Veda Skills Quantitative Aptitude Unit 16: Trigonometry
- 2 topics
- 125 MCQs
- 7 short answers
- 2 long answers
- 28 flashcards
- 2 games
- ~120 min read
- English
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Trigonometric Ratios and Identities
The six trigonometric ratios in a right triangle, standard values at 0°, 30°, 45°, 60° and 90°, the three Pythagorean identities, complementary angles, and the maximum and minimum questions of harder exams.
What you will be able to do
- I can write all six trigonometric ratios from the sides of a right triangle.
- I can find all ratios when one ratio is given.
- I can evaluate expressions using the values at 0°, 30°, 45°, 60° and 90°.
- I can simplify using the three identities and complementary angles.
- I can find the maximum and minimum of simple trigonometric expressions.
Key points
$\sin\theta$ = opposite/hypotenuse, $\cos\theta$ = adjacent/hypotenuse, $\tan\theta$ = opposite/adjacent.
$\operatorname{cosec}\theta$, $\sec\theta$, $\cot\theta$ are the reciprocals of $\sin\theta$, $\cos\theta$, $\tan\theta$.
$\tan\theta = \frac{\sin\theta}{\cos\theta}$.
$\sin 30^\circ = \frac{1}{2}$, $\sin 45^\circ = \frac{1}{\sqrt{2}}$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$; cosines run in reverse.
$\tan 30^\circ = \frac{1}{\sqrt{3}}$, $\tan 45^\circ = 1$, $\tan 60^\circ = \sqrt{3}$.
$\sin^2\theta + \cos^2\theta = 1$, $\sec^2\theta - \tan^2\theta = 1$, $\operatorname{cosec}^2\theta - \cot^2\theta = 1$.
$\sin(90^\circ - \theta) = \cos\theta$, $\tan(90^\circ - \theta) = \cot\theta$, $\sec(90^\circ - \theta) = \operatorname{cosec}\theta$.
If $\sec\theta + \tan\theta = k$, then $\sec\theta - \tan\theta = \frac{1}{k}$.
The greatest value of $a\sin\theta + b\cos\theta$ is $\sqrt{a^2 + b^2}$ and the least is $-\sqrt{a^2 + b^2}$.
Full explanation
Meena's ramp
Meena is building a ramp 13 m long that rises 5 m. The carpenter asks: "What is the angle, and how far along the ground does it run?" Both answers come from the ratios of the sides of a right triangle. Those ratios are trigonometry.
Naming the sides
Mark the angle $\theta$. The side across from it is the opposite, the side next to it (not the hypotenuse) is the adjacent, and the longest side, across from the right angle, is the hypotenuse.
$$\sin\theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos\theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan\theta = \frac{\text{opp}}{\text{adj}}$$
The other three ratios
$\displaystyle \operatorname{cosec}\theta = \frac{1}{\sin\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$
So you only ever need to find three; flip them for the rest.
One ratio gives all
If $\tan\theta = \frac{5}{12}$, draw a triangle with opposite 5 and adjacent 12. Pythagoras gives the hypotenuse 13, and now every ratio is read off: $\sin\theta = \frac{5}{13}$, $\cos\theta = \frac{12}{13}$. Learn the triples 3-4-5, 5-12-13, 8-15-17, 7-24-25.
Standard angles
Halve an equilateral triangle of side 2 to get a 30-60-90 triangle with sides $1, \sqrt{3}, 2$. Halve a unit square along its diagonal to get a 45-45-90 triangle with sides $1, 1, \sqrt{2}$. Every standard value is a ratio of these sides.
The three identities
Divide $\text{opp}^2 + \text{adj}^2 = \text{hyp}^2$ by $\text{hyp}^2$, by $\text{adj}^2$ and by $\text{opp}^2$:
$\displaystyle \sin^2\theta + \cos^2\theta = 1, \quad 1 + \tan^2\theta = \sec^2\theta, \quad 1 + \cot^2\theta = \operatorname{cosec}^2\theta$
A useful consequence: $(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1$.
Complementary angles
In a right triangle the two acute angles add to 90°, and the opposite side of one is the adjacent side of the other. So
$\displaystyle \sin(90^\circ - \theta) = \cos\theta, \quad \tan(90^\circ - \theta) = \cot\theta, \quad \sec(90^\circ - \theta) = \operatorname{cosec}\theta$
That is why $\frac{\sin 37^\circ}{\cos 53^\circ} = 1$ and $\tan 10^\circ \tan 80^\circ = 1$.
