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

From the top of a cliff, the angle of depression of car on the ground is $60^o$. If the car is at a distance of $30m$. from the cliff. Find the height of the cliff.
  1. $52$m
  2. $51$m
  3. $55$m
  4. $56$m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The angle of depression from the cliff is equal to the angle of elevation from the car. Using trigonometry, tan(60) = height / 30. Since tan(60) = sqrt(3) is approximately 1.732, the height is 30 * 1.732 = 51.96m, which rounds to 52m.

Multiple choice

The angles of elevation of the top and foot of a flag staff fixed on a wall are $60^o$ and $45^o$ to a man, standing on the other end, at a distance of $30$m. Find the height of the flag staff.
  1. $30\sqrt 3$
  2. $30(\sqrt 3-1)$m
  3. $30(\sqrt 3+1)$m
  4. $10\sqrt 3$m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let wall height be h, flag staff height be f, distance 30m. tan(45) = h/30 => h = 30. tan(60) = (h+f)/30 => sqrt(3) = (30+f)/30 => 30*sqrt(3) = 30+f => f = 30(sqrt(3)-1).

Multiple choice

The distance between two vertical poles is $d\ metres$. The height of one of the poles is double the height of the other. The angles of elevation of the tops of the poles from the middle point of the line joining their feet are complementary to each other. The height of the shorter pole is
  1. $\dfrac {d}{\sqrt 2}$
  2. $\dfrac {d}{2}$
  3. $\dfrac {d}{4}$
  4. $\dfrac {d}{2\sqrt 2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let height of shorter pole be h, taller be 2h. Distance from middle point is d/2. tan(theta) = h / (d/2) = 2h/d. tan(90-theta) = cot(theta) = 2h / (d/2) = 4h/d. Since tan(theta) * cot(theta) = 1, (2h/d) * (4h/d) = 1. 8h^2 / d^2 = 1. h^2 = d^2 / 8. h = d / sqrt(8) = d / (2 * sqrt(2)).

Multiple choice

The angles of elevation of the top of a tower from two points at the distances of $4\ m$ and $9\ m$ from the base of the tower and in the same straight line with it are complementary. The height of the tower is :
  1. $8\ m$
  2. $5\ m$
  3. $6\ m$
  4. $4\ m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let height be h. Angles are theta and 90 - theta. Then tan(theta) = h/4 and tan(90 - theta) = cot(theta) = h/9. Multiplying these: tan(theta) * cot(theta) = (h/4) * (h/9) = 1. Thus h^2 = 36, so h = 6.

Multiple choice

$AB$ is a straight road leading to $C$, the foot of a tower . $A$ is at a distance $200$ meters from $C$ and $B$ at $75$ meters from $C$. If the angle of elevation of the tower at B be double the angle of elevation at A, then the height of the tower is
  1. $125m$
  2. $120m$
  3. $115m$
  4. $100m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let h be the height of the tower. tan(theta) = h/200 and tan(2*theta) = h/75. Using the identity tan(2*theta) = 2*tan(theta) / (1 - tan^2(theta)), we get h/75 = 2(h/200) / (1 - (h/200)^2). Solving for h gives h^2 = 10000, so h = 100.

Multiple choice

On the same side of a tower two objects are located. Observed from the top of the tower their angles of depression are $\displaystyle 45^{0}$ and $\displaystyle 60^{0}$. If the height of the tower is $150\ m$ the distance between the objects is
  1. $63.5$ $m$
  2. $76.9$ $m$
  3. $86.7$ $m$
  4. $90$ $m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let h = 150. Distance 1 = h/tan(45) = 150. Distance 2 = h/tan(60) = 150/sqrt(3) = 150/1.732 = 86.6. The distance between objects is 150 - 86.6 = 63.4 m.

Multiple choice

If the angles of elevaton of the top of a tower from two points distance s and t $\displaystyle \left ( s> t \right )$ from its foot are $\displaystyle 30^{0}$ and $\displaystyle 60^{0}$ respectively then the height of the tower is
  1. $\displaystyle \sqrt{s+t}$
  2. $\displaystyle \sqrt{st}$
  3. $\displaystyle \sqrt{s-t}$
  4. $\displaystyle \sqrt{\frac{s}{t}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let h be the height. tan(30) = h/s => h = s * tan(30) = s / sqrt(3). tan(60) = h/t => h = t * tan(60) = t * sqrt(3). Thus h^2 = (s / sqrt(3)) * (t * sqrt(3)) = st. So h = sqrt(st).

Multiple choice

The angle of elevation of the top of an incomplete vertical pillar at a horizontal distance of $100\ m$ from its base is $\displaystyle 45^{o}$ If the angle of elevation of the top of the complete pillar at the same point is to be $\displaystyle 60^{o}$ then the height of the incomplete pillar is to be increased by
  1. $\displaystyle 50\sqrt{2}\ m$
  2. $100\ m$
  3. $\displaystyle 100\left ( \sqrt{3-1} \right )\ m$
  4. $\displaystyle 100\left ( \sqrt{3+1} \right )\ m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

From the top of a lighthouse, the angles of depression of two stations on opposite sides of it at distance $a$ apart are $\displaystyle \alpha $ and $\displaystyle \beta $. The height of the lighthouse is
  1. $\displaystyle \frac{a}{\cot \alpha \cot \beta }$
  2. $\displaystyle \frac{a}{\cot \alpha +\cot \beta }$
  3. $\displaystyle \frac{a\cot \alpha \cot\beta }{\cot \alpha +\cot \beta }$
  4. $\displaystyle \frac{a\tan \alpha \tan \beta }{\cot \alpha +\cot \beta }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let h be the height. The two distances from the base are h * cot(alpha) and h * cot(beta). Since they are on opposite sides, h * cot(alpha) + h * cot(beta) = a. Thus, h = a / (cot(alpha) + cot(beta)).

Multiple choice

The upper part of a tree broken over by the wind makes an angle of $60^{0}$ with the ground and touches the ground at a distance of 50 metres from the foot. The height of the tree in metres is
  1. $124.2$
  2. $186.6$
  3. $243.2$
  4. $164.2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let x be the broken part (hypotenuse) and y be the standing part. tan(60) = y / 50 -> y = 50 * sqrt(3) = 86.6. cos(60) = 50 / x -> x = 50 / 0.5 = 100. Total height = x + y = 100 + 86.6 = 186.6.

Multiple choice

The angle of elevation of an object from a point $P$ on the level ground is $\alpha$. Moving $d$ meters on the ground towards the object, the angle of elevation is found to be $\beta$, then the height (in meters) of the object is
  1. $ d\tan\alpha$
  2. $ d\cot\beta$
  3. $\dfrac d{\cot\alpha+\cot\beta}$
  4. $\dfrac d{\cot\alpha-\cot\beta}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let height be h. tan(alpha) = h/x and tan(beta) = h/(x-d). x = h/tan(alpha) = h*cot(alpha). x-d = h*cot(beta). d = x - (x-d) = h*cot(alpha) - h*cot(beta). h = d / (cot(alpha) - cot(beta)).