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 angle of elevation of the top of a tower standing on a horizontal plane from point A is $\alpha$. After walking a distance d towards the foot of the tower, the angle of elevation is found to be $\beta$. The height of the tower is
  1. $\dfrac{d \,sin\, \alpha\, sin\, \beta}{sin\, (\beta - \alpha)}$
  2. $\dfrac{d \,sin\, (\beta - \alpha)}{sin\, \alpha\,sin\, \beta}$
  3. $\dfrac{d \,sin\,\alpha\, sin\, \beta}{sin\, (\alpha\,- \beta)}$
  4. $\dfrac{d\, sin\, (\alpha\, - \beta)}{sin\, \alpha\,sin \, \beta}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let h be the height. From the geometry, h = x * tan(alpha) and h = (x-d) * tan(beta). Solving for x and then h gives h = (d * tan(alpha) * tan(beta)) / (tan(alpha) - tan(beta)). Using sine and cosine identities, this simplifies to (d * sin(alpha) * sin(beta)) / sin(beta - alpha).

Multiple choice

True or False DE is a tower standing on a horizontal plane and ABCD is a straight line in the plane. The height of the tower substends an angle $\theta$ at A, 2$\theta$ at B and 3$\theta$ at C. If AB and BC be respectively 50 metres and 20 metres then the height of the tower and the distance CD are $\dfrac{25}{2} \sqrt{7} m$ and $17.5$ m.

  1. True

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

This is a classic trigonometry problem. Using the cotangent rule for the angles theta, 2theta, and 3theta at distances 50, 20, and x, the derived height and distance values match the provided statement.

Multiple choice

A man observes a tower AB of height h from a point P on the ground. He moves a distanced towards the foot of the tower and finds that the angle of elevation is doubled. He further moves a distance $3d/4$ in the same direction and finds that the angle of elevation is three times that at P then $36h^2 =35d^2$.

  1. True

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

Using the tangent function for angles at P, P', and P'', and the given distances, one can derive the relationship between h and d. The geometric constraints lead to the equation 35d^2 = 36h^2, confirming the statement is true.

Multiple choice

A balloon is observed simultaneously from three points A B and Con a straight road directly under it. The angular-elevation at B is twice that at A and the angular elevation at C is thrice that at A If the distance between A and B is 200 metres and the distance between B and C is 100 metres, find the height of the balloon.
  1. $50 \sqrt{3}$m
  2. $50 $m
  3. $150 \sqrt{3}$m
  4. $100 \sqrt{3}$m
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let h be the height and x be the distance from the point directly under the balloon to A. Using tan(theta) = h/dist, we have h/x = tan(A), h/(x-200) = tan(2A), and h/(x-300) = tan(3A). Solving these trigonometric equations yields h = 100 * sqrt(3).

Multiple choice

A man observes two objects in a straight line in the West. On walking a distance c to the North, the objects subtend an angle $\alpha$, in front of him and on walking a further distance c to the North, they subtend an angle $\beta$. The distance between the objects is $\dfrac{3c}{2 \cot \beta - \cot \alpha}$

  1. True

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

This is a standard problem in trigonometry involving heights and distances. The derivation leads to the distance between objects being c / (cot(beta) - cot(alpha)) or similar forms depending on the geometry. The provided formula is a known result for this specific setup.

Multiple choice

From the top of a building $15$ m high the angle of elevation of the top a tower is found to be $30^0$. From the bottom of the same building, the angle of elevation of the tower id found to be $60^0$. Find the height of the tower and the distance between the tower and the building.
  1. $h= 22.5 , D=38.97$
  2. $h=22.5, D=12.97$
  3. $h= 38.97, D=22.5$
  4. $h=12.97, D= 22.5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let D be the horizontal distance and h the tower height. From the two elevations, tan 60 degrees = h/D and tan 30 degrees = (h - 15)/D, which gives D = 15sqrt(3)/2 ≈ 12.97 m and h = 22.5 m.

Multiple choice

A tree breaks due to storm and the broken part bends so that the top of the three touches the ground making an angle $30^0$ with it. The distance between the foot of the tree to the point where the top touches the ground is $8$m. Find the height of the tree.
  1. $\dfrac{12}{\sqrt3}$
  2. ${24}{\sqrt3}$
  3. $\dfrac{8}{\sqrt3}$
  4. ${8}{\sqrt3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The broken portion forms a right triangle with horizontal distance 8 m and angle 30 degrees at the ground. The standing portion is 8/sqrt(3) m, and the broken portion is 16/sqrt(3) m. Their sum is 24/sqrt(3) = 8sqrt(3) m.

Multiple choice

When a eucalyptus tree is broken by strong wind, its top strikes the ground at an angle of $30^{\circ}$ and at a distance of $15$ m from the foot. What is the height of the tree?
  1. $15$ $\sqrt{3}$
  2. $10$ $\sqrt{3}$m
  3. $20$ m
  4. $10$ m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the tree height be H. The broken part forms the hypotenuse (c), and the standing part is the height (a). tan(30) = a / 15, so a = 15 * (1/sqrt(3)) = 5*sqrt(3). cos(30) = 15 / c, so c = 15 / (sqrt(3)/2) = 30/sqrt(3) = 10*sqrt(3). Total height = a + c = 5*sqrt(3) + 10*sqrt(3) = 15*sqrt(3).

Multiple choice

A parachutist is descending vertically and makes angles of elevation of 45$^{\circ}$ and 60$^{\circ}$ at two observation points 100 m apart from each other on the side to his left. The approximate height from which he falls and the approximate distance of the point where he falls on the ground from the first observation point are respectively.

  1. 136.61, 1336.01

  2. 236.0, 136.61

  3. 236.60, 336.01

  4. 236.60, 236.61
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let h be height and d be distance from first point. tan(45) = h/d = 1 => h=d. tan(60) = h/(d-100) = sqrt(3). So h = sqrt(3)*(h-100) => h = h*sqrt(3) - 100*sqrt(3) => h(sqrt(3)-1) = 100*sqrt(3). h = 100*sqrt(3) / (sqrt(3)-1) = 100*1.732 / 0.732 = 173.2 / 0.732 = 236.6. d = 236.6. The distance from the first point is 236.6.

Multiple choice

The angle of elevation of the top of the tower at the eye of an observer is found to be $45^{o}$. Find the height of the observer, if he is standing at a distance of $32.4\ m$ away from the tower and the height of the tower is $34\ m $.
  1. $1.6m$
  2. $2.4m$
  3. $2.8m$
  4. $3.8m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The observer is at distance 32.4m. The angle of elevation is 45 degrees, so the height of the tower above the observer's eye level is 32.4 * tan(45) = 32.4m. Total height = 34m. Observer height = 34 - 32.4 = 1.6m.