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 building $h$ metres, the angle of depression of an object on the ground is $\alpha$, the distance of the object from the foot of the building is
  1. $h\cot\alpha$
  2. $h\tan\alpha$
  3. $h\cos\alpha$
  4. $h\sin\alpha$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In a right triangle formed by the building height (h), the ground distance (d), and the line of sight, the angle of depression equals the angle of elevation. Thus, tan(alpha) = h / d, which implies d = h / tan(alpha) = h * cot(alpha).

Multiple choice

The horizontal distance between two towers is $60\ m$ and angular depression of the top of the first as seen from the second, which is $150\ m$ in height, is $30^{0}$. The height of the first tower is
  1. $(150+20\sqrt{3})\ m$
  2. $(150+15\sqrt{3})\ m$
  3. $(150-20\sqrt{5})\ m$
  4. $(150-20\sqrt{3})\ m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let h1 be the height of the first tower and h2 = 150 be the height of the second. The angle of depression is 30 degrees. tan(30) = (150 - h1) / 60. 1/sqrt(3) = (150 - h1) / 60. 60/sqrt(3) = 150 - h1. 20*sqrt(3) = 150 - h1. h1 = 150 - 20*sqrt(3).

Multiple choice

$A$ flag staff stands upon the top of a building. $A$t a distance of 40 $m$. the angles of elevation of the tops of the flag staff and building are $60^{ }$ and $30^{0}$ then the height of the flag staff in metres is
  1. $40\sqrt{3}$
  2. $\displaystyle \frac{40}{\sqrt{3}}$
  3. $\displaystyle \frac{160}{\sqrt{3}}$
  4. $\displaystyle \frac{80}{\sqrt{3}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let h be the height of the building and H be the height of the flag staff. The height of the top of the building is 40 * tan(30) = 40/sqrt(3). The height of the top of the flag staff is 40 * tan(60) = 40 * sqrt(3). The flag staff height is 40 * sqrt(3) - 40/sqrt(3) = (120 - 40) / sqrt(3) = 80/sqrt(3).

Multiple choice

$A$ and $B$ are two stations due north and south of a tower of height $25m$. The angles of depression of the stations from the top of the towe $r$ observed to be $30^{0}$ and $45^{0}$ respectively. The distance between the two stations is
  1. $25(\sqrt{3}+1)$
  2. $25(\sqrt{3}-1)m$
  3. $25\sqrt{3}m$
  4. $25(2+\sqrt{3})m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The horizontal distance to the station at 30 degrees is 25tan(60 degrees) = 25sqrt(3) m. The distance to the station at 45 degrees is 25tan(45 degrees) = 25 m. Since the stations are on opposite sides of the tower, their separation is 25(sqrt(3) + 1) m.

Multiple choice

A person walking along a straight road towards a hill observes at two points distance $\sqrt{3}$ km, the angle of elevation of the hill to be $30^{0}$ and $60^{0}$. The height of the hill is
  1. $\dfrac{3}{2}$ km
  2. $\sqrt{\dfrac{2{1}}{3{2}}}$
  3. $\displaystyle \frac{(\sqrt{3}+1)}{2}$ km
  4. $\sqrt{3}$ km
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the distance from the nearer observation point to the hill be x km. Then h = x tan 60° and h = (x + sqrt(3)) tan 30°, which gives x = sqrt(3)/2 and h = 3/2 km.

Multiple choice

A tower subtends at angle $\displaystyle\theta$ at a point P on the same level as the foot of the tower and from a point h m above P the depression of the foot of the tower is $\displaystyle\alpha$ . The height of the tower is (in m)
  1. $h$ $\displaystyle\tan \theta \tan \alpha $
  2. $h$ $\displaystyle\\cot \theta \cot \alpha $
  3. $h$ $\displaystyle \tan \theta \cot \alpha $
  4. $\displaystyle \cot \theta \tan \alpha $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let H be the height of the tower. From point P, tan(theta) = H / distance_to_tower (d), so d = H / tan(theta). From the point h meters above P, the angle of depression alpha implies tan(alpha) = h / d. Substituting d, we get tan(alpha) = h / (H / tan(theta)) = h * tan(theta) / H. Thus, H = h * tan(theta) / tan(alpha) = h * tan(theta) * cot(alpha).

