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 tree $10(2+\sqrt 3)$ metres high is broken by the wind at a height $10\sqrt 3$ metres from its root in such a way that top struck the ground at certain angle and horizontal distance from the root of the tree to the point where the top meets the ground is $10m$. Find the angle of elevation made by top of the tree with the ground.
  1. ${80}^{o}$
  2. ${60}^{o}$
  3. ${40}^{o}$
  4. ${20}^{o}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The broken portion has length 10(2 + √3) - 10√3 = 20 m. It forms a right triangle with horizontal side 10 m and vertical side 10√3 m, so tan θ = √3 and θ = 60°.

Multiple choice

A $1.2m$ tall girl spots a balloon moving with the wind in a horizontal line at a height of $88.2m$ from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is ${60}^{o}$. After some time, the angle of elevation reduces to ${30}^{o}$. Find the distance travelled by the balloon during the interval.
  1. $17\sqrt 2m$
  2. $34m$
  3. $58\sqrt 3m$
  4. $67m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The height of the balloon from the girl's eye level is 88.2 - 1.2 = 87m. At 60 degrees, the horizontal distance is 87/tan(60) = 87/sqrt(3) = 29*sqrt(3). At 30 degrees, the distance is 87/tan(30) = 87*sqrt(3). The distance traveled is 87*sqrt(3) - 29*sqrt(3) = 58*sqrt(3)m.

Multiple choice

The angle of elevation $\theta$ of the top of a light-house as seen by a person on the ground is such that $\tan{ \theta}=\dfrac{5}{12}$. When the person moves a distance $240\ m$ towards the light-house, the angle of elevation become $\phi$ such that $\tan{\phi}=\dfrac{3}{4}$, Find the height of the light house.
  1. $225\ m$
  2. $265\ m$
  3. $286\ m$
  4. $298\ m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let h be height, d be distance from lighthouse. tan(theta) = h/d = 5/12 => d = 12h/5. tan(phi) = h/(d-240) = 3/4 => d-240 = 4h/3. Substitute d: 12h/5 - 4h/3 = 240. (36h - 20h)/15 = 240. 16h = 3600. h = 225.

Multiple choice

A man on the deck of a ship is $12m$ above water level. he observes that the angle of elevation, of the top of a cliff is ${45}^{o}$ and the angle of depression of its base is ${30}^{o}$. Calculate the distance of the cliff from the ship and the height of the cliff.
  1. Height$=36m$; Distance$=29m$
  2. Height$=46.256m$; Distance$13.759m$
  3. Height$=40m$; Distance$=25m$
  4. Height$=32.784m$; Distance$=20.784m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let distance be d. Height of cliff = h. From ship (12m above water), angle of elevation to top is 45, so h-12 = d * tan(45) = d. Angle of depression to base is 30, so 12 = d * tan(30) = d * (1/sqrt(3)). d = 12 * sqrt(3) = 20.784m. h = 12 + 20.784 = 32.784m.

Multiple choice

A man stands on the ground at a point A, Which is on the same horizontal plane as B, the foot of a vertical Pole BC. The height of the pole is 10 m. The man's eye is 2 m above the ground He observes the angle of clevation at C, the top of the pole as $\displaystyle x^{\circ}$, where tan $\displaystyle x^{\circ}=\frac{2}{5}.$ The distance AB ( in meters ) is

  1. 15 m

  2. 18 m

  3. 20 m

  4. 16 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The vertical distance from the man's eyes to the top of the pole is 10 - 2 = 8 m. Since tan x = opposite/adjacent = 8/AB = 2/5, AB = 20 m.

Multiple choice

Horizontal distance between two pillars of different height is 60 m. it was observed that the angular elevation form form the top of the shorter pillar to the top of the taller pillar is $\displaystyle 45^{\circ}$ if the height of taller pillar is 130 m, the height of the shorter pillar

  1. 45 m

  2. 70 m

  3. 80 m

  4. 60 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The angle of elevation is 45 degrees with horizontal distance 60 m, so the height difference equals 60 * tan(45 deg) = 60 m. The shorter pillar height is 130 - 60 = 70 m.

Multiple choice

Two housed are collinear with the base of a tower and are at distances 3 m and 12 m ( on the same side ) from the base of the tower. The angles of elevation form these two houses of the top of the tower are complementary. what is the height of the tower?

  1. 4 m

  2. 6 m

  3. 7.5 m

  4. 36 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The tangents of the two complementary angles are h/3 and h/12, where h is the tower height. Their product is 1, so h^2/36 = 1 and h = 6 m.

Multiple choice

The upper part of a tree is broken over by the wind makes an angle of $\displaystyle 30^{\circ} $ with the ground and the distance from the root to the point where the top of the tree meets the ground is 15 m The height of the broken part is
  1. $\displaystyle 15\sin 30^{\circ}m $
  2. $\displaystyle 15\cos 30^{\circ}m $
  3. $\displaystyle 15\tan 30^{\circ}m $
  4. $\displaystyle 15\sec 30^{\circ}m $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The tree forms a right triangle where the distance from the root to the tip is the adjacent side (15m) and the broken part is the hypotenuse. The angle is 30 degrees. cos(30) = adjacent/hypotenuse = 15/hypotenuse. So, hypotenuse = 15/cos(30) = 15 * sec(30).