Greatest and least values
$a\sin\theta + b\cos\theta$ swings between $-\sqrt{a^2 + b^2}$ and $\sqrt{a^2 + b^2}$. So $3\sin\theta + 4\cos\theta$ is at most 5. For a sum like $9\tan^2\theta + 4\cot^2\theta$, AM ≥ GM gives the least value $2\sqrt{9 \times 4} = 12$.
Meena's answers
Opposite 5, hypotenuse 13: $\sin\theta = \frac{5}{13}$. The ground run is the adjacent side, $\sqrt{169 - 25} = 12$ m, and $\tan\theta = \frac{5}{12}$. A whole topic built on one triangle and Pythagoras.
Concepts and terms
A ratio of two sides of a right triangle, fixed by the angle alone and not by the size of the triangle.
An equation true for every angle, such as sin²θ + cos²θ = 1, which is Pythagoras divided by the hypotenuse squared.
Two angles adding to 90°. The two acute angles of a right triangle are complementary, so the sine of one is the cosine of the other.
sin θ and cos θ lie between −1 and 1; sec θ and $\operatorname{cosec}$ θ are never between −1 and 1; tan θ can be any number.
Did you know
The word "sine" comes from Aryabhata's "jya" (bow-string), which travelled through Arabic "jiba" and was misread in Latin as "sinus".
Memory tricks
Three basic ratios — SOH-CAH-TOA — Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent.
Sine values — √0/2, √1/2, √2/2, √3/2, √4/2 — For 0°, 30°, 45°, 60°, 90°. Read the same list backwards for cosine.
Practice MCQs with answers (68)
- A 5/12
- B 5/13
- C 13/5
- D 12/13
Show answer
Why:
AC = √(5² + 12²) = 13. $\sin\theta$ = AB/AC = 5/13.
- A 3/4
- B 5/3
- C 4/5
- D 3/5
Show answer
Why:
BC = √(5² − 3²) = 4. $\cos\theta$ = BC/AC = 4/5.
- A 8/17
- B 15/17
- C 17/8
- D 8/15
Show answer
Why:
AB = √(17² − 15²) = 8. $\tan\theta$ = AB/BC = 8/15.
- A 25/7
- B 24/7
- C 24/25
- D 7/25
Show answer
Why:
AC = √(24² + 7²) = 25. $\sec\theta$ = AC/BC = 25/7.
- A 20/21
- B 21/20
- C 20/29
- D 21/29
Show answer
Why:
BC = √(29² − 20²) = 21. $\cot\theta$ = BC/AB = 21/20.
- A 9/40
- B 9/41
- C 41/9
- D 40/41
Show answer
Why:
AB = √(41² − 40²) = 9. $\operatorname{cosec}\theta$ = AC/AB = 41/9.
- A 13/12
- B 12/13
- C 12/5
- D 5/13
Show answer
Why:
AC = √(12² + 5²) = 13. $\cos\theta$ = BC/AC = 5/13.
- A 15/8
- B 15/17
- C 8/17
- D 17/15
Show answer
Why:
BC = √(17² − 15²) = 8. $\tan\theta$ = AB/BC = 15/8.
- A 25/7
- B 7/25
- C 24/25
- D 7/24
Show answer
Why:
AB = √(25² − 24²) = 7. $\sin\theta$ = AB/AC = 7/25.
- A 4/3
- B 4/5
- C 5/4
- D 3/5
Show answer
Why:
AC = √(4² + 3²) = 5. $\operatorname{cosec}\theta$ = AC/AB = 5/4.
- A 13/5
- B 13/12
- C 12/13
- D 5/13
Show answer
Why:
Sides 5, 12, 13 (opposite, adjacent, hypotenuse). $\sin\theta$ = 5/13.
- A 3/4
- B 5/3
- C 5/4
- D 4/5
Show answer
Why:
Sides 3, 4, 5 (opposite, adjacent, hypotenuse). $\tan\theta$ = 3/4.
- A 17/8
- B 17/15
- C 15/17
- D 15/8
Show answer
Why:
Sides 15, 8, 17 (opposite, adjacent, hypotenuse). $\operatorname{cosec}\theta$ = 17/15.
- A 7/24
- B 7/25
- C 24/7
- D 24/25
Show answer
Why:
Sides 7, 24, 25 (opposite, adjacent, hypotenuse). $\cot\theta$ = 24/7.
- A 20/21
- B 29/20
- C 21/29
- D 20/29
Show answer
Why:
Sides 20, 21, 29 (opposite, adjacent, hypotenuse). $\sin\theta$ = 20/29.
- A 5
- B 1/5
- C 3
- D -5
Show answer
Why:
Divide top and bottom by $\cos\theta$: (2 × 3/4 + 1) ÷ (2 × 3/4 − 1) = 10/2 = 5.
