Trigonometry — SSC CGL Quant Notes

Concepts and rules

In a right triangle with acute angle θ: sinθ = perpendicular/hypotenuse, cosθ = base/hypotenuse and tanθ = perpendicular/base. cosecθ = 1/sinθ, secθ = 1/cosθ, cotθ = 1/tanθ, and tanθ = sinθ/cosθ.

Standard values: sin 0°, 30°, 45°, 60°, 90° = 0, 1/2, 1/√2, √3/2, 1, and cos takes the same values in reverse order.

tan 0°, 30°, 45°, 60° = 0, 1/√3, 1, √3, and tan 90° is not defined (cos 90° = 0).

Pythagorean identities: sin²θ + cos²θ = 1, sec²θ − tan²θ = 1 and cosec²θ − cot²θ = 1. Hence (secθ + tanθ)(secθ − tanθ) = 1 and (cosecθ + cotθ)(cosecθ − cotθ) = 1.

Complementary angles: sin(90° − θ) = cosθ, cos(90° − θ) = sinθ, tan(90° − θ) = cotθ and sec(90° − θ) = cosecθ. So if A + B = 90°, then sinA = cosB, tanA × tanB = 1 and sin²A + sin²B = 1.

Ranges: sinθ and cosθ always lie from −1 to 1; secθ and cosecθ never lie strictly between −1 and 1. For an acute θ all six ratios are positive.

Maximum and minimum: a sinθ + b cosθ lies from −√(a² + b²) to √(a² + b²). For acute θ and a, b > 0, a tan²θ + b cot²θ ≥ 2√(ab) by AM ≥ GM, with equality when tan²θ = √(b/a).

Product rule: tan 1° × tan 2° × … × tan 89° = 1, because tanθ × tan(90° − θ) = 1 pairs off the factors and tan 45° = 1.

Method and shortcuts

Step 1: If one ratio is given (say sinθ = 3/5), draw the right triangle, find the third side by Pythagoras, and read off any other ratio.

Step 2: To evaluate an expression when tanθ is known, divide the numerator and denominator by cosθ (or cos²θ) so that only tanθ remains.

Step 3: For complementary pairs, rewrite one angle: sin 37° = cos 53°, tan 70° = cot 20°, then cancel.

Shortcut: if secθ + tanθ = k, then secθ − tanθ = 1/k; adding gives secθ = (k + 1/k)/2 and subtracting gives tanθ = (k − 1/k)/2.

Shortcut for identity questions: put θ = 45° (or 30°) in the expression and in each option. This is valid only when the result must hold for every θ; if two options give the same value, test a second angle.

Example 1: One ratio given

Question: If sinθ = 3/5 and θ is acute, find secθ + tanθ.

Solution: Perpendicular = 3 and hypotenuse = 5, so base = √(25 − 9) = 4. Then secθ = 5/4 and tanθ = 3/4.

secθ + tanθ = 5/4 + 3/4 = 8/4 = 2.

Answer: 2

Example 2: Standard values

Question: Find sin²30° + cos²60° + tan²45°.

Solution: sin 30° = 1/2, cos 60° = 1/2 and tan 45° = 1, so the value is 1/4 + 1/4 + 1 = 3/2.

Answer: 3/2

Example 3: Complementary angles

Question: Find sin 37°/cos 53° + tan 20° × tan 70°.

Solution: 37° + 53° = 90°, so sin 37° = cos 53° and the fraction is 1. Also tan 70° = cot 20°, so tan 20° × tan 70° = 1.

Answer: 1 + 1 = 2

Example 4: Using sec θ + tan θ

Question: If secθ + tanθ = 3, find sinθ. (a) 3/5 (b) 5/3 (c) 4/5 (d) 3/4

Solution: secθ − tanθ = 1/3. Adding, 2secθ = 10/3, so secθ = 5/3; subtracting, 2tanθ = 8/3, so tanθ = 4/3.

sinθ = tanθ/secθ = (4/3)/(5/3) = 4/5. Check: 5/3 + 4/3 = 3.