Multiple choice

Flag-staff of length $d$ stands on a tower of height $h$. lf at a point on the ground the angles of elevation of the tower and the top of the flag-staff be $\alpha,\ \beta$ respectively, then $h=$
  1. $\displaystyle \dfrac{d\cot\beta}{\cot\beta-\cot\alpha}$
  2. $\displaystyle \dfrac{d\tan\beta}{\tan\alpha-\tan\beta}$
  3. $\displaystyle d[\frac{\tan\alpha+\tan\beta}{\cot\alpha-\cot\beta}]$
  4. $\displaystyle \frac{d\tan\alpha}{\tan\beta}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

$A$ man observes a tower$AB$ of height $h$ from a point $P$ on the ground. He moves a distance d' towards the foot of the tower and finds that the angle of elevation is doubled. He further moves a distance $\dfrac {3d}{4}$ in the same direction and the angle of elevation is three times that at $P$. Then $\displaystyle \frac{h^{2}}{d^{2}}=$
  1. $\dfrac {35}{9}$
  2. $\dfrac {35}{36}$
  3. $\dfrac {36}{5}$
  4. $\dfrac {36}{35}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the initial angle be theta and the initial horizontal distance be x. Using tan(2theta) and tan(3theta) with the two stated movements gives tan^2(theta) = 5/7 and d/x = 6/7. Hence h^2/d^2 = tan^2(theta)/(d/x)^2 = 35/36.

Multiple choice

A pole of height $h$ stands at one corner of a park in the shape of an equilateral triangle. If $\alpha$ is the angle which the pole subtends at the midpoint of the opposite side, the length of each side of the park is:
  1. $\displaystyle \frac{\sqrt{3}}{2}h\cot\alpha$
  2. $\displaystyle \frac{2}{\sqrt{3}}h\cot\alpha$
  3. $\displaystyle \frac{\sqrt{3}}{2}h\tan\alpha$
  4. $\displaystyle \frac{2}{\sqrt{3}}h\tan\alpha$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the side of the equilateral triangle be 'a'. The distance from the midpoint of a side to the opposite vertex is (sqrt(3)/2) * a. Using trigonometry in the right triangle formed by the pole and this distance, tan(alpha) = h / ((sqrt(3)/2) * a). Solving for 'a' gives a = (2h / sqrt(3)) * cot(alpha).

Multiple choice

$PQ$ is a vertical tower. $A,\ B,\ C$ are three points in a horizontal line through $Q$, the foot of the tower. If the angles of elevation of the top of the tower from $A,\ B,\ C$ are $\alpha,\ \beta,\ \gamma$ respectively, then $BC\cot\alpha-CA\cot\beta+AB \cot$ $\gamma=$
  1. $3$
  2. $1$
  3. $2$
  4. $0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let tower height be h. Then QA = h cot(alpha), QB = h cot(beta), QC = h cot(gamma). The expression is BC cot(alpha) - CA cot(beta) + AB cot(gamma). Substituting distances: (QC-QB) cot(alpha) - (QC-QA) cot(beta) + (QB-QA) cot(gamma). This simplifies to 0.

Multiple choice

Tower is observed from two stations $A$and $B$ where $B$ is East of $A$ at a distance $100m$. The tower is to north of $A$ and to North-West of $B$. The angles of elevation of the tower from $A$and $B$ are complementary. The height of the tower (in meters) is:
  1. $100$ $\sqrt{2}$
  2. $100(2)^{1/4}$
  3. $100(2)^{-1/4}$
  4. $100(2)^{1/3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let h be the height of the tower at O. Let A be at (0,0,0) and B at (100,0,0). The tower is at (0, y, 0). From A, tan(alpha) = h/y. From B, the tower is at North-West, so the distance from B to the base is sqrt(100^2 + y^2). tan(90-alpha) = h/sqrt(100^2 + y^2). Thus, cot(alpha) = h/sqrt(100^2 + y^2). Multiplying tan(alpha) and cot(alpha) gives 1 = h^2 / (y * sqrt(100^2 + y^2)). Solving for h leads to h = 100 * 2^(1/4).