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

$P$ and $Q$ are two points observed from the top of a building $10\sqrt{3}\ m$ height if the angles of depression of the points are complementary and $PQ=20\ m$, then the distance of $P$ from the building is : (P being the farther point)
  1. $30\ m$
  2. $75\ m$
  3. $45\ m$
  4. $40\ m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let h = 10*sqrt(3). Angles are θ and 90-θ. Distances are d1 = h/tan(θ) and d2 = h/tan(90-θ) = h*tan(θ). d1 - d2 = 20. h/tan(θ) - h*tan(θ) = 20. 10*sqrt(3) * (1/tan(θ) - tan(θ)) = 20. 1/tan(θ) - tan(θ) = 2/sqrt(3). Let x = tan(θ). 1/x - x = 2/sqrt(3) => (1-x^2)/x = 2/sqrt(3). Solving this leads to tan(θ) = 1/sqrt(3), so θ = 30 degrees. d1 = 10*sqrt(3) / (1/sqrt(3)) = 30.

Multiple choice

The angle of elevation of a cloud from a point $x$ $m$ above a lake is $\theta$ and the angle of depression of its reflection in the lake is ${45}^{o}$. The height of the cloud is
  1. $x\tan{({45}^{o}-\theta)}$
  2. $x\tan{({45}^{o}+\theta)}$
  3. $\cfrac{1}{x}\cot{({45}^{o}-\theta)}$
  4. $\cfrac{1}{x}\cot{({45}^{o}+\theta)}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let h be the height of the cloud. The distance from the point to the cloud is h-x, and the distance to the reflection is h+x. Using trigonometry, tan(theta) = (h-x)/d and tan(45) = (h+x)/d. Solving for h gives h = x * tan(45+theta).

Multiple choice

The angle of elevation of the top of a tower from a point A on the ground is $30^0$. On moving a distance of 20 metres towards the foot of the tower to a point B, the angle of elevation increases to $60^0$. The height of the tower is :
  1. $\sqrt{3}m$
  2. $5\sqrt{3}m$
  3. $10\sqrt{3}m$
  4. $20\sqrt{3}m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let h be the height. From the 60 degree angle, the distance to the foot is h/sqrt(3). From the 30 degree angle, the distance is h/tan(30) = h*sqrt(3). The difference is 20m, so h*sqrt(3) - h/sqrt(3) = 20. Solving gives h(2/sqrt(3)) = 20, so h = 10*sqrt(3).

Multiple choice

The angle of elevation of the top of a tower standing on a horizontal plane from two points on a line passing through the foot of the tower at a distance 9 ft and 16 ft respectively are complementary angles. Then the height of the tower is:

  1. 9 ft

  2. 12 ft

  3. 16 ft

  4. 144 ft
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let height be h. tan(theta) = h/9, tan(90-theta) = h/16. cot(theta) = h/16. tan(theta) * cot(theta) = 1 = (h/9) * (h/16) = h^2 / 144. h^2 = 144, h = 12.

Multiple choice

The angle elevation of the top of a tower from two points situated at distances 36 m and 64 m from its base and in the same straight line with it are complementary. What is the height of the tower?

  1. 50 m

  2. 48 m

  3. 25 m

  4. 24 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the height be h. The angles of elevation are complementary, so tan(theta) = h/36 and tan(90-theta) = h/64. Since tan(90-theta) = cot(theta) = 1/tan(theta), we have h/64 = 36/h, which leads to h^2 = 36 * 64 = 2304. Thus, h = sqrt(2304) = 48 m.

Multiple choice

An airplane when flying at a height of 5000 m from the ground passes vertically above another airplane at an instant, when the angles of elevation of the two airplane from the same point on the ground are $60^{\circ}$ and $45^{\circ}$ respectively. The vertical distance between the airplane at that instant is:
  1. $5000 (\sqrt{3}-1)m$
  2. $5000 (3-\sqrt{3})m$
  3. $5000 (1-\frac{1}{\sqrt{3}})m$

  4. 4500 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the point on the ground be at distance x from the vertical line. Height of top plane = 5000. tan(60) = 5000/x => x = 5000/sqrt(3). Let height of lower plane be h. tan(45) = h/x => h = x = 5000/sqrt(3). Vertical distance = 5000 - 5000/sqrt(3) = 5000(1 - 1/sqrt(3)).

Multiple choice

A flagstaff $5$ m high is placed on a building $25$ m high. If the flag and building be subtend equal angels on the observer at a height $30$ m, then the distance between observer and the top of flag is
  1. $\dfrac {5\sqrt {3}}{2}$
  2. $5\sqrt {\dfrac {2}{3}}$
  3. $5\sqrt {\dfrac {3}{2}}$
  4. $\dfrac {5\sqrt {2}}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the observer be at (0, 30). The building top is at (x, 25) and flag top is at (x, 30). The angles subtended are equal, implying the ratio of heights to horizontal distance is the same. Solving the geometry leads to the distance between the observer and the flag top.