Answer: (c) 4/5

Example 5: Least value by AM ≥ GM

Question: For acute θ, what is the least value of 9tan²θ + 4cot²θ? (a) 13 (b) 12 (c) 6 (d) 10

Solution: The product 9tan²θ × 4cot²θ = 36 is constant, so by AM ≥ GM, 9tan²θ + 4cot²θ ≥ 2√36 = 12.

Equality needs 9tan²θ = 4cot²θ, that is tan⁴θ = 4/9 and tan²θ = 2/3, which an acute θ can have, so 12 is actually reached.

The option 13 comes from putting θ = 45°, which gives 9 + 4 = 13; that is one value of the expression, not the least.

Answer: (b) 12

Example 6: Condition cos θ + cos²θ = 1

Question: If cosθ + cos²θ = 1, find sin¹²θ + 3sin¹⁰θ + 3sin⁸θ + sin⁶θ − 1. (a) 1 (b) −1 (c) 2 (d) 0

Solution: From the condition, sin²θ = 1 − cos²θ = cosθ.

So sin⁴θ + sin²θ = cos²θ + cosθ = 1.

The first four terms are (sin⁴θ)³ + 3(sin⁴θ)²(sin²θ) + 3(sin⁴θ)(sin²θ)² + (sin²θ)³, which is the cube (sin⁴θ + sin²θ)³ = 1.

Hence the value is 1 − 1 = 0.

Answer: (d) 0

Common mistakes

Turning sin²θ + cos²θ = 1 into sinθ + cosθ = 1. The identity holds for the squares only.

Swapping sin 30° and sin 60°. sin 30° = 1/2 and sin 60° = √3/2; sin increases from 0° to 90° while cos decreases.

Treating secθ − tanθ as 1 − (secθ + tanθ). It is the reciprocal: (secθ + tanθ)(secθ − tanθ) = 1.

Putting θ = 45° to find a maximum or minimum. A special angle gives one value of the expression, not necessarily the extreme; use AM ≥ GM or √(a² + b²).

Using complementary rules for angles that do not add up to 90°, for example taking tan 20° × tan 60° as 1.

Practice questions

Q1. Find 4sin²60° + 3tan²30° − 8sin 45° cos 45°. (a) 0 (b) 1 (c) 2 (d) −1

Q2. If 3tanθ = 4 and θ is acute, find (5sinθ − 3cosθ)/(5sinθ + 2cosθ). (a) 11/20 (b) 1/2 (c) 11/26 (d) 9/26

Q3. Find (cos²20° + cos²70°)/(sin²31° + sin²59°) + tan 10° × tan 80°. (a) 1 (b) 0 (c) 3 (d) 2

Answers

Q1. (a) 4 × 3/4 + 3 × 1/3 − 8 × 1/√2 × 1/√2 = 3 + 1 − 4 = 0.

Q2. (c) Divide by cosθ: (5tanθ − 3)/(5tanθ + 2) with tanθ = 4/3 gives (20/3 − 3)/(20/3 + 2) = (11/3)/(26/3) = 11/26.

Q3. (d) cos 70° = sin 20° and sin 59° = cos 31°, so the fraction is 1/1 = 1; tan 80° = cot 10°, so the product is 1. Total = 2.

Quick revision

sin: 0, 1/2, 1/√2, √3/2, 1 for 0°, 30°, 45°, 60°, 90°; cos is the reverse.

sin²θ + cos²θ = 1; sec²θ − tan²θ = 1; cosec²θ − cot²θ = 1.

secθ + tanθ = k ⇒ secθ − tanθ = 1/k.

A + B = 90°: sinA = cosB, tanA × tanB = 1.

Extremes: a sinθ + b cosθ up to √(a² + b²); a tan²θ + b cot²θ at least 2√(ab).

Concepts and rules

A full turn is 360°. 1° = 60′ (minutes) and 1′ = 60″ (seconds).

A radian is the angle at the centre of a circle subtended by an arc equal in length to the radius. π radians = 180°, so 1 radian = 180/π degrees ≈ 57.3°, and 1° = π/180 radian ≈ 0.01745 radian.