- A 8/39
- B 39/8
- C 5/3
- D -9/32
Show answer
Why:
Divide top and bottom by $\cos\theta$: (3 × 5/12 + 2) ÷ (4 × 5/12 − 1) = 39/8.
- A 2
- B 1/11
- C 7/5
- D 5/7
Show answer
Why:
Divide top and bottom by $\cos\theta$: (1 × 4/3 + 1) ÷ (2 × 4/3 − 1) = 7/5.
- A 2
- B 1/2
- C 3/2
- D 1
Show answer
Why:
Put in the standard values: 1/2 + 1/2 = 1.
- A 1
- B 3/2
- C 2
- D 1/2
Show answer
Why:
Put in the standard values: 3/4 + 1/4 = 1.
- A 3
- B 2
- C 3/2
- D 4
Show answer
Why:
Put in the standard values: 3 − 2 × 1/2 = 2.
- A 2
- B 1/2
- C 1
- D 3/2
Show answer
Why:
Put in the standard values: 2 × 1/4 + 3 × 1/3 − 1/2 = 1.
- A 10/3
- B 8/3
- C 7/6
- D 5/3
Show answer
Why:
Put in the standard values: 4 × 1/3 + 4/3 − 2 × 1/2 = 5/3.
- A 3
- B 4
- C 5/2
- D 6
Show answer
Why:
Put in the standard values: 4 − 3 + 2 = 3.
- A 2
- B 1
- C 1/2
- D 3/2
Show answer
Why:
Put in the standard values: 3 × 1/3 + 1 − 1 = 1.
- A 7/2
- B 2
- C 5/2
- D 5
Show answer
Why:
Put in the standard values: 2 × 1/4 + 3 × 4/3 − 2 × 1 = 5/2.
- A 4
- B 3
- C 3/2
- D 2
Show answer
Why:
Put in the standard values: 1/2 + 1/2 + 1 = 2.
- A 2
- B 3/2
- C 4
- D 3
Show answer
Why:
Put in the standard values: 2 − 1 + 2 × 1/2 = 2.
- A 1
- B √3/3
- C 0.5
- D √3
Show answer
Why:
(√3 − 1/√3) ÷ (1 + 1) = (2/√3) ÷ 2 = 1/√3 = √3/3. (It is tan 30°.)
- A 0.75
- B 0.5
- C 1
- D √3/2
Show answer
Why:
(√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1. (It is sin 90°.)
- A 30°
- B 45°
- C 90°
- D 60°
Show answer
Why:
sin θ = √3/2, so θ = 60°.
- A 30°
- B 90°
- C 60°
- D 45°
Show answer
Why:
tan θ = 1/√3, so θ = 30°.
- A 90°
- B 45°
- C 30°
- D 60°
Show answer
Why:
cos²θ = 1/2, cos θ = 1/√2, θ = 45°.
- A 60°
- B 0°
- C 45°
- D 30°
Show answer
Why:
tan θ = 1, so θ = 45°.
- A 15°
- B 45°
- C 60°
- D 30°
Show answer
Why:
2θ = 60°, so θ = 30°.
- A 12/25
- B 37/25
- C 24/25
- D 49/25
Show answer
Why:
Square: 1 + 2 $\sin\theta\cos\theta$ = 49/25, so $\sin\theta\cos\theta$ = (49/25 − 1)/2 = 12/25.
- A 229/169
- B 60/169
- C 120/169
- D 289/169
Show answer
Why:
Square: 1 + 2 $\sin\theta\cos\theta$ = 289/169, so $\sin\theta\cos\theta$ = (289/169 − 1)/2 = 60/169.
- A 3/4
- B 3/5
- C 1/2
- D 5/4
Show answer
Why:
$(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1$, so $\sec\theta - \tan\theta$ = 1/2.
- A 5/3
- B 4/3
- C 1/3
- D 4/5
Show answer
Why:
$(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1$, so $\sec\theta - \tan\theta$ = 1/3. Adding and halving: $\sec\theta$ = 5/3, so $\tan\theta$ = 4/3 and $\sin\theta$ = $\tan\theta$ ÷ $\sec\theta$ = 4/5.
- A 21/20
- B 29/20
- C 21/29
- D 2/5
Show answer
Why:
$(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1$, so $\sec\theta - \tan\theta$ = 2/5. Adding and halving: $\sec\theta$ = 29/20, so $\tan\theta$ = 21/20.
- A 26°
- B 29°
- C 32°
- D 34°
Show answer
Why:
Complementary: (3A) + (A - 26°) = 90°, so 4A = 116°, A = 29°.
- A 33°
- B 72°
- C 36°
- D 41°
Show answer
Why:
Complementary: (2A) + (A - 18°) = 90°, so 3A = 108°, A = 36°.
