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 ladder leaning against a vertical wall is inclined at an angle $\alpha$ to the horizontal. The top of the ladder touches the parapet. On moving its foot $a$ feet away from the wall, the ladder now stands inclined at an angle $\beta$ to the horizon, and its top now touching a window. Then the distance of the parapet from the window is $a \cot [(\alpha+\beta)/2]$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a standard trigonometric identity related to a ladder sliding down a wall. The derivation confirms the formula.

Multiple choice

A tree standing on a horizontal plane is leaning towards east. At two points situated at distances $a$ and $b$ exactly due West of it, the angles of elevation of the top are respectively $\alpha$ and $\beta$. then the height of the top from the ground is $\dfrac { \left( b-a \right) \tan { \alpha } \tan { \beta } }{ \tan { \alpha } -\tan { \beta } }$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a standard trigonometric application problem. For a leaning tree, the height formula derived from the cotangent rule matches the expression provided.

Multiple choice

PQ is a vertical tower. P is the foot, Q the top of the tower, A, B, C are three points in the horizontal plane through P. The angles of elevation of Q from A, B, C are equal and each is equal to $\theta$. The sides of the triangle ABC are a, b, c and the area of the triangle ABC is $\Delta$. The height of the tower is
  1. (abc) tan $\theta/ \, 4\,\Delta$
  2. (abc) sin $\theta/ \, 4\,\Delta$
  3. (abc) cot $\theta/ \, 4\,\Delta$

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

In triangle ABC, the circumradius R = abc / 4 * Delta. The height h = R * tan(theta). Thus h = (abc * tan(theta)) / 4 * Delta.

Multiple choice

A flag-staff 5 m high stands on a building 25 m high. To an observer at a height of 30 m the flag-staff and the building subtend equal angles. The distance of the observer from the top of the flag-staff is
  1. $\dfrac{5}{2}$
  2. $5\sqrt{\dfrac{3}{2}}$
  3. $5\sqrt{\dfrac{2}{3}}$

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

Let the observer be at height 30m. The building is 25m, flag-staff is 5m (total 30m). Observer is at the same level as the top of the flag-staff. The geometry involves equal angles subtended at the observer's position.

Multiple choice

True or False The angle of elevation of the top of a tower from a point A due South of the tower is a and from B due East of tower is $\beta$. if AB = d, then the height of the tower is $\dfrac{d}{\sqrt{(cot^2 \alpha + cot^2 \beta)}}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let height be h. From point A, cot(alpha) = distance_A / h. From point B, cot(beta) = distance_B / h. Since A and B are perpendicular (South and East), d^2 = distance_A^2 + distance_B^2. Thus d^2 = h^2 * (cot^2(alpha) + cot^2(beta)). Solving for h gives the formula.

Multiple choice

True or False The top of a tower is observed from three points A, B, C on a straight line leading to the tower. If the angles of elevation are $\theta, 2\theta,3\theta$ from them, then $\dfrac{AB}{BC} = \dfrac{cot \theta -cot 2 \theta}{ cot 2 \theta - cot 3 \theta}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the cotangent rule for heights and distances, if a tower of height h is observed at angles theta, 2theta, 3theta, the horizontal distances are h*cot(theta), h*cot(2theta), h*cot(3theta). AB = h(cot(theta) - cot(2theta)) and BC = h(cot(2theta) - cot(3theta)). The ratio AB/BC is indeed the expression given.

Multiple choice

True or False Top of a mountain is observed from A and B at the sea level. If N is the point vertically below P and $\angle NAB = \alpha, \angle NBA= \beta, \angle NAP = \theta, \angle NBP = \phi$, then $cot \phi sin \beta = cot \theta sin \alpha$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using trigonometry in the right triangles formed by the mountain peak P and the ground points A, B, and N, we have PN = AN tan(theta) = BN tan(phi). Since AN = PN cot(theta) and BN = PN cot(phi), applying the sine rule in triangle ABN gives AN/sin(beta) = BN/sin(alpha). Substituting the expressions for AN and BN yields PN cot(theta)/sin(beta) = PN cot(phi)/sin(alpha), which simplifies to cot(theta) sin(alpha) = cot(phi) sin(beta).

Multiple choice

From a light house the angle of depression of two ships on opposite sides of the light house are observed to be 30 and 45. If the height of the light house be 100 metres, then the distance between the ships of the line joining them passes through the foot of the light house is 100 $\sqrt{3 - 1}$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The horizontal distances from the foot of the lighthouse are 100 cot 30° = 100sqrt(3) metres and 100 cot 45° = 100 metres. Since the ships are on opposite sides, their separation is 100sqrt(3) + 100 = 100(sqrt(3) + 1) metres, not 100sqrt(3 - 1).

Multiple choice

The angles of elevation of the top of a tower from two points on the ground at distances a metres and b metres from the base of the tower and in the same straight line are complementary. The height of the tower is $\sqrt{ab}$ metres.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If the angles are complementary (theta and 90-theta), the height h satisfies tan(theta) = h/a and tan(90-theta) = h/b. Since tan(90-theta) = cot(theta) = 1/tan(theta), we have h/b = a/h, which leads to h^2 = ab, or h = sqrt(ab).

Multiple choice

An aeroplane is flying horizontally at a height of 3150 m above a horizontal plane ground. At a particular instant it passes another plane vertically below it. At this instant, the angles of elevation of the planes from a point on the ground are $30^o$ and $60^o$.Hence, the distance between the two planes at that instant is :

  1. 1050 m

  2. 2100 m

  3. 4200 m

  4. 5250 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the point on the ground be O. The height of the plane is 3150. For the 60-degree angle, the base distance is 3150/tan(60) = 3150/sqrt(3) = 1050*sqrt(3). For the 30-degree angle, the base distance is 3150/tan(30) = 3150*sqrt(3). The distance between the planes is the difference in their heights if they were at the same horizontal distance, but here they are vertically aligned. The distance between the two planes is 3150 - (3150/3) = 2100m.

Multiple choice

If $\alpha , \beta , \gamma$ be the three angles given by $\cos \alpha=\dfrac{a}{b+c}, \cos \beta =\dfrac{b}{c+a}$ and $\cos \gamma =\dfrac{c}{a+b}$ where $a,b,c$ are the sides of a $\triangle ABC$, then $\tan \dfrac{\alpha}{2} \tan \dfrac{\beta}{2} \tan \dfrac{\gamma}{2}$ is equal to
  1. $\cot \dfrac{A}{2} \cot \dfrac{B}{2} \cot \dfrac{C}{2}$
  2. $\csc \dfrac{A}{2} \csc \dfrac{B}{2} \csc \dfrac{C}{2}$
  3. $\tan \dfrac{A}{2} \tan \dfrac{B}{2} \tan \dfrac{C}{2}$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the half-angle formulas and the given cosine definitions, the expression simplifies using the properties of a triangle where cos(alpha) = a/(b+c). This is a standard trigonometric identity derivation in triangle geometry.