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CBSE · Class 10 · Mathematics

Introduction to Trigonometry 8.2: Trigonometric Ratios

5 concepts · 40 practice questions with answers · Free to read

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The six ratios in a right triangle

From daily life

A ramp’s steepness is “rise ÷ run”. Trigonometry gives such side ratios a name for every angle.

Simple definition

In a right triangle with acute angle A, the side facing A is the opposite side, the side touching A is the adjacent side, and the side facing the right angle is the hypotenuse. The main ratios are sin A, cos A, tan A; their reciprocals are cosec A, sec A, cot A. For a fixed A they do not change with the size of the triangle.

Formula: sin A = opposite ÷ hypotenuse; cos A = adjacent ÷ hypotenuse; tan A = opposite ÷ adjacent

ABCadjacent 4opposite 3hypotenuse 5

In the figure: For angle A in a 3-4-5 triangle: opposite 3, adjacent 4, hypotenuse 5.

Live figure

The six ratios in a right triangle

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Rule 1: Name the sides from angle A — The names depend on which acute angle you stand at.

  • Easy: Hypotenuse faces the right angle and is always the longest.
  • Medium: Opposite faces angle A (BC in the figure).
  • Exam-type: Adjacent touches A but is not the hypotenuse (AB here).

Rule 2: Writing the six ratios — Three ratios use the hypotenuse or its reciprocal; tan and cot use only the two legs.

  • Easy: sin A = 3/5, cos A = 4/5, tan A = 3/4.
  • Medium: cosec A = 5/3, sec A = 5/4, cot A = 4/3.
  • Exam-type: At the other angle C the roles swap: sin C = 4/5, cos C = 3/5.
Remember

A ratio depends only on the angle, not on how big the triangle is.

Common mistake

Mixing up opposite and adjacent; decide them from the angle in question each time.

Memory clue

"SOH-CAH-TOA": Sin = Opposite/Hypotenuse, Cos = Adjacent/Hypotenuse, Tan = Opposite/Adjacent.

Practice questions

Easy
  1. 1In △XFC, right-angled at F, XF = 9 cm, FC = 40 cm and XC = 41 cm. Find cosec X.
  2. 2In △GLC, right-angled at L, GL = 6 cm, LC = 8 cm and GC = 10 cm. Find cot G.
  3. 3In △DSX, right-angled at S, DS = 10 cm, SX = 24 cm and DX = 26 cm. Find cosec D.
Medium
  1. 4In △XZY, right-angled at Z, XZ = 9 cm and XY = 41 cm. Find cosec X.
  2. 5In △KGL, right-angled at G, KG = 7 cm and KL = 25 cm. Find cosec K.
  3. 6In △BHY, right-angled at H, BH = 4 cm and HY = 3 cm. Find sin Y.
Challenging
  1. 7In △XWZ, right-angled at W, XW = 20 cm and XZ = 29 cm. Find sin X.
  2. 8In △EUQ, right-angled at U, sin E = 40/41. Find tan Q.
Show answers (try them yourself first)
  1. 141/40
  2. 23/4
  3. 313/12
  4. 441/40
  5. 525/24
  6. 64/5
  7. 721/29
  8. 89/40

Finding other ratios from a given one

From daily life

If you know a ramp rises 3 for every 5 of slope length, you can work out how far it runs without measuring.

Simple definition

One known ratio fixes the triangle’s shape. Draw a right triangle, write two sides as multiples of a common k, find the third side with Pythagoras, then read off every other ratio. The k cancels.

ABC4k3k5ksin A = 3/5cos A = 4/5tan A = 3/4

In the figure: Given sin A = 3/5: sides 3k and 5k, third side 4k, so cos A = 4/5 and tan A = 3/4.

Rule 1: Build the triangle — Put the numerator on the leg or hypotenuse the ratio names.

  • Easy: sin A = 3/5: opposite 3k, hypotenuse 5k, adjacent 4k.
  • Medium: tan A = 5/12: opposite 5k, adjacent 12k, hypotenuse √(25 + 144)k = 13k.
  • Exam-type: Then sin A = 5/13, cos A = 12/13.

Rule 2: Read off the rest — Use opposite, adjacent and hypotenuse in the right places.

  • Easy: cos A = 8/17: opposite = √(289 − 64) = 15, so sin A = 15/17, tan A = 15/8.
  • Medium: Check: 8² + 15² = 64 + 225 = 289 = 17².
  • Exam-type: The common factor k always cancels.
Remember

Check that your three sides satisfy Pythagoras before reading ratios.