- A 19°
- B 32°
- C 27°
- D 22°
Show answer
Why:
Complementary: (4A) + (A - 20°) = 90°, so 5A = 110°, A = 22°.
- A 20°
- B 17°
- C 30°
- D 25°
Show answer
Why:
Complementary: (2A + 10°) + (3A - 20°) = 90°, so 5A = 100°, A = 20°.
- A 2
- B 1
- C 0
- D 37/53
Show answer
Why:
cos 53° = sin 37°, so the ratio is 1.
- A 0
- B √3
- C 1
- D 1/√3
Show answer
Why:
tan 75° = cot 15°, tan 65° = cot 25°; each pair multiplies to 1.
- A 0
- B 2
- C 1/2
- D 1
Show answer
Why:
sin 70° = cos 20°, so the sum is sin²20° + cos²20° = 1.
- A 0
- B 2
- C −1
- D 1
Show answer
Why:
sec 59° = $\operatorname{cosec}$ 31°, so the difference is 0.
- A −1
- B 0
- C 1
- D 1/2
Show answer
Why:
The product includes cos 90° = 0.
- A $\cot^2\theta$
- B $\sin^2\theta$
- C $\tan^2\theta$
- D $1$
Show answer
Why:
sin²θ × 1/cos²θ = tan²θ.
- A $0$
- B $1$
- C $4\sin\theta\cos\theta$
- D $2$
Show answer
Why:
The cross terms cancel: 2(sin²θ + cos²θ) = 2.
- A $2\operatorname{cosec}\theta$
- B $2\sin\theta$
- C $\operatorname{cosec}\theta$
- D $2\sec\theta$
Show answer
Why:
Numerator sin²θ + 1 + 2cos θ + cos²θ = 2(1 + cos θ); dividing by sin θ(1 + cos θ) gives 2/sin θ.
- A $\sec^2\theta - \tan^2\theta$
- B $\sec^2\theta + \tan^2\theta$
- C $2\tan^2\theta$
- D $1$
Show answer
Why:
(sec²θ − tan²θ)(sec²θ + tan²θ) = 1 × (sec²θ + tan²θ).
- A $\operatorname{cosec}\theta - \cot\theta$
- B $1 - \tan\theta$
- C $\sec\theta - \tan\theta$
- D $\sec\theta + \tan\theta$
Show answer
Why:
Multiply inside by (1 − sin θ)/(1 − sin θ): √((1 − sin θ)²/cos²θ) = (1 − sin θ)/cos θ = sec θ − tan θ.
- A $\tan\theta = 5$
- B $\sec\theta = \frac{5}{4}$
- C $\cos\theta = \frac{1}{5}$
- D $\sin\theta = \frac{5}{4}$
Show answer
Why:
sin θ cannot exceed 1. tan θ can be any number; sec θ ≥ 1 is fine.
- A $2$
- B $0$
- C $1$
- D $4$
Show answer
Why:
Square: tan²θ + cot²θ + 2 = 4, so the answer is 2.
- A 4
- B 5
- C 25
- D 7
Show answer
Why:
Greatest value = √(3² + 4²) = √25 = 5. (Not 7: sin and cos cannot both be 1 at once.)
- A 169
- B 17
- C 13
- D 12
Show answer
Why:
Greatest value = √(5² + 12²) = √169 = 13. (Not 17: sin and cos cannot both be 1 at once.)
- A −23
- B −15
- C 0
- D −17
Show answer
Why:
The range of a sin θ + b cos θ is from −√(a² + b²) to √(a² + b²); here √(64 + 225) = 17, so the least is −17.
- A 12
- B 36
- C 13
- D 6
Show answer
Why:
AM ≥ GM: 9tan²θ + 4cot²θ ≥ 2√(9 × 4) = 12, reached when 9tan²θ = 4cot²θ, that is tan²θ = 2/3.
- A 400
- B 40
- C 41
- D 20
Show answer
Why:
AM ≥ GM: 16tan²θ + 25cot²θ ≥ 2√(16 × 25) = 40, reached when 16tan²θ = 25cot²θ, that is tan²θ = 5/4.
- A 2
- B 0.25
- C 0.5
- D 1
Show answer
Why:
sin θ cos θ = ½ sin 2θ, and sin 2θ is at most 1, so the greatest value is ½ = 0.5, reached at θ = 45°.
- A 300°
- B 120°
- C 135°
- D 150°
Show answer
Why:
π rad = 180°, so 5π/6 = 5 × 30° = 150°.
