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

Two light posts are of equal height. A person standing mid-way between the line joining their feet, observes the elevation of the posts to be $30^{0}$. After walking $12$ metres towards one of them, he observes that the same post now subtends an angle of $60^{0}$. Find the distance between them.
  1. $18\ m$
  2. $26\ m$
  3. $36\ m$
  4. $24\ m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let height be h. Midpoint distance is d. tan(30) = h/d, so h = d/sqrt(3). After moving 12m, distance to one post is d-12, other is d+12. tan(60) = h/(d-12). sqrt(3) = (d/sqrt(3)) / (d-12). 3(d-12) = d. 3d - 36 = d. 2d = 36. d = 18. Total distance = 2d = 36m.

Multiple choice

The angle of elevation of a tower from a point $A$ due south of it, is $x$, from a point $B$ due east of $A$, is $y$. If $AB=l,$ then the height $h$ of the tower is given by
  1. $\displaystyle \frac { l }{ \sqrt { \cot ^{ 2 }{ y } -\cot ^{ 2 }{ x }  }  } $
  2. $\displaystyle \frac { l }{ \sqrt { \tan ^{ 2 }{ y } -\tan ^{ 2 }{ x }  }  } $
  3. $\displaystyle \frac { 2l }{ \sqrt { \cot ^{ 2 }{ y } -\cot ^{ 2 }{ x }  }  } $

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

Let the tower height be h. Distance to point A = h * cot(x). Distance to point B = h * cot(y). Since A and B are due south and east, triangle OAB is right-angled at A (where O is the base of the tower). By Pythagoras, OA^2 + AB^2 = OB^2. (h*cot(x))^2 + l^2 = (h*cot(y))^2. l^2 = h^2(cot^2(y) - cot^2(x)). h = l / sqrt(cot^2(y) - cot^2(x)).

Multiple choice

Two points at distance x and y from the base point are on the same side of the line passing through the base pf a tower. The angle of elevation from these two points to the top of the tower are complementary. Then, the height of the tower is :
  1. $\sqrt{x}$
  2. $\sqrt{y}$
  3. $\sqrt{xy}$
  4. $\sqrt{\displaystyle{\frac{x}{y}}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the height be h. The angles are theta and 90-theta. Then tan(theta) = h/x and tan(90-theta) = cot(theta) = h/y. Multiplying these gives tan(theta) * cot(theta) = (h/x) * (h/y), so 1 = h^2 / (xy), which means h = sqrt(xy).

Multiple choice

$P$ is a point on the segment joining the feet of two vertical poles of heights $a$ and $b$. The angles of elevation of the tops of the poles from $P$ are $45^0$ each. Then, the square of the distance between the tops of the poles is
  1. $\displaystyle \frac {a^2\, +\, b^2}{2}$
  2. $a^2\, +\, b^2$
  3. $2(a^2\, +\, b^2)$
  4. $4(a^2\, +\, b^2)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the poles be at (0,0) and (d,0) with heights a and b. P is at (x,0). tan(45) = a/x = 1, so x=a. tan(45) = b/(d-x) = 1, so d-x=b, d=a+b. The tops are at (0,a) and (a+b,b). Distance squared = (a+b-0)^2 + (b-a)^2 = (a+b)^2 + (b-a)^2 = a^2 + 2ab + b^2 + b^2 - 2ab + a^2 = 2(a^2 + b^2).

Multiple choice

After covering a distance of $10$ meters from the flagpole, Mike lies on the ground, and measures an angle of ${70}^{o}$ from the ground to the base of the ball at the top of the flagpole. Find the approximate height of the flagpole from the ground to the base of the ball at the top of the flagpole.
  1. $3m$
  2. $9m$
  3. $27m$
  4. $29m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Height = Distance * tan(angle). Height = 10 * tan(70 degrees). tan(70) is approx 2.747. Height = 10 * 2.747 = 27.47m, which is approximately 27m.

Multiple choice

From the top of a building $h$ metre high, the angle of depression of an object on the ground has a measure $\theta$. The distance of the object from the building is
  1. $h\cos { \theta } $ metre
  2. $h\sin { \theta } $ metre
  3. $\tan { \theta } $ metre
  4. $h\cot { \theta } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice

The top of a hill observed from the top and bottom of a building of height $h$ is at angles of elevation $\alpha$ and $\beta$ respectively. The hehill is:ight of the
  1. $\cfrac { h\cot { \beta } }{ \cot { \beta } -\cot { \alpha } } $
  2. $\cfrac { h\cot { \alpha } }{ \cot { \alpha } -\cot { \beta } } $
  3. $\cfrac { h\tan { \alpha } }{ \tan { \alpha } -\tan { \beta } } $

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

Using standard trigonometric relations for the height of a hill observed from the top and bottom of a building, the height of the hill is given by h * cot(alpha) / (cot(alpha) - cot(beta)).

Multiple choice

A house subtends a right angle at the window of the opposite house and the angle of elevation of the window from the bottom of the first house is $60$. If the distance between the two houses be $6$m, then the height of the first house is?
  1. $6\sqrt{3}$m
  2. $8\sqrt{3}$m
  3. $4\sqrt{3}$m

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the height of the first house be H. The window is at height h. From the bottom of the first house, the angle of elevation to the window is 60 degrees, so tan(60) = h/6, meaning h = 6 * sqrt(3). The house subtends 90 degrees at the window, implying the height of the house above the window is 6 * tan(30) = 6 * (1/sqrt(3)) = 2 * sqrt(3). Total height = 6 * sqrt(3) + 2 * sqrt(3) = 8 * sqrt(3) m.

Multiple choice

At a distance $2h$ $m$ from the foot of a tower of height $h$ $m$, the top of the tower and a pole at the top of the tower subtend equal angles. Height of the pole should be
  1. $\dfrac { 5h }{ 3 } m$
  2. $\dfrac { 4h }{ 3 } m$
  3. $\dfrac { 7h }{ 5 } m$
  4. $\dfrac { 3h }{ 2 } m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the pole height be x. The tower height is h. Angle at distance 2h is tan(theta) = h / 2h = 0.5. The pole is at the top, so the total height is h + x. The angle subtended by the pole is the difference between the angle to the top of the pole and the angle to the top of the tower. Setting these equal leads to the result x = 5h/3.

Multiple choice

Two men standing on opposite sides of a flag staff measure the angles of the top of the flagstaff as $30^{\circ}$ and$\, 60^{\circ}$. If the height of the flagstaff is $18$ m the distance between the men is :
  1. $24$ m
  2. $\displaystyle24\sqrt{3}m$
  3. $\displaystyle\frac{24}{\sqrt{3}}m$
  4. $31.2$ m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let h be the height (18m) and d be the distance. The distance is h * (cot 30 + cot 60). This is 18 * (sqrt(3) + 1/sqrt(3)) = 18 * (3/sqrt(3) + 1/sqrt(3)) = 18 * (4/sqrt(3)) = 72/sqrt(3) = 24 * sqrt(3).