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Trigonometric Ratios

Key Concepts

Solved Examples: 38

Practice Questions: 5

Solved Examples

Question 1: In a right-angled triangle, the hypotenuse is 25 cm and one of the legs is 24 cm. Find sin θ, where θ is the angle opposite the shorter leg.

Using Pythagoras’ theorem,
Other side² = 25² − 24²
= 625 − 576
= 49
Other side = 7 cm
Therefore,
sin θ = Opposite ÷ Hypotenuse
= 7/25

Question 2: A right-angled triangle has legs measuring 9 cm and 12 cm. Find cos θ, where θ is the angle opposite the 9 cm side.

Hypotenuse² = 9² + 12²
= 81 + 144
= 225
Hypotenuse = 15 cm
Adjacent side to θ = 12 cm
Therefore,
cos θ = Adjacent ÷ Hypotenuse
= 12 ÷ 15
= 4/5

Question 3: The sides of a right-angled triangle are in the ratio 5 : 12 : 13. Find tan θ, where θ is opposite the smallest side.

Opposite side = 5
Adjacent side = 12
Therefore,
tan θ = Opposite ÷ Adjacent
= 5/12

Question 4: In a right-angled triangle, one leg is x cm, the other is 2x cm and the hypotenuse is x√5 cm. Find sec θ, where θ is adjacent to the shorter leg.

Adjacent side = x
Hypotenuse = x√5
Therefore,
sec θ = Hypotenuse ÷ Adjacent
= x√5 ÷ x
= √5

Question 5: The perimeter of a right-angled triangle is 56 cm. Two of its sides are 7 cm and 24 cm. Find cot θ, where θ is opposite the shorter leg.

Third side
= 56 − (7 + 24)
= 25 cm
Since
7² + 24² = 25²,
the triangle is right-angled.
Opposite side = 7 cm
Adjacent side = 24 cm
Therefore,
cot θ = Adjacent ÷ Opposite
= 24/7

Question 6: In a right-angled triangle, tan θ = 5/12 and the hypotenuse is 26 cm. Find the lengths of the other two sides.

Since,
tan θ = Opposite ÷ Adjacent = 5/12
Let the opposite side = 5x and the adjacent side = 12x.
Using Pythagoras’ theorem,
(5x)² + (12x)² = 26²
25x² + 144x² = 676
169x² = 676
x² = 4
x = 2
Opposite side = 10 cm
Adjacent side = 24 cm

Question 7: In a right-angled triangle, sin θ = 8/17. If the opposite side is 24 cm, find the lengths of the adjacent side and the hypotenuse.

Since,
sin θ = Opposite ÷ Hypotenuse
8/17 = 24 ÷ Hypotenuse
Hypotenuse
= (24 × 17) ÷ 8
= 51 cm
Using Pythagoras’ theorem,
Adjacent²
= 51² − 24²
= 2601 − 576
= 2025
Adjacent = 45 cm

Question 8: In a right-angled triangle, cot θ = 15/8. If the perimeter of the triangle is 80 cm, find the lengths of all three sides.

Since,
cot θ = Adjacent ÷ Opposite = 15/8
Let the adjacent side = 15x and the opposite side = 8x.
Hypotenuse
= √[(15x)² + (8x)²]
= √289x²
= 17x
Perimeter
= 15x + 8x + 17x
= 40x
40x = 80
x = 2
Adjacent side = 30 cm
Opposite side = 16 cm
Hypotenuse = 34 cm

Question 9: In a right-angled triangle, tan θ = 3/4, the hypotenuse is 15 cm less than twice the adjacent side. Find the lengths of all three sides.

Since, tan θ = 3/4
Let
Opposite = 3x
Adjacent = 4x
Hypotenuse = 5x (Pythagorean triplet)
Given,
5x = 8x − 15
3x = 15
x = 5
Opposite = 15 cm
Adjacent = 20 cm
Hypotenuse = 25 cm

Question 10: In a right-angled triangle, sin θ = 12/13. If the sum of the hypotenuse and the opposite side is 100 cm, find the lengths of all three sides.

Since, sin θ = Opposite ÷ Hypotenuse = 12/13
Let
Opposite = 12x
Hypotenuse = 13x
Given,
12x + 13x = 100
25x = 100
x = 4
Opposite = 48 cm
Hypotenuse = 52 cm
Adjacent
= √(52² − 48²)
= √400
= 20 cm

Question 15: The area of a right-angled triangle is 270 cm² and one of its legs is 15 cm. Find cos θ, where θ is opposite the shorter leg.

