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 $\delta _{1}$ and $\delta _{2}$ are the angles of dip observed in two vertical planes at right angles to each other and $\delta $ is the true value of dip then :
  1. $tan^{2}\delta =tan^{2}\delta _{1}+tan^{2}\delta _{2}$
  2. $cot^{2}\delta =cot^{2}\delta _{1}+cot^{2}\delta _{2}$
  3. $tan^{2}\delta =\dfrac{tan^{2}\delta _{1}+tan^{2}\delta _{2}}{tan^{2}\delta _{1}tan^{2}\delta _{2}}$
  4. $cot^{2}\delta =1+cot^{2}\delta _{1}cos^{2}\delta _{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The relation between the true dip delta and the apparent dips delta1 and delta2 in two mutually perpendicular vertical planes is given by cot^2(delta) = cot^2(delta1) + cot^2(delta2). This is a standard result in geomagnetism.

Multiple choice

The maximum distance between the transmitting and receiving TV towers is $D$. If the heights of both transmitting and receiving towers are doubled, then the maximum distance between them becomes
  1. $2D$
  2. $\sqrt{2}D$
  3. $4D$
  4. $D/2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The distance D is proportional to the square root of the height (D is proportional to sqrt(h)). If heights are doubled, the new distance is proportional to sqrt(2h) = sqrt(2) * sqrt(h), which is sqrt(2) * D.

Multiple choice

At certain distance from a transmitting tower, a receiver tower of height $20\ m$ is used to receive direct signal. Another tower is installed beyond the first, along the same line of sight to receive the signals from the same transmitter. Its height is $44$% more than the first receiving tower. Then, the separation between the two receiving towers is (Radius of Earth = $6400\ km $)
  1. $6.4\ km$
  2. $3.2\ km$
  3. $1.6\ km$
  4. $0.8\ km$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The distance d to the horizon is sqrt(2Rh). d1 = sqrt(2 * 6400 * 0.02) = sqrt(256) = 16 km. h2 = 20 * 1.44 = 28.8 m. d2 = sqrt(2 * 6400 * 0.0288) = sqrt(368.64) = 19.2 km. Separation = d2 - d1 = 19.2 - 16 = 3.2 km.

Multiple choice

A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in LOS mode?. Given that radius of the earth is $6.4 \times 10^{6}m$

  1. 45.5 km

  2. 25.5 km

  3. 85.2 km

  4. 33.5 km
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The maximum distance d = sqrt(2*R*h1) + sqrt(2*R*h2). d = sqrt(2 * 6.4*10^6 * 32) + sqrt(2 * 6.4*10^6 * 50) = sqrt(409.6*10^6) + sqrt(640*10^6) = 20200 + 25300 = 45500 m = 45.5 km.

Multiple choice

A TV tower has a height of $75\ m$. The maximum distance and area up to which this TV transmission can be received is (Radius of the Earth = $6400\ km$)
  1. $31\ km$, $3018\ {km}^{2}$
  2. $30\ km$, $3000\ {km}^{2}$
  3. $28\ km$, $2800\ {km}^{2}$
  4. $25\ km$, $2500\ {km}^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Distance d = sqrt(2Rh), where R = 6400 km and h = 0.075 km. d = sqrt(2 * 6400 * 0.075) = sqrt(960) approx 30.98 km. Area A = pi * d^2 = 3.14 * 960 approx 3014 sq km.

Multiple choice

If $\displaystyle\frac{\sec\theta - \tan\theta}{\sec\theta+\tan\theta} = \displaystyle\frac{36}{49}$, find $\displaystyle\frac{cosec\theta - \sec\theta}{cosec\theta + \sec\theta}$
  1. $\displaystyle\frac{71}{97}$
  2. $\displaystyle\frac{84}{85}$
  3. $\displaystyle\frac{84}{97}$
  4. $\displaystyle\frac{71}{85}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

(sec - tan)/(sec + tan) = 36/49. Using (sec^2 - tan^2) = 1, we have (sec - tan)^2 = 36/49, so sec - tan = 6/7. Then sec + tan = 49/6 / 7 = 7/6. Solving these gives sec = (6/7 + 7/6)/2 = 85/84 and tan = (7/6 - 6/7)/2 = 13/84. Using sin/cos and 1/cos, we find cosec = 85/13. The expression (cosec - sec)/(cosec + sec) = (85/13 - 85/84) / (85/13 + 85/84) = (1/13 - 1/84) / (1/13 + 1/84) = (84-13)/(84+13) = 71/97.

