Quantitative Aptitude
Trigonometry
435 Questions
Trigonometry Questions
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$8 \sqrt { 3 } \mathrm { mt }$
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$8 \sqrt { 2 } \mathrm { mt }$
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$8 \sqrt { 5 } \mathrm { mt }$
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$8 \mathrm { mt }$
A
Correct answer
Explanation
The angle of depression equals the angle of elevation from the boat to the cliff top. Using tan(60) = height / distance, we get sqrt(3) = 24 / distance. Distance = 24 / sqrt(3) = 8 * sqrt(3).
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$\dfrac { \pi }{ 2 } $
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$\dfrac { \pi }{ 6 } $
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$\dfrac { \pi }{ 12 } $
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$\dfrac { \pi }{ 18 } $
C
Correct answer
Explanation
Let tower height be h. tan(alpha) = h/x and tan(2*alpha) = h/(x-2h). Substituting x = h/tan(alpha) into the second equation: tan(2*alpha) = h / (h/tan(alpha) - 2h) = tan(alpha) / (1 - 2*tan(alpha)). Using tan(2*alpha) = 2*tan(alpha) / (1 - tan^2(alpha)), we solve for tan(alpha).
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$2h^{2} = a^{2} tan^{2} \theta$
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$a^{2} = 2h^{2}tan^{2} \theta$
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$a^{2} = 2h^{2}cot^{2} \theta$
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$2a^{2} = h^{2}cot^{2} \theta$
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$10$
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$30$
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$10\sqrt { 3 } $
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$30\sqrt { 3 } $
C
Correct answer
Explanation
The angle of depression is 60 degrees, so the angle of elevation from the ship is 60 degrees. tan(60) = height / distance. sqrt(3) = 30 / distance. distance = 30 / sqrt(3) = 10 * sqrt(3).
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$\displaystyle 30^{\circ}$
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$\displaystyle 45^{\circ}$
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$\displaystyle 60^{\circ}$
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any acute angle
B
Correct answer
Explanation
The angle of elevation is given by the tangent function, where tan(theta) = height / distance. Since the height equals the distance, tan(theta) = 1, which means theta = 45 degrees.
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$r cos \dfrac {\beta}{2}sec\alpha$
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$r cos \beta sec\dfrac {\alpha}{2}$
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$r sin \dfrac {\alpha}{2}cosec\beta$
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$r sin\beta cosec\dfrac {\alpha}{2}$
D
Correct answer
Explanation
Using trigonometry in the right triangle formed by the observer, the center of the balloon, and the ground, the height h = d * sin(beta). The angle alpha subtended at the eye relates to the radius r as sin(alpha/2) = r / d. Thus d = r / sin(alpha/2). Substituting d gives h = r * sin(beta) * cosec(alpha/2).
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$173.2 m$
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$196 m$
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$144 m$
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$123.2 m$
A
Correct answer
Explanation
Height = distance * tan(60). tan(60) = sqrt(3) approx 1.732. Height = 100 * 1.732 = 173.2 m.
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$ \displaystyle 250\sqrt{3} $ meter
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$ \displaystyle \frac{500}{\sqrt{3}} $
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$ \displaystyle 500\sqrt{3} $ meter
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250 meter
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$\displaystyle \sqrt{d_{1}d_{2}}$
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$\displaystyle \sqrt{d_{1}/d_{2}}$
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$\displaystyle d_{1}d_{2}$
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$\displaystyle d_{1}/d_{2}$
A
Correct answer
Explanation
If the height is h, the two tangent values are h/d1 and h/d2. Complementary angles have tangent values whose product is 1, so h^2/(d1d2) = 1 and h = sqrt(d1d2).
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$1.098$m
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$2.098$m
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$3.098$m
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None of these
A
Correct answer
Explanation
Let h = 1.5m. Distance d1 = h / tan(30) = 1.5 * sqrt(3) approx 2.598m. Distance d2 = h / tan(45) = 1.5 * 1 = 1.5m. The distance moved is d1 - d2 = 2.598 - 1.5 = 1.098m.
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$\sin { \theta } $
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$\cos { \theta } $
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$\tan { \theta } $
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$\cot { \theta } $
B
Correct answer
Explanation
In a right-angled triangle formed by the ladder, wall, and ground, the distance between the foot of the ladder and the wall is the adjacent side to angle theta, and the ladder is the hypotenuse. The ratio of adjacent/hypotenuse is cos(theta).
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$15$ m
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$22.5$ m
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$30$ m
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$7.5$ m
A
Correct answer
Explanation
The difference in height between the two poles is 80 - 65 = 15 m. If the line joining the tops makes a 45-degree angle with the horizontal, the triangle formed by the height difference and the horizontal distance is an isosceles right triangle. Thus, the horizontal distance equals the height difference, which is 15 m.
A
Correct answer
Explanation
This is a known identity in trigonometry regarding the elevation of a body moving at constant speed in a horizontal line.
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$a = b\, tan \dfrac{\alpha + \beta}{2}$
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$a = b\, cot \dfrac{\alpha + \beta}{2}$
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$a \, tan \dfrac{\alpha - \beta}{2}$
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None
A
Correct answer
Explanation
Using trigonometry, the ladder length L satisfies L cos(alpha) = x and L sin(alpha) = y. After sliding, L cos(beta) = x + a and L sin(beta) = y - b. By manipulating these equations, one can derive the relationship between the shift distances and the angles.