Area = ½ × Base × Height
270 = ½ × 15 × Height
Height = 36 cm
Hypotenuse
= √(15² + 36²)
= √1521
= 39 cm
Adjacent side = 36 cm
Therefore,
cos θ = 36/39
= 12/13

Question 16: A right-angled triangle has legs measuring 9 cm and 12 cm. Find all six trigonometric ratios for the angle opposite the 9 cm side.

Using Pythagoras’ theorem,
Hypotenuse² = 9² + 12²
= 81 + 144
= 225
Hypotenuse = 15 cm
Therefore,
sin θ = 9/15 = 3/5
cos θ = 12/15 = 4/5
tan θ = 9/12 = 3/4
cosec θ = 15/9 = 5/3
sec θ = 15/12 = 5/4
cot θ = 12/9 = 4/3

Question 17: The area of a right-angled triangle is 210 cm² and one of its legs is 20 cm. Find all six trigonometric ratios for the angle opposite the shorter leg.

Area = ½ × Base × Height
210 = ½ × 20 × Height
Height = 21 cm
Hypotenuse
= √(20² + 21²)
= √841
= 29 cm
Therefore,
sin θ = 20/29
cos θ = 21/29
tan θ = 20/21
cosec θ = 29/20
sec θ = 29/21
cot θ = 21/20

Question 18: If cos θ = 8/17, find cot θ.

Since cos θ = Adjacent/Hypotenuse
Adjacent side = 8
Hypotenuse = 17
Opposite side
= √(17² − 8²)
= √225
= 15
Therefore,
cot θ
= Adjacent ÷ Opposite
= 8/15

Question 19: If cosec θ = 17/8, find sec θ.

Since cosec θ = Hypotenuse/Opposite
Opposite side = 8
Hypotenuse = 17
Adjacent side
= √(17² − 8²)
= √225
= 15
Therefore,
sec θ = Hypotenuse ÷ Adjacent
= 17/15

Question 20: The perimeter of a right-angled triangle is 120 cm. If tan θ = 3/4, find all three sides and sec θ.

Since, tan θ = 3/4
Let the opposite side = 3x and the adjacent side = 4x.
Hypotenuse
= √[(3x)² + (4x)²]
= 5x
Perimeter
= 3x + 4x + 5x
= 12x
12x = 120
x = 10
Sides are
30 cm, 40 cm and 50 cm.
Therefore,
sec θ
= Hypotenuse ÷ Adjacent
= 50/40
= 5/4
So, Sides = 30 cm, 40 cm, 50 cm; sec θ = 5/4

Question 21: The area of a right-angled triangle is 336 cm². If sin θ = 7/25, find the lengths of all three sides.

Since, sin θ = 7/25
Let
Opposite side = 7x
Hypotenuse = 25x
Adjacent side
= √[(25x)² − (7x)²]
= 24x
Area
= ½ × 7x × 24x
= 84x²
Given,
84x² = 336
x² = 4
x = 2
Sides are
14 cm, 48 cm and 50 cm.
So, 14 cm, 48 cm, 50 cm

Question 22: In a right-angled triangle, cos θ = 12/13. If the difference between the hypotenuse and the adjacent side is 10 cm, find tan θ.

Let
Adjacent side = 12x
Hypotenuse = 13x
Given,
13x − 12x = 10
x = 10
Adjacent side = 120 cm
Hypotenuse = 130 cm
Opposite side
= √(130² − 120²)
= 50 cm
Therefore,
tan θ
= 50/120
= 5/12
So tan θ = 5/12

Question 23: A right-angled triangle has an area of 750 cm² and tan θ = 5/12. Find the perimeter of the triangle.

Since,
tan θ = 5/12
Let
Opposite side = 5x
Adjacent side = 12x
Area
= ½ × 5x × 12x
= 30x²
Given,
30x² = 750
x² = 25
x = 5
Sides are
25 cm, 60 cm and 65 cm.
Perimeter
= 25 + 60 + 65
= 150 cm

Question 24: The hypotenuse of a right-angled triangle is 85 cm. If tan θ = 8/15, find the lengths of the other two sides.

Since,
tan θ = 8/15
The corresponding Pythagorean triple is
8 : 15 : 17
Let the sides be
8x, 15x and 17x.
17x = 85
x = 5
Therefore,
Opposite side = 40 cm
Adjacent side = 75 cm

Question 25: In a right-angled triangle, tan θ = 4/3. Evaluate the numerical value of the rational expression (3 sin θ − 2 cos θ) ÷ (3 sin θ + 2 cos θ)

Given tan θ = 4/3
Since, tan θ = Opposite ÷ Adjacent
So, Opposite side = 4
Adjacent side = 3
Using Pythagoras’ theorem:
Hypotenuse = √(4² + 3²)
= √25
= 5
Therefore,
sin θ = Opposite ÷ Hypotenuse = 4/5 and
cos θ = Adjacent ÷ Hypotenuse = 3/5
Substitute these values into the given expression:
(3 sin θ − 2 cos θ) ÷ (3 sin θ + 2 cos θ)
= [3(4/5) − 2(3/5)] ÷ [3(4/5) + 2(3/5)]
= [(12/5) − (6/5)] ÷ [(12/5) + (6/5)]
= (6/5) ÷ (18/5)
= (6/5) × (5/18)
= 6/18
= 1/3
∴ The numerical value of the expression is 1/3.