- A $\frac{3\pi}{4}$
- B $\frac{3\pi}{2}$
- C $\frac{2\pi}{3}$
- D $\frac{5\pi}{6}$
Show answer
Why:
135 × π/180 = 3π/4.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
Both equal √3/2, and 60° = 90° − 30°, so R explains A.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
R is true, but sin and cos are not 1 at the same θ. The greatest value is √(9 + 16) = 5, so A is false.
- A 1 only
- B 2 only
- C Both 1 and 2
- D Neither 1 nor 2
Show answer
Why:
1 + 1 = 2, true. 1/2 + √3/2 ≈ 1.37 ≠ 1, false.
- A 1 only
- B 2 only
- C Both 1 and 2
- D Neither 1 nor 2
Show answer
Why:
The identity is sec²θ − tan²θ = 1, so 1 is false; 2 is a true identity.
Important short questions (4)
Opposite = √(289 − 64) = 15. sin θ = 15/17, tan θ = 15/8, $\operatorname{cosec}$ θ = 17/15, sec θ = 17/8, cot θ = 8/15.
Numerator 3/2 − 2/√3, denominator 3/2 + 2/√3; multiply by 2√3: (3√3 − 4)/(3√3 + 4) = (43 − 24√3)/11.
Put t = tan θ: t²/(t − 1) − 1/(t(t − 1)) = (t³ − 1)/(t(t − 1)) = (t² + t + 1)/t = 1 + t + 1/t = 1 + (sin²θ + cos²θ)/(sin θ cos θ) = 1 + sec θ $\operatorname{cosec}$ θ.
A + B = 90°, A − B = 30°, so A = 60°, B = 30°.
Long answer questions (1)
(a) Sides 3, 4, 5: sin θ = 3/5, cos θ = 4/5.
(b) 9/25 + 16/25 = 25/25 = 1.
(c) Divide by cos θ: (2 × 3/4 − 1)/(2 × 3/4 + 1) = (1/2)/(5/2) = 1/5.
Flashcards (15)
Play this topic
- Ratio check
Heights and Distances
Angles of elevation and depression, towers, shadows, ladders, broken trees, two observation points, two buildings, flagstaffs and moving objects, all solved with tan 30°, 45° and 60° and drawn to scale.
What you will be able to do
- I can draw the right triangle hidden in a heights-and-distances question.
- I can find a height or a distance from one angle of elevation or depression.
- I can solve two-angle problems: two points, two buildings, a flagstaff on a building.
- I can solve ladder and broken-tree questions using sin and cos.
- I can find speeds of moving objects from changing angles.
Key points
The angle of elevation is measured up from the horizontal; the angle of depression is measured down from the horizontal.
The angle of depression from A to B equals the angle of elevation from B to A (alternate angles).
Height = distance × tan θ; distance = height × cot θ.
At 45° the height equals the distance; at 60° the height is √3 times it; at 30° the distance is √3 times the height.
Two points on the same side at 60° and 30°, a apart: height = a√3/2.
A ladder of length L at angle θ to the ground reaches height L sin θ, with its foot L cos θ from the wall.
A tree broken so that its top meets the ground at 30°, at distance d from the foot, was d√3 tall.
Use √3 ≈ 1.732 only when the question asks for a decimal.
Full explanation
Arjun's water tank
Arjun stands 20 m from the village water tower and looks up at its top. His protractor app shows 60°. Without climbing, he can find the height. A few minutes later he walks back and the angle drops to 30°. How far did he walk?
Elevation
The angle of elevation is the angle between the horizontal and the line of sight to an object above you. Tower, ground and line of sight make a right triangle, so
$$\tan\theta = \frac{\text{height}}{\text{distance}}$$
Depression
The angle of depression is measured down from the horizontal at the observer's eye to an object below. Because the horizontal at the top is parallel to the ground, the depression from the top equals the elevation from the bottom.
The three working values
| θ | 30° | 45° | 60° |
|---|---|---|---|
| $\tan\theta$ | $\frac{1}{\sqrt{3}}$ | 1 | $\sqrt{3}$ |
At 45° height = distance. At 60° height = $\sqrt{3}$ × distance. At 30° distance = $\sqrt{3}$ × height.
Two points on one side
From C the angle is 60°, from D (farther, a metres behind) it is 30°. Then $BC = \frac{h}{\sqrt{3}}$ and $BD = h\sqrt{3}$, so
$$h\sqrt{3} - \frac{h}{\sqrt{3}} = a \Rightarrow h = \frac{a\sqrt{3}}{2}$$
Ladders and broken trees
When the slanted length is involved, use sin and cos. A ladder of length $L$ at $\theta$ reaches $L\sin\theta$ up the wall, with its foot $L\cos\theta$ out.
