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

From a point on a horizontal plane, the elevation of the top of a hill is $45^o$. The elevation becomes $75^o$ after walking a distance of 500 m up a slope of inclined at an angle of $15^o$ to the horizon. The height of the hill is
  1. $500 \sqrt{6}$
  2. $500 \sqrt{3}$
  3. $250 \sqrt{6}$
  4. $250 \sqrt{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using trigonometry, the height h = 500 * sin(15) + (500 * cos(15) * tan(75)) / (tan(75) - tan(45)). This simplifies to 250 * sqrt(6).

Multiple choice

If a tower subtends angles $\theta$, $2\theta$ and $3\theta$ at three points $A, B$ and $C$ respectively, lying on the same side of a horizontal line through the foot of the tower, $\dfrac{AB}{BC}$ is equal to?
  1. $\displaystyle \frac{\sin 2 \theta}{\sin \theta} $
  2. $\displaystyle \frac{\sin 3 \theta}{\sin 2 \theta} $
  3. $\displaystyle \frac{\sin 3 \theta}{\sin \theta} $
  4. $\displaystyle \frac{\cot \theta - \cot 2 \theta}{\cot 2 \theta- \cot 3 \theta} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

In a triangle two sides are of length 20 and 21 and the sine of angle between them is equal to 0.6 then the third side is equal to
  1. $13\ or\ \sqrt{1513}$
  2. $14\ or\ \sqrt{1315}$
  3. $15\ or\ \sqrt{1531}$
  4. $17\ or\ \sqrt{1531}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Area = 1/2 * a * b * sin(C). 1/2 * 20 * 21 * 0.6 = 126. Using Law of Cosines: c^2 = a^2 + b^2 - 2ab*cos(C). sin(C) = 0.6, so cos(C) = +/- 0.8. c^2 = 20^2 + 21^2 - 2*20*21*(+/- 0.8) = 400 + 441 - 840*(+/- 0.8) = 841 - 672 = 169 (c=13) or 841 + 672 = 1513 (c=sqrt(1513)).

Multiple choice

In $\Delta$ $ABC$ the sides opposite to angles $A, B, C$ are denoted by $a, b, c$ respectively. Then $\dfrac { \cot { \dfrac { A }{ 2 } } +\cot { \dfrac { B }{ 2 } } +\cot { \dfrac { C }{ 2 } } }{ \cot { A } +\cot { B } +\cot { C } } =?$
  1. $\displaystyle \frac { 4{ s }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } $
  2. $\displaystyle \frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 2s } $
  3. $\displaystyle \frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 3s } $

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

Using standard trigonometric identities in a triangle, the sum of cot(A/2) is s/r and the sum of cot(A) is (a^2+b^2+c^2)/(4*Area). The ratio simplifies to the given expression involving s and side lengths.

Multiple choice

lf in $\Delta$ $ABC$ $\displaystyle \cot\frac{A}{2}$ : $\displaystyle \cot\frac{B}{2}$ : $\displaystyle \cot\frac{C}{2}=1:4:15$, then the greatest angle of triangle is
  1. $60^{0}$
  2. $90^{0}$
  3. $120^{0}$
  4. $135^{0}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given cot(A/2) : cot(B/2) : cot(C/2) = 1 : 4 : 15. This implies tan(A/2) : tan(B/2) : tan(C/2) = 1 : 1/4 : 1/15. Using the identity for triangles, this leads to angles where the largest angle is 120 degrees.

Multiple choice

In $\Delta$ $ABC$ the sides opposite to angles $A, B, C$ are denoted by $a, b, c$ respectively. If $\cot\dfrac{A}{2}=30,\ \displaystyle \cot\dfrac{B}{2}=50,\ \displaystyle \cot\frac{C}{2}=70$ then put the sides $a,\ b,\ c$ in ascending order
  1. $b,a,c$
  2. $a,b,c$
  3. $c,b,a$
  4. $a,c,b$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In a triangle, cot(A/2) = sqrt(s(s-a)/(s-b)(s-c)). Larger cotangent values imply smaller angles. Given cot(A/2)=30, cot(B/2)=50, cot(C/2)=70. A/2 > B/2 > C/2, so A > B > C. In any triangle, larger angles are opposite larger sides. Thus a > b > c. Ascending order: c, b, a.

Multiple choice

A vertical pole consists of two parts, the lower part being one third of the whole. At a point in the horizontal plane through the base of the pole and distance 20m from it, the upper part of the pole subtends an angle whose tangent is 1/2. The possible heights of the pole are -
  1. $20m$ and $20\sqrt{3}$m
  2. $20m$ and $60m$
  3. $16m$ and $48m$

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

Let the height of the pole be H. The lower part is H/3 and the upper part is 2H/3. Let the distance from the base be 20m. The angle of elevation to the top is alpha and to the top of the lower part is beta. We are given tan(alpha - beta) = 1/2. Using the tangent subtraction formula, (tan(alpha) - tan(beta)) / (1 + tan(alpha)tan(beta)) = 1/2, where tan(alpha) = H/20 and tan(beta) = (H/3)/20 = H/60. Solving this quadratic equation for H yields 20m and 60m.

Multiple choice

If $\displaystyle \frac{a}{\sin A}=K,$ then the area of $\displaystyle \Delta ABC$ in terms of K and sines of the angles is -
  1. $\displaystyle \frac{K^{2}}{4}\sin A\sin B\sin C$
  2. $\displaystyle \frac{K^{2}}{2}\sin A\sin B\sin C$
  3. $\displaystyle 2K^{2}\sin A\sin B\sin C\left ( A+B \right )$

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

By the Law of Sines, a/sinA = b/sinB = c/sinC = 2R = K. Area = 1/2 * b * c * sinA = 1/2 * (K sinB) * (K sinC) * sinA = K^2/2 * sinA * sinB * sinC.

Multiple choice

If the median AD of a triangle ABC divides the angle $\displaystyle \angle BAC$ in the ratio 1 :2, then $\dfrac{\sin B}{\sin C}$ is equal to
  1. $2 \cos \dfrac{A}{3}$
  2. $\dfrac12 \: \sec \dfrac A3$
  3. $\dfrac12 \: \sin \dfrac A3$
  4. $2 \: \mathrm{cosec} \dfrac A3 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the sine rule in triangles ABD and ACD, and the fact that AD bisects angle A into A/3 and 2A/3, we can relate the sides and angles. The ratio sin(B)/sin(C) simplifies to (AC/AB) * (sin(angle ADB)/sin(angle ADC)). Applying the angle bisector theorem and sine rule properties leads to the result 1/2 sec(A/3).