Question 26: The diagonals of a rhombus measure 12 cm and 16 cm. If θ is the acute angle between a side of the rhombus and its shorter diagonal, find the value of sec θ.

The diagonals of a rhombus bisect each other at right angles.
Therefore, half of the shorter diagonal is 12 ÷ 2 = 6 cm
Half of the longer diagonal is 16 ÷ 2 = 8 cm
These two half-diagonals and a side of the rhombus form a right-angled triangle.
The side of the rhombus is the hypotenuse
Hypotenuse = √(6² + 8²)
= √(36 + 64)
= √100
= 10 cm
Since θ is the angle between the side and the shorter diagonal
Adjacent side = 6 cm
Hypotenuse = 10 cm
Therefore,
sec θ = Hypotenuse ÷ Adjacent
= 10/6
= 5/3
∴ sec θ = 5/3

Question 27: In a right-angled triangle ABC, the right angle is at B. If AB = 7 cm and BC = 24 cm, find the numerical value of sin A × cos C.

Since ∠B = 90°, AC is the hypotenuse.
Using Pythagoras’ theorem
AC² = AB² + BC²
= 7² + 24²
= 49 + 576
= 625
AC = 25 cm
For angle A sin A = Opposite ÷ Hypotenuse
= BC ÷ AC
= 24/25
For angle C cos C = Adjacent ÷ Hypotenuse
= BC ÷ AC
= 24/25
Therefore, sin A × cos C
= 24/25 × 24/25
= 576/625
∴ sin A × cos C = 576/625.

Question 28: In a right-angled triangle, if 3 sin θ = 4 cos θ, find the values of tan θ and sec θ.

Given 3 sin θ = 4 cos θ
Divide both sides by cos θ
3(sin θ ÷ cos θ) = 4
or, 3 tan θ = 4
∴ tan θ = 4/3
Let the opposite side = 4x and the adjacent side = 3x.
Using Pythagoras’ theorem
Hypotenuse = √[(4x)² + (3x)²]
= √(16x² + 9x²)
= √25x²
= 5x
∴ sec θ = Hypotenuse ÷ Adjacent
= 5x ÷ 3x
= 5/3
∴ tan θ = 4/3 and sec θ = 5/3

Question 29: In a right-angled triangle, the adjacent side to angle θ is a and the opposite side is b. Evaluate the expression (2a sin θ − b cos θ) ÷ (2a sin θ + b cos θ)

Let the hypotenuse be h.
Since sin θ = b/h and cos θ = a/h
Substitute these values
(2a sin θ − b cos θ) ÷ (2a sin θ + b cos θ)
= [2a(b/h) − b(a/h)] ÷ [2a(b/h) + b(a/h)]
= [(2ab/h) − (ab/h)] ÷ [(2ab/h) + (ab/h)]
= (ab/h) ÷ (3ab/h)
= 1/3
∴ The value of the expression is 1/3.

Question 30: If tan θ = x/√(a² − x²), evaluate the expression x cosec θ + √(a² − x²) sec θ

Given tan θ = x/√(a² − x²)
Therefore, Opposite = x
Adjacent = √(a² − x²)
Hypotenuse = √[x² + (√(a² − x²))²]
= √(x² + a² − x²)
= a
Hence, cosec θ = a/x and
sec θ = a/√(a² − x²)
Substituting these values
x cosec θ + √(a² − x²) sec θ
= x × a/x + √(a² − x²) × a/√(a² − x²)
= a + a
= 2a

Question 31: In a right-angled triangle, cos θ = 2mn/(m² + n²). Find the value of cosec θ + cot θ.

Given cos θ = 2mn/(m² + n²)
∴ Adjacent = 2mn
Hypotenuse = m² + n²
Opposite = √[(m² + n²)² − (2mn)²]
= √(m⁴ + 2m²n² + n⁴ − 4m²n²)
= √(m⁴ − 2m²n² + n⁴)
= √(m² − n²)²
= m² − n²
Hence, cosec θ + cot θ
= (m² + n²)/(m² − n²) + 2mn/(m² − n²)
= (m² + 2mn + n²)/(m² − n²)
= (m + n)²/[(m − n)(m + n)]
= (m + n)/(m − n)
∴ cosec θ + cot θ = (m + n)/(m − n).