A tree broken at height $s$ whose top meets the ground $d$ from the foot at angle $\theta$: $s = d\tan\theta$, broken part $= \frac{d}{\cos\theta}$; the full height is their sum.
Two buildings
From the roof of a building of height $h$, the top of a taller tower is seen at elevation $\alpha$ and its foot at depression $\beta$. Draw the horizontal from the eye: it splits the tower into $h$ (below) and $d\tan\alpha$ (above), where $d = h\cot\beta$.
Moving objects
A plane flying level at height $h$ is seen at 60°, and $t$ seconds later at 30°. It covered $h\cot 30^\circ - h\cot 60^\circ = \frac{2h}{\sqrt{3}}$, so its speed is $\frac{2h}{\sqrt{3},t}$.
Arjun's answers
Height $= 20\tan 60^\circ = 20\sqrt{3} \approx 34.6$ m. At 30° he is $20\sqrt{3} \times \sqrt{3} = 60$ m from the foot, so he walked $60 - 20 = 40$ m. One triangle per angle, and tan does the rest.
Concepts and terms
The straight line from the observer's eye to the object. With the horizontal it makes the angle of elevation or depression.
Tower, ground and line of sight form a right triangle with the right angle at the foot of the tower.
The horizontal at the top is parallel to the ground, so the angle of depression at the top equals the angle of elevation at the bottom.
When two angles are given, write tan for each triangle and eliminate the unknown distance.
Did you know
In 1856 the Great Trigonometrical Survey of India announced the height of Peak XV (Mount Everest) as 8840 m, worked out from angles of elevation taken about 170 km away. Today's value, 8849 m, is only 9 m more.
Memory tricks
Elevation vs depression — Elevation: eyes up. Depression: eyes down. Both from the horizontal. — Never measure from the vertical tower.
The 60-30 shortcut — Gap times √3 over 2 — Angles 60° and 30° from two points a apart give height a√3/2.
Practice MCQs with answers (57)
- A 20√3
- B 20√3/3
- C 10√3
- D 40
Show answer
Why:
h = 20 × tan 60° = 20√3 m.
- A 30√3
- B 30
- C 10√3
- D 60
Show answer
Why:
h = 30 × tan 45° = 30 m.
- A 90
- B 45√3
- C 15√3
- D 22.5
Show answer
Why:
h = 45 × tan 30° = 15√3 m.
- A 22.5
- B 30√3
- C 15
- D 45
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Why:
h = 15√3 × tan 60° = 45 m.
- A 50√3/3
- B 100
- C 50√3
- D 25
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Why:
h = 50 × tan 30° = 50√3/3 m.
- A 60
- B 30√3
- C 30
- D 10√3
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Why:
d = 30 ÷ tan 30° = 30 × cot 30° = 30√3 m.
- A 40√3
- B 20√3
- C 20
- D 60
Show answer
Why:
d = 20√3 ÷ tan 60° = 20√3 × cot 60° = 20 m.
- A 12√3
- B 24
- C 6√3
- D 12
Show answer
Why:
d = 12 ÷ tan 45° = 12 × cot 45° = 12 m.
- A 8
- B 8√3
- C 24
- D 16√3
Show answer
Why:
d = 8√3 ÷ tan 60° = 8√3 × cot 60° = 8 m.
- A 90°
- B 60°
- C 30°
- D 45°
Show answer
Why:
tan θ = 10√3/10 = √3, so θ = 60°.
- A 30°
- B 60°
- C 45°
- D 90°
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Why:
tan θ = 15/15 = 1, so θ = 45°.
- A 90°
- B 45°
- C 60°
- D 30°
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Why:
tan θ = 6/6√3 = √3/3, so θ = 30°.
- A 5√3
- B 5
- C 10√3
- D 10
Show answer
Why:
h = 10 sin 60° = 5√3 m.
- A 4√3
- B 6
- C 12
- D 6√3
Show answer
Why:
d = 12 cos 60° = 6 m.
- A 8
- B 4√3
- C 4
- D 8√3/3
Show answer
Why:
h = 8 sin 30° = 4 m.
- A 14√3
- B 14
- C 7
- D 7√3
Show answer
Why:
d = 14 cos 30° = 7√3 m.
- A 60√3
- B 60
- C 20√3
- D 120
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Why:
Depression at the top = elevation at the bottom = 30°. d = 60 × cot 30° = 60√3 m.
- A 150
- B 75
- C 75√3
- D 25√3
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Why:
Depression at the top = elevation at the bottom = 45°. d = 75 × cot 45° = 75 m.
- A 50√3
- B 150
- C 50
- D 100√3
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Why:
Depression at the top = elevation at the bottom = 60°. d = 50√3 × cot 60° = 50 m.
