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

A flag staff of height $'h'$ stands in the centre of a rectangular field and subtends angle of $15^{o}$ and $45^{o}$ at the mid point of its sides. Find the distance between them
  1. $h\sqrt{2+\sqrt{3}}$
  2. $2h\sqrt{2+\sqrt{3}}$
  3. $4h\sqrt{2+\sqrt{3}}$
  4. $3h\sqrt{2+\sqrt{3}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the distance from the center to the midpoints of the sides be x and y. Then h/x = tan(15) and h/y = tan(45). The distance between the midpoints is sqrt(x^2 + y^2). Using trigonometric identities for tan(15) = 2 - sqrt(3), the calculation leads to the result.

Multiple choice

From a window (h metres high above the ground) of a house in a street, the angles of elevation and depression of the top and foot of another house on the opposite side of the street are $\alpha$ and $\beta$ respectively. Is this possible where the height of the opposite house is : $h(1+\tan\,\alpha\,\cot \beta )$ metres.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using trigonometry, the height of the opposite house is h + h*tan(alpha)*cot(beta). The expression h(1 + tan(alpha)*cot(beta)) is equivalent.

Multiple choice

A person standing on the ground observes the angle of elevation of the top of a water to be ${ 30 }^{ \circ }$. On walking a distance one unit in a certain direction he finds the elevation of the top to be the same as before. He then walks a distance $\quad \dfrac { 5 }{ 3 } $unit at right to his former direction, and finds that the elevation of the top has doubled. The height of the tower can be.
  1. $\sqrt { \dfrac { 5 }{ 16 } } $
  2. $\sqrt { \dfrac { 85 }{ 48 } } $
  3. $\sqrt { \dfrac { 6 }{ 5 } } $
  4. $\sqrt { \dfrac { 48 }{ 85 } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

A circus artist is climbing from the ground along a rope stretched from the top of a vertical pole and tied to a peg at the ground. The height of the pole is 12 m and the angle made by the rope with the ground level is $30^{\circ}$ .Calculate the distance covered by the artist in climbing to the top of the pole.

  1. 12

  2. 13

  3. 24

  4. 26
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The rope forms the hypotenuse of a right triangle where the pole is the opposite side to the 30 degree angle. Using sin(30) = height / hypotenuse, we get 0.5 = 12 / hypotenuse, so the hypotenuse is 24 m.

Multiple choice

A tree is broken by the wind. Its top struck the ground at an angle ${30}^{o}$ at a distance of $30m$ from its foot. The whole height of the tree is:
  1. $10\sqrt 3m$
  2. $20\sqrt 3m$
  3. $40\sqrt 3m$
  4. $30\sqrt 3m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the tree height be H = x + y, where x is the standing part and y is the broken part. tan(30) = x/30 => x = 30 * (1/sqrt(3)) = 10*sqrt(3). cos(30) = 30/y => y = 30 / (sqrt(3)/2) = 60/sqrt(3) = 20*sqrt(3). Total height = 10*sqrt(3) + 20*sqrt(3) = 30*sqrt(3).

Multiple choice

An aeroplane is observed at the same time by two anti-aircraft batteries distant $6000m$ apart to be at elevations of ${30}^{o}$ and ${45}^{o}$ respectively. Assuming that the aeroplane is travelling directly towards the two batteries, find its height and its horizontal distance from the nearer battery.
  1. $3000(\sqrt 3+1)m$; $3000(\sqrt 3+1)m$
  2. $4000(\sqrt 2+1)m$; $3000(\sqrt 3+1)m$
  3. $3000(\sqrt 3+1)m$; $4000(\sqrt 2+1)m$
  4. $4000(\sqrt 2+1)m$; $4000(\sqrt 2+1)m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

From the top of a cliff $200$ metres high, the angles of depression of the top and bottom of a tower are observed to be ${30}^{o}$ and ${60}^{o}$. Find the height of the tower and calculate the distance between them.
  1. Height:$=156$; Distance$=119.7m$
  2. Height$=133\cfrac{1}{3}$; Distance$=115.46m$
  3. Height$=220$; Distance$=112.76m$

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

Cliff height = 200. Let tower height be h and distance be d. From top of cliff to tower top: tan(30) = (200 - h) / d. From top of cliff to tower bottom: tan(60) = 200 / d. d = 200 / sqrt(3) = 115.47m. 200 - h = d * tan(30) = (200 / sqrt(3)) * (1 / sqrt(3)) = 200 / 3 = 66.67. h = 200 - 66.67 = 133.33m.

Multiple choice

The height of a tower is half the height of the flagstaff at its top. The angle of elevation of the top of the tower as seen from a distance of $10$ metres from its foot is ${30}^{o}$. Find the angle of elevation of the top of the flagstaff from the same point.
  1. Angle of elevation is ${35}^{o}$
  2. Angle of elevation is ${52}^{o}$
  3. Angle of elevation is ${55}^{o}$
  4. Angle of elevation is ${60}^{o}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let tower height be h, flagstaff height be 2h. From 10m away, tan(30) = h/10, so h = 10 * tan(30) = 10 / sqrt(3). The flagstaff is on top, so total height is 3h. Tan(theta) = 3h / 10 = 3 * (10 / sqrt(3)) / 10 = 3 / sqrt(3) = sqrt(3). Theta = arctan(sqrt(3)) = 60 degrees.