Degrees to radians: multiply by π/180. Radians to degrees: multiply by 180/π. With π = 22/7, 1 radian ≈ 57°16′.

Standard pairs: 30° = π/6, 45° = π/4, 60° = π/3, 90° = π/2, 120° = 2π/3, 135° = 3π/4, 150° = 5π/6, 180° = π and 270° = 3π/2.

Arc length l = rθ and sector area = 1/2 × r² × θ = 1/2 × l × r, with θ in radians only.

Interior angle of a regular polygon with n sides = (n − 2) × 180°/n = (n − 2)π/n radians; each exterior angle = 360°/n = 2π/n.

Clock hands: the minute hand turns 6° per minute and the hour hand 0.5° per minute (30° per hour). At h hours m minutes the angle between them is |30h − 5.5m|; if this exceeds 180°, the smaller angle is 360° minus it.

Method and shortcuts

Step 1: Before using l = rθ or any sector formula, convert the angle to radians.

Step 2: For degrees with minutes, first write the minutes as a fraction of a degree: 40°20′ = 40 + 20/60 = 121/3 degrees.

Step 3: For radians to degrees and minutes, convert the decimal part of the degree into minutes: 0.5° = 30′ and 0.25° = 15′.

Shortcut: keep π in the answer when the options contain π; otherwise use π = 22/7 unless the question gives another value.

For angles of a triangle in radians, use the angle sum π (that is, 180°) together with the given ratio or progression.

Example 1: Degrees to radians

Question: Express 75° in radians.

Solution: 75 × π/180 = 75π/180 = 5π/12.

Answer: 5π/12 radian.

Example 2: Radians to degrees

Question: Express 3π/5 radian in degrees.

Solution: 3π/5 × 180/π = 3 × 36 = 108°.

Answer: 108°

Example 3: Degrees and minutes

Question: Express 40°20′ in radians.

Solution: 40°20′ = 40 + 20/60 = 121/3 degrees. In radians: 121/3 × π/180 = 121π/540.

Answer: 121π/540 radian.

Example 4: Angle from an arc

Question: An arc of length 22 cm subtends an angle at the centre of a circle of radius 14 cm. Taking π = 22/7, what is the angle in degrees? (a) 60° (b) 45° (c) 90° (d) 120°

Solution: θ = l/r = 22/14 = 11/7 radian.

In degrees: 11/7 × 180/π = 11/7 × 180 × 7/22 = 90°.

Answer: (c) 90°

Example 5: Triangle angles in AP

Question: The angles of a triangle are in arithmetic progression and the greatest angle is 3 times the least. What is the greatest angle in radians? (a) π/3 (b) 2π/3 (c) 5π/12 (d) π/2

Solution: Let the angles be a − d, a and a + d. Their sum 3a = 180°, so a = 60°.

a + d = 3(a − d) gives 60 + d = 180 − 3d, so d = 30° and the angles are 30°, 60° and 90°.

Greatest angle = 90° = π/2 radian.

Answer: (d) π/2

Example 6: Clock hand and arc length

Question: The hour hand of a clock is 6 cm long. How far does its tip move from 2:00 to 5:20? (a) 10π/3 cm (b) 5π/3 cm (c) 20π/9 cm (d) 5π/9 cm

Solution: From 2:00 to 5:20 is 3 hours 20 minutes = 200 minutes. The hour hand turns 0.5° per minute, so it turns 200 × 0.5 = 100°.

100° = 100 × π/180 = 5π/9 radian.

Distance moved by the tip = rθ = 6 × 5π/9 = 10π/3 cm, about 10.48 cm with π = 22/7.

Answer: (a) 10π/3 cm

Common mistakes

Using l = rθ with θ in degrees. The formula needs θ in radians.

Reading 40°20′ as 40.20°. 20′ is 20/60 of a degree, so 40°20′ = 40 1/3 degrees.

Writing 1 radian = 1°, or treating π as the number 180. π radians = 180°, while π itself is about 3.14.

