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

If the sides of a triangle are proportional to the cosine of the opposite angles, then the triangle is ___________.

  1. Right angled

  2. equilateral

  3. obtuse angled

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

By the Law of Cosines, cos(A) = (b^2 + c^2 - a^2) / (2bc). If sides are proportional to the cosines of opposite angles, a/cos(A) = b/cos(B) = c/cos(C). This condition is satisfied in an equilateral triangle where a=b=c and A=B=C=60 degrees.

Multiple choice

The angles of depression of two points from the top of the tower are $\displaystyle 30^{\circ}$ and $\displaystyle 60^{\circ}$ If the height of the tower is 30 m then find the maximum possible distance between the two points (in m)
  1. $\displaystyle 20\sqrt{3}$
  2. $\displaystyle 30\sqrt{3}$
  3. $\displaystyle 40\sqrt{3}$
  4. $\displaystyle 50\sqrt{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The distance of the two points from the base of the tower are d1 = 30/tan(30) = 30*sqrt(3) and d2 = 30/tan(60) = 30/sqrt(3) = 10*sqrt(3). The maximum distance between the two points occurs when they are on opposite sides of the tower, which is d1 + d2 = 30*sqrt(3) + 10*sqrt(3) = 40*sqrt(3).

Multiple choice

A ladder 20 m long is placed against a vertical wall of height 10 m, determine the distance between foot of the ladder and the wall and also the inclination of the ladder with the horizontal.
  1. $10\sqrt3, 30 ^\circ$
  2. $20\sqrt2, 20 ^\circ$
  3. $10\sqrt6, 60 ^\circ$
  4. $6\sqrt10, 10$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Pythagoras, distance^2 + 10^2 = 20^2, so distance^2 = 300, distance = 10*sqrt(3). Sin(theta) = 10/20 = 1/2, so theta = 30 degrees.

Multiple choice

The value of $\dfrac{\sin30^0+\tan45^0-\text{cosec}60^0}{\sec30^0+\cos60^0+\cot45^0}$ is:
  1. $\dfrac{43-24\sqrt3}{11}$
  2. $\dfrac{27-16-24\sqrt3}{27-16}$
  3. $\dfrac{43+24\sqrt3}{11}$
  4. $\text {both a and b}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Substituting values: sin30=1/2, tan45=1, cosec60=2/sqrt(3), sec30=2/sqrt(3), cos60=1/2, cot45=1. The numerator is (3/2 - 2/sqrt(3)) and the denominator is (3/2 + 2/sqrt(3)). Simplifying gives (3sqrt(3)-4)/(3sqrt(3)+4). Rationalizing results in (43-24sqrt(3))/11.

Multiple choice

If $\displaystyle x\cos 60^{\circ}+y\cos 0^{\circ}=3: : : and: : : 4x\sin 30^{\circ}-y\cot 45^{\circ}=2 $, then what is the value of $x$?
  1. $-1$
  2. $0$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Substituting the trigonometric values cos(60) = 1/2, cos(0) = 1, sin(30) = 1/2, and cot(45) = 1 into the given equations yields the system of linear equations: x/2 + y = 3 and 2x - y = 2. Solving this system by substituting y = 2x - 2 into the first equation gives x/2 + 2x - 2 = 3, which simplifies to 5x/2 = 5, resulting in x = 2.

Multiple choice

The angle of elevation of the top of a hill from each of the vertices$A,\ B,\ C$ of a horizontal triangle is $\alpha$. The height of the hill is where $a=BC,\ b=CA,\ c=AB$
  1. $\displaystyle \frac{1}{2} atan \alpha cosec A$
  2. $c\tan\alpha cosec C$
  3. $b\tan\alpha cosecB$
  4. $a \tan\alpha cosecA$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The hill's projection on the horizontal plane is equidistant from A, B, and C, so its distance from each vertex is the triangle's circumradius R. Since R = a/(2 sin A), the height is R tan alpha = (1/2)a tan alpha cosec A. This matches option A.

Multiple choice

A stationary balloon is observed from $3$ points $A,\ B$ and $C$ on the plane ground and is found that its angle of elevation from each point is $\alpha$. If $\angle ABC=\beta$ and $AC=b$, the height of the balloon is
  1. ${\dfrac{b}{2}\tan\alpha}\cdot\text{cosec }\beta$
  2. $\displaystyle \frac{b}{2}\tan\alpha\cdot\sin\beta$
  3. $\displaystyle \frac{b}{2}\cot\alpha\cdot\text{cosec }\beta$
  4. $\displaystyle \frac{b}{2}\cot\alpha\cdot\sin\beta$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let h be the height of the balloon. The distance from the projection of the balloon on the ground to each point A, B, C is r = h * cot(alpha). Points A, B, C lie on a circle of radius r. In triangle ABC, by the law of sines, b / sin(beta) = 2r. Substituting r, we get b / sin(beta) = 2 * h * cot(alpha). Solving for h gives h = b * tan(alpha) / (2 * sin(beta)). Note that 1/sin(beta) is cosec(beta).