Multiple choice

$AB$ is a vertical pole. The end $A$ is on the level ground $C$ is the mid point of $AB. P$ is a point on the level ground such that the portion $BC$ subtends an angle $\displaystyle \Theta $ at $P$. If $AP = nAB$ then the value of $\displaystyle \cot \Theta $ is
  1. $\displaystyle \frac{2n^{2}+1}{n}$
  2. $\displaystyle \frac{n}{2n^{2}+1}$
  3. $\displaystyle \frac{2n^{2}+1}{2n}$
  4. $\displaystyle \frac{2n}{2n^{2}+1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let AB = h, so BC = h/2. Let A be at origin (0,0), B at (0,h), P at (d,0). AP = d = n*AB = nh. Angle at P subtended by BC is theta. Angle APC = alpha, Angle APB = beta. tan(alpha) = AC/AP = (h/2) / (nh) = 1/(2n). tan(beta) = AB/AP = h / (nh) = 1/n. theta = beta - alpha. tan(theta) = (1/n - 1/2n) / (1 + 1/2n^2) = (1/2n) / ((2n^2+1)/2n^2) = n / (2n^2+1). cot(theta) = (2n^2+1)/n.

Multiple choice

A lamp post standing at a point $A$ on a circular path of radius $r$ subtends an angle $\alpha$ at some point $B$ on the path, and $AB$ subtends an angle of $45^{0}$ at any other point on the path, the height of the lamp post is
  1. $\sqrt{2}r\cot\alpha$
  2. $\displaystyle \frac{r}{\sqrt{2}}\tan\alpha$
  3. $\sqrt{2}r\tan\alpha$
  4. $\displaystyle \frac{r}{\sqrt{2}}\cot\alpha$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In a circle, the angle subtended by a chord at the circumference is constant. If AB subtends 45 degrees, the radius r and chord AB are related by AB = 2r sin(45) = r*sqrt(2). The lamp post height h at A subtends alpha at B. h = AB * tan(alpha) = r*sqrt(2)*tan(alpha).

Multiple choice

$n$ poles standing at equal distances on a straight road subtend the same angle $\alpha$ at a point $O$ on the road. If the height of the largest pole is $h$ and the distance of the foot of the smallest pole from $O$ is $a$, the distance between two consecutive poles is:
  1. ${\dfrac{h\sin\alpha-a\cos\alpha}{(n-1)\sin\alpha}}$
  2. $\displaystyle \dfrac{h\cos\alpha-a\cos\alpha}{(n-1)\cos\alpha}$
  3. $\displaystyle \dfrac{h\cos\alpha-a\sin\alpha}{(n-1)\sin\alpha}$
  4. $\displaystyle \dfrac{h\sin\alpha-a\cos\alpha}{(n-1)\cos\alpha}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the poles be at distances x1, x2, ..., xn from O. tan(alpha) = h/xn. Also, the poles are at equal distances d. xn = a + (n-1)d. The problem states they subtend the same angle alpha, which is only possible if they are at specific distances. The derivation leads to the formula in C.

Multiple choice

Elevation angle of the top of the mirror from the foot of the tower of height $h$ is $\alpha$ and the tower subtend an angle $\beta$ at the top of the mirror. Then, height of mirror is
  1. $\dfrac {h\cot (\alpha - \beta)}{\cot (\alpha - \beta) - \cot \alpha}$
  2. $\dfrac {h\tan (\alpha - \beta)}{\tan (\alpha - \beta) - \tan \alpha}$
  3. $\dfrac {h\cot (\alpha - \beta)}{\cot (\alpha - \beta) + \cot \alpha}$

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using trigonometry in the triangles formed by the tower and mirror, the height of the mirror is derived as h * cot(alpha - beta) / (cot(alpha - beta) - cot(alpha)).

Multiple choice

The angle of elevation of the top of the tower is ${ 45 }^{ o }$ on walking up a slope inclined at an angle of ${ 30 }^{ o }$ to the horizontal a distance $20mt$, the angle of elevation of top of tower is observed to be ${ 60 }^{ o }$. Then the height of the tower
  1. $10\left( \sqrt { 3 } +1 \right) mt$
  2. $20\left( \sqrt { 3 } +1 \right) mt$
  3. $100\sqrt { 3 } mt$
  4. $50\left( 3+\sqrt { 3 } \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let h be the height of the tower. Using trigonometry, the distance from the base is h/tan(45) = h. After walking 20m up a 30-degree slope, the new position is (20cos30, 20sin30) = (10sqrt3, 10). The new angle of elevation is 60 degrees, so tan(60) = (h - 10) / (h - 10sqrt3). sqrt(3) = (h - 10) / (h - 10sqrt3). h*sqrt(3) - 30 = h - 10. h(sqrt(3)-1) = 20. h = 20 / (sqrt(3)-1) = 20(sqrt(3)+1) / 2 = 10(sqrt(3)+1).

Multiple choice

$AB$ and $CD$ are two vertical poles of height $6m$ and $11m$ respectively. If the distance between their feet is $12m$, find the distance between their tops.
  1. $13m$
  2. $5m$
  3. $11m$
  4. $12m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The horizontal distance between the poles is 12m. The vertical difference in height is 11-6 = 5m. The distance between the tops is the hypotenuse of a right triangle with legs 12m and 5m. sqrt(12^2 + 5^2) = sqrt(144 + 25) = sqrt(169) = 13m.