Common mistake

Putting the numbers on the wrong sides, e.g. taking 3 as the adjacent side when sin A = 3/5 says it is the opposite side.

Memory clue

"Draw, k, Pythagoras, read."

Practice questions

Easy
  1. 1If sin A = 12/13 for an acute angle A, find cos A.
  2. 2If cos A = 24/25 for an acute angle A, find sin A.
  3. 3If cos A = 21/29 for an acute angle A, find sin A.
Medium
  1. 4If cos A = 21/29 for an acute angle A, find sec A.
  2. 5If cosec A = 25/24 for an acute angle A, find tan A.
  3. 6If cos A = 7/25 for an acute angle A, find cosec A.
Challenging
  1. 7If sin A = 4/5 for an acute angle A, find the value of cot A + 3 tan A.
  2. 8If cos A = 15/17 for an acute angle A, find the value of 4 cos A − 2 cot A.
Show answers (try them yourself first)
  1. 15/13
  2. 27/25
  3. 320/29
  4. 429/21
  5. 524/7
  6. 625/24
  7. 719/4
  8. 8−15/68

Reciprocal ratios and relations

From daily life

Speed = distance ÷ time and “time per distance” are flip sides of one idea. These ratio pairs are likewise flips of each other.

Simple definition

The reciprocal pairs are sin A and cosec A, cos A and sec A, tan A and cot A; each pair multiplies to 1. Also tan A = sin A ÷ cos A and cot A = cos A ÷ sin A.

Formula: cosec A = 1/sin A; sec A = 1/cos A; cot A = 1/tan A; tan A = sin A/cos A

sin A × cosec A = 1cos A × sec A = 1tan A × cot A = 1tan A = sin A ÷ cos A

In the figure: Each reciprocal pair multiplies to 1; tan is sin divided by cos.

Rule 1: Flip the fraction — The reciprocal swaps numerator and denominator.

  • Easy: sin A = 3/5 gives cosec A = 5/3.
  • Medium: cos A = 4/5 gives sec A = 5/4.
  • Exam-type: tan A = 3/4 gives cot A = 4/3.

Rule 2: Use the quotient relation — tan A is sin A divided by cos A, so the hypotenuse cancels.

  • Easy: sin A = 3/5, cos A = 4/5: tan A = (3/5) ÷ (4/5) = 3/4.
  • Medium: If sin A = 5/13 and cos A = 12/13, then cot A = 12/5.
  • Exam-type: Check: tan A × cot A = (5/12) × (12/5) = 1.
Remember

cos, cot and cosec are not reciprocals of sin, tan and sec; the pairs are sin–cosec, cos–sec, tan–cot.

Common mistake

Taking sec A as the reciprocal of sin A; sec goes with cos.

Memory clue

"The co- ones are not flipped partners: sin–cosec, cos–sec, tan–cot."

Practice questions

Easy
  1. 1Which expression equals sec A?
  2. 2Which expression equals cosec A?
  3. 3For an acute angle A, sin A = 9/41 and cos A = 40/41. Find sec A.
Medium
  1. 4For an acute angle A, sin A = 5/13 and cos A = 12/13. Find cosec A.
  2. 5For an acute angle A, sin A = 12/13 and cos A = 5/13. Find tan A.
  3. 6For an acute angle A, sin A = 21/29 and cos A = 20/29. Find tan A.
Challenging
  1. 7If cot A = 20/21 for an acute angle A, find the value of tan A + sec A.
  2. 8If sec A = 25/24 for an acute angle A, find the value of sin A × cos A.
Show answers (try them yourself first)
  1. 11 / cos A
  2. 21 / sin A
  3. 341/40
  4. 413/5
  5. 512/5
  6. 621/20
  7. 75/2
  8. 8168/625

Finding sides using a ratio

From daily life

To find how long a sloping rope must be to reach a given height, you use a known ratio instead of measuring.

Simple definition

If one side and a ratio are known, write the ratio as a fraction of sides and solve for the unknown. Pick the ratio that links the known side and the unknown side. The third side comes from Pythagoras.

ABCBC = xAC = 20sin A = 3/5x = 20 × 3/5 = 12

In the figure: sin A = 3/5 and AC = 20: BC = 20 × 3/5 = 12.

Rule 1: Pick the right ratio — Known side and unknown side decide whether to use sin, cos or tan.

  • Easy: sin A = 3/5, AC = 20: BC = 20 × 3/5 = 12.
  • Medium: cos A = 4/5, AC = 15: AB = 15 × 4/5 = 12.
  • Exam-type: tan A = 5/12, AB = 24: BC = 24 × 5/12 = 10.

