
  3. $\dfrac{x}{r}$
  4. $\dfrac{r}{x}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By using Pythagoras theorem, we will find the value of hypotenuse.
$x^{2}+y^{2}=r^{2}$
So, cosec $\theta$ = $\dfrac{hypotenuse}{opposite\space\ side }$= $\dfrac{r}{y}$

cosec $\theta$ = $\dfrac{r}{y}$

So, option B is correct.

Multiple choice trigonometric equations trigonometric functions trigonometry maths

If $\alpha$ is the angle of first quadrant such that $co\sec ^{ 4 }{ \alpha  }=17+\cot ^{ 4 }{ \alpha  } $, then what is the value of $\sin{\alpha}$?

  1. $\cfrac{1}{3}$
  2. $\cfrac{1}{4}$
  3. $\cfrac{1}{9}$
  4. $\cfrac{1}{16}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$ cosec^{4}\alpha -cot^{4}\alpha = 17$

 (As $ cosec^{2}\alpha -cot^{2}\alpha=1) $

$ \Rightarrow (cosec^{2}\alpha -cot^{2}\alpha )(cosec^{2}\alpha +cot^{2}\alpha ) = 17 $ 

$ \Rightarrow cosec^{2}\alpha +cot^{2}\alpha = 17...(1) $

$ cosec^{2}\alpha -cot^{2}\alpha = 1...(2) $

then $ (1) + (2) \Rightarrow 2cosec^{2}\alpha = 18 $

$ \Rightarrow sin^{2}\alpha = \dfrac{1}{9}\Rightarrow \boxed{sin\,\alpha = \dfrac{1}{3}} $ $ \left ( \because \alpha \,in\,1st\,quadrant \right ) $ 
Multiple choice trigonometric equations trigonometric functions trigonometry maths

The cosine of the obtuse angle formed by the medians from the vertices of the acute angles of an isosceles right angled triangle is

  1. $- 2 / 3$
  2. $- 4 / 5$
  3. $- 3 / 5$
  4. $- 3 / 4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In an isosceles right triangle, placing vertices at (0,0), (a,0), and (0,a), the medians from the acute angles are calculated. The cosine of the angle between them is -2/3.

Multiple choice trigonometric equations trigonometric functions trigonometry maths

In an isosceles $\triangle ABC$, if the altitudes intersect on the inscribed circle then cosine of the vertical angle $'A'$ is :

  1. $\cfrac{1}{9}$
  2. $\cfrac{1}{3}$
  3. $\cfrac{2}{3}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If the altitudes of an isosceles triangle intersect on the inscribed circle, the geometry dictates that cos(A) = 1/9.

Multiple choice mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

Find the radian measure corresponding to the degree $-47^{o}30'$

  1. $\dfrac {-19\ \pi}{72}rad$
  2. $\dfrac {19\ \pi}{72}rad$
  3. $\dfrac {13\ \pi}{72}rad$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$-47^{o} 30'$
$\Rightarrow - (47+ \dfrac{30}{60}) (\because 1^{o} =60')$
$\Rightarrow  -\left( 47+ \dfrac{1}{2} \right)$
$-\left( \dfrac{95}{2} \right)$
Radian measure $\Rightarrow \dfrac{\pi}{180} \times \dfrac{-95}{2}$
$\Rightarrow \pi x - \dfrac{19}{72} \Rightarrow - \dfrac{19 \pi}{72}$ radian
Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

In $\Delta ABC$, if $\angle A+\angle B=90^{\circ}$, cot $B=\dfrac{3}{4}$, then the value of tan A is :

  1. $\dfrac{4}{5}$
  2. $\dfrac{3}{4}$
  3. $\dfrac{4}{3}$
  4. $\dfrac{3}{5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\angle A+\angle B={ 90 }^{ \circ  }\ \angle B={ 90 }^{ \circ  }-\angle A$

$\cot { B } =\dfrac { 3 }{ 4 } \ \cot { \left( { 90 }^{ \circ  }-\angle A \right)  } =\dfrac { 3 }{ 4 } \ \tan { A } =\dfrac { 3 }{ 4 } $

Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

If a straight line in space is equally inclined to the co-ordinate axes, the cosine of its angle of inclination to any of the axes is 

  1. $\dfrac{1}{3}$
  2. $\dfrac{1}{2}$
  3. $\dfrac{1}{\sqrt{3}}$
  4. $\dfrac{1}{\sqrt{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If a line makes equal angles alpha with all three coordinate axes, then the direction cosines satisfy cos^2(alpha) + cos^2(alpha) + cos^2(alpha) = 1. This simplifies to 3 cos^2(alpha) = 1, meaning cos(alpha) = 1 / sqrt(3), so option C is correct.

Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

Find the value of, $\dfrac {4}{3}\cot^{2}30^{o}+\cot^{2}60^{o}-2\csc ^{2}60^{o}-\dfrac {3}{4}\tan^{2}30^{o}$

  1. $10/3$
  2. $11/3$
  3. $4$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac { 4 }{ 3 } { \cot }^{ 2 }30+{ \cot }^{ 2 }60-2{ csc }^{ 2 }60-\dfrac { 3 }{ 4 } { \tan }^{ 2 }30$

$\Rightarrow \dfrac { 4 }{ 3 } { \left( \sqrt { 3 }  \right)  }^{ 2 }+{ \left( \dfrac { 1 }{ \sqrt { 3 }  }  \right)  }^{ 2 }-2\times { \left( \dfrac { 2 }{ \sqrt { 3 }  }  \right)  }^{ 2 }-\dfrac { 3 }{ 4 } \times { \left( \dfrac { 1 }{ \sqrt { 3 }  }  \right)  }^{ 2 }$
$\Rightarrow 4+\dfrac { 1 }{ 3 } -\dfrac { 8 }{ 3 } -\dfrac { 1 }{ 4 } $
$\Rightarrow \dfrac { 15 }{ 4 } -\dfrac { 7 }{ 3 } $
$=\dfrac { 17 }{ 12 } $
None of these.

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

The sine of angle formed by the lateral face ADC and plane of the base ABC of the tetrahedron ABCD where $\displaystyle a\equiv (3, -2, 1); B\equiv (3, 1, 5); C\equiv (4, 0, 3)and D\equiv (1, 0, 0)is$

  1. $\displaystyle \frac{2}{\sqrt{29}}$
  2. $\displaystyle \frac{5}{\sqrt{29}}$
  3. $\displaystyle \frac{3\sqrt3}{\sqrt{29}}$
  4. $\displaystyle \frac{-2}{\sqrt{29}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\overrightarrow { AD } =-2\hat { i } +2\hat { j } -\hat { k } ,\overrightarrow { Ac } =\hat { i } +2\hat { j } +2\hat { k } ,\overrightarrow { AB } =3\hat { j } +4\hat { k } : \ \overrightarrow { n _{ 1 } } =\overrightarrow { AD } \times \overrightarrow { AC } =\begin{vmatrix} \hat { i }  & \hat { j }  & \hat { k }  \ -2 & 2 & -1 \ 1 & 2 & 2 \end{vmatrix}=6\hat { i } +3\hat { j } -6\hat { k } =3\left( 2\hat { i } +\hat { j } -2\hat { k }  \right) \ \overrightarrow { n _{ 2 } } =\overrightarrow { AC } \times \overrightarrow { AB } =\begin{vmatrix} \hat { i }  & \hat { j }  & \hat { k }  \ 1 & 2 & 2 \ 0 & 3 & 4 \end{vmatrix}=2\hat { i } -4\hat { j } +3\hat { k } : \ \left| \overrightarrow { n _{ 1 } } \times \overrightarrow { n _{ 2 } }  \right| =3\begin{vmatrix} \hat { i }  & \hat { j }  & \hat { k }  \ 2 & 1 & -2 \ 2 & -4 & 3 \end{vmatrix}=3\left( 5\hat { i } -10\hat { j } -10\hat { k }  \right) \ \sin  \theta =\dfrac { 5 }{ \sqrt { 29 }  } \left( \because \sin  \theta =\dfrac { \left| \overrightarrow { n _{ 1 } } \times \overrightarrow { n _{ 2 } }  \right|  }{ \left| \overrightarrow { n _{ 1 } }  \right| \left| \overrightarrow { n _{ 2 } }  \right|  }  \right) $