Multiple choice

The length of the sides of a right-angled triangle are in geometric progression. What is the ratio of the sines of its acute angles?
  1. $1$
  2. $\sqrt {3}$
  3. $\dfrac{\sqrt{5} + 1}{2}$
  4. $\sqrt{\dfrac{\sqrt 5 + 1}{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let sides be a/r, a, ar. By Pythagoras, (a/r)^2 + a^2 = (ar)^2. Dividing by a^2 gives 1/r^2 + 1 = r^2. Let x = r^2, then 1/x + 1 = x, so x^2 - x - 1 = 0. Solving for x gives r^2 = (1+sqrt(5))/2. The sines of the angles are 1/r and 1/r^2. The ratio is r = sqrt((1+sqrt(5))/2).

Multiple choice

From the top of a tower, the angle of depression of a point on the ground is $60^o$ . If the distance of this point from the tower is $ \dfrac {1}{\sqrt{3}+1}$ m, then the height of the tower is :
  1. $ \dfrac {4 \sqrt{3}}{2}$ m
  2. $ \dfrac { \sqrt{3} + 3}{2}$ m
  3. $ \dfrac {3 - \sqrt{3}}{2}$ m
  4. $ \dfrac { \sqrt{3} }{2}$ m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The height h = distance * tan(60). Given distance = 1 / (sqrt(3) + 1), h = (1 / (sqrt(3) + 1)) * sqrt(3). Rationalizing the denominator: sqrt(3) * (sqrt(3) - 1) / (3 - 1) = (3 - sqrt(3)) / 2.

Multiple choice

The top of a pole, placed against a wall at an angle $\alpha$ with the horizon just touches the coping and when its foot is moved a meter away from the wall and its angle of inclination is $\beta$, it rests on the sill of a window, the vertical distance of the sill from the coping is
  1. $a \sin\left ( \dfrac{\alpha +\beta }{2} \right )$
  2. $-a \cot \left ( \dfrac{\alpha +\beta }{2} \right )$
  3. $a \cot \left ( \dfrac{\alpha +\beta }{2} \right )$
  4. $a \tan \left ( \dfrac{\alpha +\beta }{2} \right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is a classic trigonometry problem involving a ladder or pole leaning against a wall, where the vertical displacement is calculated using the difference in heights at two different angles.

Multiple choice

If the length of sides of a right triangle are in A.P., then the sines of the acute angle are
  1. $\dfrac{3}{5}, \dfrac{4}{5}$
  2. $\sqrt{\dfrac{2}{3}}$ , $\sqrt{\dfrac{1}{3}}$
  3. $ \sqrt{\dfrac{\sqrt{5}-1}{2}}$ , $\sqrt{\dfrac{\sqrt{5}-1}{2}}$
  4. $ \sqrt{\dfrac{\sqrt{3}-1}{2}}$ , $\sqrt{\dfrac{\sqrt{3}-1}{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the sides be a-d, a, a+d. By Pythagoras, (a-d)^2 + a^2 = (a+d)^2. a^2 - 2ad + d^2 + a^2 = a^2 + 2ad + d^2. a^2 = 4ad, so a = 4d. Sides are 3d, 4d, 5d. Sines of acute angles are 3/5 and 4/5.

Multiple choice

From the top of the tree, a man observes the angle of depression of a point which is at a distance of $40\ m$ from the foot is $75^{0}$. The height of the tree is :
  1. $40\sqrt{3}\ m$
  2. $40(2+\sqrt{3})\ m$
  3. $21\sqrt{3}\ m$
  4. $3\sqrt{21}\ m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Height h = 40 * tan(75 degrees). tan(75) = tan(45+30) = (1 + 1/sqrt(3)) / (1 - 1/sqrt(3)) = (sqrt(3)+1)/(sqrt(3)-1) = (sqrt(3)+1)^2 / 2 = (3+1+2sqrt(3))/2 = 2 + sqrt(3). Height = 40(2 + sqrt(3)).

Multiple choice

In a right angled triangle, $\cot{\theta} =$
  1. $\dfrac{hypotenuse}{perpendicular}$
  2. $\dfrac{perpendicular}{hypotenuse}$
  3. $\dfrac{perpendicular}{base}$
  4. $\dfrac{base}{perpendicular}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a right-angled triangle, cot(theta) is defined as the ratio of the adjacent side (base) to the opposite side (perpendicular).