Rule 2: Find the third side — After one new side, Pythagoras gives the last one.

  • Easy: In the last example AC = √(24² + 10²) = √676 = 26.
  • Medium: With sin A = 3/5, AC = 20, BC = 12: AB = √(400 − 144) = 16.
  • Exam-type: Check: cos A = 16/20 = 4/5.
Remember

Multiply when the unknown is the top of the fraction, divide when it is the bottom.

Common mistake

Dividing when you should multiply, e.g. BC = 20 ÷ (3/5) instead of 20 × 3/5.

Memory clue

"Ratio = part ÷ part; to find a part, multiply."

Practice questions

Easy
  1. 1In △UNV, right-angled at N, cos U = 3/5 and UN = 6 cm. Find UV (in cm).
  2. 2In △NVP, right-angled at V, sin N = 20/29 and NP = 58 cm. Find VP (in cm).
  3. 3In △TCW, right-angled at C, sin T = 20/29 and CW = 60 cm. Find TW (in cm).
Medium
  1. 4In △MUY, right-angled at U, tan M = 20/21 and UY = 80 cm. Find MU (in cm).
  2. 5In △UAM, right-angled at A, sin U = 7/25 and AM = 63 cm. Find UM (in cm).
  3. 6In △WBS, right-angled at B, tan W = 9/40 and BS = 45 cm. Find WB (in cm).
Challenging
  1. 7In △RBM, right-angled at B, cos R = 8/17 and RM = 51 cm. Find RB (in cm).
  2. 8In △FPD, right-angled at P, tan F = 3/4 and FP = 16 cm. Find the perimeter of △FPD (in cm).
Show answers (try them yourself first)
  1. 110
  2. 240
  3. 387
  4. 484
  5. 5225
  6. 6200
  7. 724
  8. 848

Possible values and checking statements

From daily life

A ladder cannot be shorter than the height it reaches. That is why some ratio values are impossible.

Simple definition

The hypotenuse is the longest side, so for an acute angle sin A and cos A lie between 0 and 1, sec A and cosec A are at least 1, and tan A and cot A can be any positive number. A statement outside these limits is impossible.

012sin, cossec, cosec: 1 or moretan, cot: any positive number

In the figure: sin and cos stay up to 1; sec and cosec start at 1; tan and cot can be larger.

Rule 1: Is the value possible? — Compare with the limits for that ratio.

  • Easy: sin A = 4/3 is impossible, because 4/3 is more than 1.
  • Medium: tan A = 12 is possible; sec A = 3/4 is impossible, being less than 1.
  • Exam-type: cosec A = 5/4 is possible, since it is at least 1.

Rule 2: Do two values fit one angle? — Build the triangle from one value and test the other.

  • Easy: sin A = 3/5 and cos A = 4/5 fit the 3-4-5 triangle: possible.
  • Medium: sin A = 3/5 and tan A = 1 do not fit: the 3-4-5 triangle gives tan A = 3/4.
  • Exam-type: A ratio of an acute angle is never negative: cot A = −2 is impossible.
Remember

Compare sin and cos with 1, and sec and cosec with 1, before doing any other work.

Common mistake

Thinking tan A cannot exceed 1; it can be 12 or 100 for a steep angle.

Memory clue

"sin, cos stay under one; sec, cosec stay above one."

Practice questions

Easy
  1. 1Is it possible to have cos A = 14/13 for an acute angle A?
  2. 2Is it possible to have cos A = 10 for an acute angle A?
  3. 3Is it possible to have sin A = 7/8 for an acute angle A?
Medium
  1. 4Is it possible to have tan A = 11/12 for an acute angle A?
  2. 5△FPZ is right-angled at P with FP = 15 cm, PZ = 8 cm and FZ = 17 cm. △HKQ is right-angled at K with ∠H = ∠F and HK = 30 cm. Find sec H.
  3. 6△WYT is right-angled at Y with WY = 7 cm, YT = 24 cm and WT = 25 cm. △NUZ is right-angled at U with ∠N = ∠W and NU = 21 cm. Find tan N.
Challenging
  1. 7In △VQK, right-angled at Q, VQ = 5 cm, QK = 12 cm and VK = 13 cm. Find the value of tan V × tan K.
  2. 8In △ZBT, right-angled at B, ZB = 8 cm, BT = 15 cm and ZT = 17 cm. Find the value of sin Z × sin T.
Show answers (try them yourself first)
  1. 1No
  2. 2No
  3. 3Yes
  4. 4Yes
  5. 517/15
  6. 624/7
  7. 71
  8. 8120/289