- A 80
- B 40
- C 40√3/3
- D 40√3
Show answer
Why:
Depression at the top = elevation at the bottom = 30°. d = 40 × cot 30° = 40√3 m.
- A 10√3
- B 20√3
- C 20√3/3
- D 20
Show answer
Why:
BC = h cot 60°, BD = h cot 30°; BD − BC = 20 gives h = 20 ÷ (cot 30° − cot 60°) = 10√3 m.
- A 20√3
- B 20
- C 60
- D 80
Show answer
Why:
h = 20√3 (from the two triangles); BC = h cot 60° = 20 m.
- A 40√3/3
- B 40(√3 + 1)
- C 20(√3 + 1)
- D 40
Show answer
Why:
BC = h cot 45°, BD = h cot 30°; BD − BC = 40 gives h = 40 ÷ (cot 30° − cot 45°) = 20(√3 + 1) m.
- A 30(√3 + 3)
- B 30√3
- C 30
- D 15(√3 + 3)
Show answer
Why:
BC = h cot 60°, BD = h cot 45°; BD − BC = 30 gives h = 30 ÷ (cot 45° − cot 60°) = 15(√3 + 3) m.
- A 25√3
- B 50√3/3
- C 50
- D 50√3
Show answer
Why:
BC = h cot 60°, BD = h cot 30°; BD − BC = 50 gives h = 50 ÷ (cot 30° − cot 60°) = 25√3 m.
- A 20
- B 10√3
- C 10√3/3
- D 20√3/3
Show answer
Why:
Standing part = 10 tan 30° = 10√3/3; broken part = 10 ÷ cos 30° = 20√3/3; total = 10√3 m.
- A 12
- B 12√3
- C 6(√3 + 2)
- D 6√3
Show answer
Why:
Standing part = 6 tan 60° = 6√3; broken part = 6 ÷ cos 60° = 12; total = 6(√3 + 2) m.
- A 18√3
- B 9
- C 18
- D 27
Show answer
Why:
Standing part = 9√3 tan 30° = 9; broken part = 9√3 ÷ cos 30° = 18; total = 27 m.
- A 20(√3 + 1)
- B 20(√3 + 2)
- C 20(√3 − 1)
- D 20√3
Show answer
Why:
Gap = 20 cot 45° = 20. Part above eye = 20 tan 60° = 20√3. Tower = 20 + 20√3 = 20(√3 + 1) m.
- A 45
- B 60
- C 75
- D 30
Show answer
Why:
Gap = 15 cot 30° = 15√3. Part above eye = 15√3 tan 60° = 45. Tower = 15 + 45 = 60 m.
- A 10
- B 70
- C 40
- D 80
Show answer
Why:
Gap = 30 cot 60° = 10√3. Part above eye = 10√3 tan 30° = 10. Tower = 30 + 10 = 40 m.
- A 8√3
- B 48
- C 48√3
- D 24√3
Show answer
Why:
d = 24 cot 30° = 24√3 m.
- A 20√3
- B 40√3
- C 40
- D 60
Show answer
Why:
h cot 30° + h cot 60° = 80, so h = 80 ÷ (√3 + √3/3) = 20√3 m.
- A 50√3
- B 50
- C 100
- D 25
Show answer
Why:
h cot 45° + h cot 45° = 100, so h = 100 ÷ (1 + 1) = 50 m.
- A 30
- B 30(3 − √3)
- C 30(√3 − 1)
- D 60(√3 − 1)
Show answer
Why:
h cot 30° + h cot 45° = 60, so h = 60 ÷ (√3 + 1) = 30(√3 − 1) m.
- A 40(√3 − 1)
- B 20√3
- C 20
- D 20(√3 − 1)
Show answer
Why:
Building = 20 tan 45° = 20; building + flag = 20 tan 60° = 20√3; flagstaff = 20(√3 − 1) m.
- A 20√3
- B 40√3
- C 30√3
- D 10√3
Show answer
Why:
Building = 30 tan 30° = 10√3; building + flag = 30 tan 60° = 30√3; flagstaff = 20√3 m.
- A 6(3 − √3)
- B 6√3
- C 36
- D 18
Show answer
Why:
Building = 18 tan 30° = 6√3 m.
- A 360
- B 1440
- C 720
- D 200
Show answer
Why:
Distance = h(cot 30° − cot 60°) = 1500√3 × 2/√3 = 3000 m in 15 s = 200 m/s = 720 km/h.
- A 216
- B 864
- C 120
- D 432
Show answer
Why:
Distance = h(cot 30° − cot 60°) = 600√3 × 2/√3 = 1200 m in 10 s = 120 m/s = 432 km/h.