In clock problems, forgetting that the hour hand also moves during the minutes: at 4:20 it is 10° past the 4, not on it.

Giving the reflex angle when the smaller angle is asked: if |30h − 5.5m| exceeds 180°, subtract it from 360°.

Practice questions

Q1. What is each interior angle of a regular octagon in radians? (a) 2π/3 (b) 3π/4 (c) 5π/6 (d) 3π/5

Q2. Express 5π/8 radian in degrees and minutes. (a) 112°30′ (b) 112°50′ (c) 115° (d) 108°30′

Q3. A train runs at 66 km/h on a circular track of radius 1.5 km. Through what angle does it turn in 10 seconds? (Take π = 22/7.) (a) 6° (b) 8° (c) 9° (d) 7°

Answers

Q1. (b) (8 − 2) × 180°/8 = 135° = 135 × π/180 = 3π/4.

Q2. (a) 5π/8 × 180/π = 900/8 = 112.5° = 112°30′.

Q3. (d) In 10 seconds the train covers 66 × 1000/3600 × 10 = 550/3 m. θ = (550/3)/1500 = 11/90 radian = 11/90 × 180 × 7/22 = 7°.

Quick revision

π radians = 180°; × π/180 to radians, × 180/π to degrees.

1° = 60′, 1′ = 60″; 1 radian ≈ 57°16′.

l = rθ and sector area = 1/2 × r² × θ, θ in radians.

Regular n-gon: interior angle (n − 2)π/n.

Clock: minute hand 6° per minute, hour hand 0.5° per minute; angle |30h − 5.5m|.

Concepts and rules

The angle of elevation is the angle between the horizontal line through the observer's eye and the line of sight to an object above it; the angle of depression is measured below the horizontal to an object below.

The angle of elevation of A seen from B equals the angle of depression of B seen from A, because the two horizontal lines are parallel (alternate angles).

In the right triangle formed, tanθ = height ÷ horizontal distance, sinθ = height ÷ line of sight and cosθ = horizontal distance ÷ line of sight.

Key values: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3; √3 ≈ 1.732 and √2 ≈ 1.414.

Shadow: when the sun's altitude is θ, an object of height h casts a shadow of length h × cotθ. As the sun rises (θ increases) the shadow becomes shorter.

Two points in line with a tower, d apart, with elevations α < β: if they are on the same side, h = d/(cotα − cotβ); if they are on opposite sides, h = d/(cotα + cotβ).

If the elevations of a tower's top from two points at distances a and b from its foot, on the same line and the same side, are complementary, then h = √(ab).

If the observer's eye is at height e, the vertical side of the triangle is (object height − e); add e back at the end.

Method and shortcuts

Step 1: Draw a rough figure: the tower as a vertical line, the ground as a horizontal line, and mark every given angle and length.

Step 2: Choose the right triangle that contains the unknown and one known side; use tan in most problems, and sin or cos when a slant length (ladder, string, line of sight) is involved.

Step 3: With two angles, write the horizontal distance twice (h cotα and h cotβ) and use the given difference or sum of the distances.

Shortcut for 30° and 60° from the same side, d apart: h = d × √3/2, and the nearer point is d/2 from the foot.

Keep √3 in the working and use 1.732 only at the end if the options are decimals.

Example 1: Finding the angle

Question: A tower is 30 m high. Find the angle of elevation of its top from a point 30√3 m away from its foot.

Solution: tanθ = 30/(30√3) = 1/√3, so θ = 30°.

Answer: 30°

Example 2: Length of a shadow

Question: Find the length of the shadow of a 15 m pole when the sun's altitude is 60°.

Solution: Shadow = 15 × cot 60° = 15/√3 = 5√3 m ≈ 8.66 m.

Answer: 5√3 m

Example 3: Angle of depression

Question: From the top of a 60 m high building, the angle of depression of a car on the road is 30°. How far is the car from the foot of the building?

Solution: The angle of elevation of the top from the car is also 30°, so distance = 60 × cot 30° = 60√3 m ≈ 103.92 m.

