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 wall makes an angle ( \theta ) with the horizontal ground such that ( \cos \theta = \frac{5}{13} ). If the height of the top of the ladder from the foot of the wall is 18 m, then what is the distance (in m) of the foot of the ladder from the wall?

  1. 19.5

  2. 7.5

  3. 13

  4. 18
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Ladder forms right triangle with wall. Given cosθ = 5/13, and opposite = 18m. Using sin²θ = 1 - cos²θ: sinθ = √(1 - 25/169) = √(144/169) = 12/13. Since tanθ = opposite/adjacent = sinθ/cosθ = (12/13)/(5/13) = 12/5. Let adjacent = x, then 18/x = 12/5, so x = 18 × 5/12 = 7.5m.

Multiple choice

The angle of elevation of the top of a tall building from the points M and N at the distances of 72m and 128m, respectively, from the base of the building and in the same straight line with it, are complementary. The height of the building (in m ) is:

  1. 96

  2. 90

  3. 80

  4. 84
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the angles of elevation from M and N be α and (90°-α) since they are complementary. Then tan α = h/72 and tan(90°-α) = cot α = h/128. Using cot α = 1/tan α gives h/128 = 72/h, so h² = 72×128 = 9216, hence h = 96m. The answer matches option A.

Multiple choice

Points P and Q are on the opposite sides of a tower. The angle of elevation of the top of the tower from point P and Q are 60∘ and 30∘ respectively and the distance between P and Q is 64√3 m. What is the height of the tower (in m)?

  1. 48

  2. 54

  3. 35

  4. 64
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let tower height = h. From point P: h/DP = tan60° = √3, so DP = h/√3. From point Q: h/DQ = tan30° = 1/√3, so DQ = h√3. Since P and Q are opposite sides: DP + DQ = h/√3 + h√3 = 64√3. Solving: h(1/√3 + √3) = 64√3. h(4/√3) = 64√3, so h = 48m. Options B (54m) and D (64m) don't satisfy the tangent relationships.

Multiple choice

The angles of elevation of the tops of two pillars of heights h and 2h from a point P on the line joining the feet of the two pillars are complementary. If the distances of the foot of the pillars from the point P are x and y respectively, then which one of the following is correct?

  1. 2h2 = x2y

  2. 2h2 = xy2

  3. 2h2 = xy

  4. 2h2 = x2 y2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the angles of elevation be θ and (90° - θ) since they're complementary. For pillar of height h: tan(θ) = h/x. For pillar of height 2h: tan(90° - θ) = cot(θ) = 2h/y. Since tan(θ) × cot(θ) = 1, we have (h/x) × (2h/y) = 1, which gives 2h² = xy. Option A has x²y which is dimensionally inconsistent. Option B has xy² which is also incorrect. Option D has x²y² which gives the wrong dimensions. The key is recognizing that complementary angles give tan(θ) × cot(θ) = 1.

Multiple choice

The distance between the two trees of height 961 cm and 841 cm is ‘y’ meters. If two angles of elevation of their respective top from the bottom of the other are complementary to each other then the value of ‘y’ in cm is :

  1. 6.77 cm

  2. 7.88 cm

  3. 8.99 cm

  4. 10.11 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When angles of elevation are complementary, if α + β = 90°, then tanα × tanβ = 1. This gives h1/y × h2/y = 1, so y² = h1 × h2 = 961 × 841 = 808801. Therefore y = √808801 = 899 cm = 8.99 m. This elegant property emerges from tan(90-α) = cotα = 1/tanα.

Multiple choice

If ( \cot (P+Q) = 0 ) and ( \tan (P-Q) = \frac{1}{\sqrt{3}} ), then what is the value of ( \angle P ) and ( \angle Q )? ( (0^{\circ} < P, Q \le 90^{\circ}) )

  1. 25 degree, 65 degree

  2. 45 degree, 45 degree

  3. 60 degree, 30 degree

  4. 15 degree, 75 degree
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

From cot(P+Q) = 0, we get P+Q = 90 degrees. From tan(P-Q) = 1/sqrt(3), we get P-Q = 30 degrees. Solving these equations: P = (90+30)/2 = 60 degrees, Q = (90-30)/2 = 30 degrees.

Multiple choice

$ABC$ is a triangular park with $AB=AC=100$ cm. $A$ clock tower is situated at the midpoint of $BC$. The angles of elevation of the top of the tower from $A$ and $B$ are $cot ^{-1}3.2$ and $cosec ^{-1} 2.6$. The height of the tower is
  1. $25$ mt
  2. $50$ mt
  3. $100$ mt
  4. $50\sqrt{2}$ mt
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From cot(theta) = 3.2, the horizontal distance from A is 3.2h. From cosec(phi) = 2.6, sin(phi) = 5/13, so the horizontal distance from B is 2.4h. These form a 3-4-5 right triangle with AB = 100, giving h = 25 in the stated length units.

Multiple choice

In a cubical hall $ABCDPQRS$ with each side $10m,\ G$ is the center of the wall $BCRQ$ and $T$ is the mid point of the side $AB$. The angle of elevation of $G$ at the point $T$ is
  1. $\displaystyle \sin^{-1}(\frac{1}{\sqrt{3}})$
  2. $\displaystyle \cos^{-1}(\frac{1}{\sqrt{3}})$
  3. $\displaystyle \tan^{-1}(\frac{1}{\sqrt{3}})$
  4. $\displaystyle \cot^{-1}(\frac{1}{\sqrt{3}})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In a cube of side 10, let T be (5, 0, 0) and G be (10, 5, 5). The vector TG = (5, 5, 5). The projection on the floor is (5, 5, 0) with length sqrt(50). The height is 5. Tan(theta) = 5 / sqrt(50) = 1/sqrt(2). The angle is sin^-1(1/sqrt(3)).

Multiple choice

A straight line passes through the point $(1,1,1)$ makes an angle $60^0$ with the positive direction of z-axis and the cosine of the angles made by it with the positive directions of the y-axis and x-axis are in the ratio $\sqrt{3}:1$. What is the acute angle between two possible positions of the line ?
  1. $90^0$
  2. $60^0$
  3. $45^0$
  4. $30^0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the direction cosines be (l, m, n). Given n = cos(60) = 1/2. Since l^2 + m^2 + n^2 = 1, l^2 + m^2 = 3/4. Given m/l = sqrt(3), so m = sqrt(3)l. Substituting, l^2 + 3l^2 = 3/4, 4l^2 = 3/4, l^2 = 3/16, l = sqrt(3)/4, m = 3/4. Two possible lines have direction ratios (sqrt(3)/4, 3/4, 1/2) and (-sqrt(3)/4, 3/4, 1/2). The dot product gives cos(theta) = (-3/16 + 9/16 + 1/4) = 0. Thus, the angle is 90 degrees.