- A 40
- B 60
- C 20
- D 80
Show answer
Why:
20√3 cot 30° − 20√3 cot 60° = 60 − 20 = 40 m.
- A 30√3
- B 30(√3 − 1)
- C 30
- D 60(√3 − 1)
Show answer
Why:
30 cot 30° − 30 cot 45° = 30√3 − 30 = 30(√3 − 1) m.
- A 30
- B 33
- C 31.5
- D 28.5
Show answer
Why:
Above eye level = 30 tan 45° = 30; add 1.5: 31.5 m.
- A 87.2
- B 82.4
- C 84
- D 85.6
Show answer
Why:
Above eye level = 28√3 tan 60° = 84; add 1.6: 85.6 m.
- A 86.6
- B 28.87
- C 173.2
- D 43.3
Show answer
Why:
h = 50 tan 60° = 50√3 = 86.6 m.
- A 69.28
- B 34.64
- C 30
- D 103.92
Show answer
Why:
h = 60 tan 30° = 20√3 = 34.64 m.
- A 60(√3 − 1)
- B 60√3
- C 60(√3 + 1)
- D 60
Show answer
Why:
60 cot 45° + 60 cot 30° = 60 + 60√3 = 60(√3 + 1) m.
- A 100√3/3
- B 400√3/3
- C 100√3
- D 200√3/3
Show answer
Why:
100 cot 30° − 100 cot 60° = 100√3 − 100/√3 = 200/√3 = 200√3/3 m.
- A increases
- B decreases
- C stays the same
- D first increases, then decreases
Show answer
Why:
Closer means a steeper line of sight: the height is fixed and the distance shrinks, so tan θ grows.
- A 90°
- B 45°
- C 30°
- D 60°
Show answer
Why:
tan θ = 1, so θ = 45°.
- A the angle between the cliff and the line of sight
- B twice the angle of elevation
- C the angle of elevation of the top from the boat
- D 90° minus the angle of elevation from the boat
Show answer
Why:
The horizontal at the top is parallel to the ground; the two angles are alternate angles.
- A 90°
- B 60°
- C 45°
- D 30°
Show answer
Why:
tan θ = 1/√3, so θ = 30°.
- A one-third as long
- B unchanged
- C half as long
- D √3 times as long
Show answer
Why:
Shadow = h cot θ: h√3 becomes h/√3, which is one-third.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
h = 30 tan 45° = 30, and R is exactly the reason.
- A Both A and R are true, and R is the correct explanation of A
- B Both A and R are true, but R is not the correct explanation of A
- C A is true but R is false
- D A is false but R is true
Show answer
Why:
Depression is measured from the horizontal, so A is false. R is true (alternate angles).
- A 1 only
- B 2 only
- C Both 1 and 2
- D Neither 1 nor 2
Show answer
Why:
10 sin 60° = 5√3, true. The foot is 10 cos 60° = 5 m, so 2 is false.
- A 1 only
- B 2 only
- C Both 1 and 2
- D Neither 1 nor 2
Show answer
Why:
Height = 20√3/2 = 10√3, so 1 is false. tan 45° = 1, so 2 is true.
Important short questions (3)
L = 60 ÷ sin 60° = 120/√3 = 40√3 m.
75 cot 30° = 75√3 ≈ 129.9 m.
h√3 − h/√3 = 40, so h = 40√3/2 = 20√3 m.
Long answer questions (1)
(a) Depression 60° of the bottom: d = 60 cot 60° = 20√3 m.
(b) Depression 30° of the top: drop = 20√3 × tan 30° = 20 m, so the post is 60 − 20 = 40 m.
(c) 60 − 40 = 20 m.
Flashcards (13)
Play this topic
- Angle check
Veda Skills Quantitative Aptitude Unit 16 — frequently asked questions
Opposite = √(289 − 64) = 15. sin θ = 15/17, tan θ = 15/8, $\operatorname{cosec}$ θ = 17/15, sec θ = 17/8, cot θ = 8/15.
Numerator 3/2 − 2/√3, denominator 3/2 + 2/√3; multiply by 2√3: (3√3 − 4)/(3√3 + 4) = (43 − 24√3)/11.
Put t = tan θ: t²/(t − 1) − 1/(t(t − 1)) = (t³ − 1)/(t(t − 1)) = (t² + t + 1)/t = 1 + t + 1/t = 1 + (sin²θ + cos²θ)/(sin θ cos θ) = 1 + sec θ $\operatorname{cosec}$ θ.
L = 60 ÷ sin 60° = 120/√3 = 40√3 m.
75 cot 30° = 75√3 ≈ 129.9 m.
h√3 − h/√3 = 40, so h = 40√3/2 = 20√3 m.
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