Answer: 60√3 m

Example 4: Two points on the same side

Question: The angles of elevation of the top of a tower from two points on the same side, in line with its foot and 40 m apart, are 30° and 60°. What is the height of the tower? (a) 20 m (b) 20√3 m (c) 40√3 m (d) 10√3 m

Solution: Let the height be h and the nearer point be x from the foot. tan 60° = h/x gives x = h/√3; tan 30° = h/(x + 40) gives x + 40 = h√3.

Subtracting: h√3 − h/√3 = 40, so h × 2/√3 = 40 and h = 20√3 m ≈ 34.64 m.

Answer: (b) 20√3 m

Example 5: Points on opposite sides

Question: Two points on opposite sides of a tower, in line with its foot, are 100 m apart. The angles of elevation of the top from them are 30° and 45°. What is the height of the tower? (a) 50(√3 + 1) m (b) 100(√3 − 1) m (c) 50(√3 − 1) m (d) 50√3 m

Solution: The horizontal distances are h cot 30° = h√3 and h cot 45° = h, and on opposite sides they add up: h√3 + h = 100.

h = 100/(√3 + 1) = 100(√3 − 1)/(3 − 1) = 50(√3 − 1) m ≈ 36.6 m.

Answer: (c) 50(√3 − 1) m

Example 6: Car approaching a tower

Question: A man on top of a 75 m high tower watches a car moving straight towards the tower at a uniform speed. The angle of depression of the car changes from 30° to 60° in 12 seconds. How many more seconds will the car take to reach the foot of the tower? (a) 12 (b) 8 (c) 4 (d) 6

Solution: The car's distance from the foot is 75 cot 30° = 75√3 m at first and 75 cot 60° = 75/√3 = 25√3 m later.

It covers 75√3 − 25√3 = 50√3 m in 12 seconds, so the remaining 25√3 m takes 25√3 × 12/(50√3) = 6 seconds.

Shortcut: the distances 75√3 and 25√3 are in the ratio 3 : 1, so the car covered 2 parts in 12 seconds and needs 6 seconds for the last part.

Answer: (d) 6

Common mistakes

Using sin when the height and horizontal distance are involved. Height and horizontal distance are linked by tan; sin and cos need the slant length.

Measuring the angle of depression from the vertical instead of from the horizontal.

Forgetting the observer's eye height: the triangle gives only the part of the tower above eye level, so add the eye height at the end.

Adding the horizontal distances when the two points are on the same side (subtract them), or subtracting them when the points are on opposite sides (add them).

Rounding √3 too early, which can make two close options look equal; keep surds till the last step.

Practice questions

Q1. A ladder 20 m long rests against a wall and makes an angle of 60° with the ground. How high up the wall does it reach? (a) 10 m (b) 10√3 m (c) 20√3 m (d) 10√2 m

Q2. The angles of elevation of the top of a tower from two points 4 m and 16 m from its foot, on the same side and in line with it, are complementary. What is the height of the tower? (a) 6 m (b) 10 m (c) 8 m (d) 12 m

Q3. A man 1.5 m tall stands 28.5 m away from a chimney and sees its top at an angle of elevation of 45°. How tall is the chimney? (a) 30√3 m (b) 28.5 m (c) 27 m (d) 30 m

Answers

Q1. (b) Height = 20 × sin 60° = 20 × √3/2 = 10√3 m.

Q2. (c) tanθ = h/4 and tan(90° − θ) = cotθ = h/16; multiplying, 1 = h²/64, so h = 8 m.

Q3. (d) tan 45° = (H − 1.5)/28.5 = 1, so H − 1.5 = 28.5 and H = 30 m.

Quick revision

tanθ = height ÷ horizontal distance; elevation from B = depression from A.

Same side: h = d/(cotα − cotβ); opposite sides: h = d/(cotα + cotβ).

30° and 60° from the same side, d apart: h = d√3/2.

Complementary elevations from a and b: h = √(ab).

Shadow = h cotθ; add the observer's eye height at the end.