Question 32: If b sin θ = a cos θ, evaluate the expression (sin θ − cos θ) ÷ (sin θ + cos θ)

Given b sin θ = a cos θ
Dividing both sides by b cos θ
sin θ ÷ cos θ = a/b
∴ tan θ = a/b
Now divide the numerator and denominator of the given expression by cos θ
(sin θ − cos θ) ÷ (sin θ + cos θ)
= (tan θ − 1) ÷ (tan θ + 1)
Substituting tan θ = a/b
= (a/b − 1) ÷ (a/b + 1)
= [(a − b)/b] ÷ [(a + b)/b]
= (a − b)/(a + b)
∴ (sin θ − cos θ) ÷ (sin θ + cos θ) = (a − b)/(a + b).

Question 33: In a right-angled triangle, the adjacent side to angle θ is p and the opposite side is q. Evaluate the expression (3p sin θ − 2q cos θ) ÷ (3p sin θ + 2q cos θ)

Given Adjacent = p and Opposite = q
∴ sin θ = q ÷ Hypotenuse and
cos θ = p ÷ Hypotenuse
Substitute these values
(3p sin θ − 2q cos θ) ÷ (3p sin θ + 2q cos θ)
= [3p(q/h) − 2q(p/h)] ÷ [3p(q/h) + 2q(p/h)]
= [(3pq/h) − (2pq/h)] ÷ [(3pq/h) + (2pq/h)]
= (pq/h) ÷ (5pq/h)
= 1/5
∴ The value of the expression is 1/5.

Question 34: In a right-angled triangle, the adjacent side to angle θ is p and the opposite side is q. Evaluate the expression (p sin θ + q cos θ) ÷ (p cos θ + q sin θ)

Given Adjacent = p and Opposite = q
Let the hypotenuse be h.
∴ sin θ = q/h and cos θ = p/h
Substitute these values
(p sin θ + q cos θ) ÷ (p cos θ + q sin θ)
= [p(q/h) + q(p/h)] ÷ [p(p/h) + q(q/h)]
= [(pq/h) + (pq/h)] ÷ [(p²/h) + (q²/h)]
= (2pq/h) ÷ [(p² + q²)/h]
= 2pq/(p² + q²)
∴ The value of the expression is 2pq/(p² + q²).

Question 35: In a right-angled triangle ABC, the right angle is at B. If AB = 8 cm and BC = 15 cm, find the value of sin A ÷ sin C.

Since ∠B = 90°, AC is the hypotenuse.
For angle A
Opposite side = BC = 15 cm
∴ sin A = 15/AC
For angle C:
Opposite side = AB = 8 cm
∴ sin C = 8/AC
Hence, sin A ÷ sin C
= (15/AC) ÷ (8/AC)
= 15/8
∴ sin A ÷ sin C = 15/8.

Question 36: In a right-angled triangle ABC, the right angle is at B. If sin A = 5/13, find the value of tan C.

Given sin A = 5/13
∴ pposite side to A = 5
Hypotenuse = 13
Using Pythagoras’ theorem
Adjacent side to A
= √(13² − 5²)
= √(169 − 25)
= √144
= 12
Since the adjacent side to A is the opposite side to C, and the opposite side to A is the adjacent side to C
Opposite side to C = 12
Adjacent side to C = 5
∴ tan C = Opposite ÷ Adjacent
= 12/5
∴ tan C = 12/5.

Question 37: In a right-angled triangle, tan θ = 3/4. Which of the following three trigonometric ratios has the greatest value: sin θ, cos θ or tan θ?

Given tan θ = 3/4
Take Opposite side = 3 and Adjacent side = 4
Hypotenuse = √(3² + 4²) = 5
Therefore,
sin θ = 3/5
cos θ = 4/5
tan θ = 3/4
Comparing the three values
sin θ = 0.6
cos θ = 0.8
tan θ = 0.75
Hence, cos θ > tan θ > sin θ
∴ cos θ has the greatest value.

Question 38: An isosceles triangle has equal sides of 13 cm each and a base of 10 cm. If θ is the angle between the base and one of the equal sides, find the value of tan θ.

In an isosceles triangle, the perpendicular drawn from the vertex to the base bisects the base.
∴ Half of the base = 10 ÷ 2 = 5 cm
The perpendicular and half of the base form a right-angled triangle with hypotenuse 13 cm.
Using Pythagoras’ theorem:
Height = √(13² − 5²)
= √(169 − 25)
= √144
= 12 cm
For angle θ
Opposite side = 12 cm
Adjacent side = 5 cm
∴ tan θ = Opposite ÷ Adjacent
= 12/5
∴ tan θ = 12/5.

What are Trigonometric Ratios?

Trigonometric ratios are relationships between the sides of a right-angled triangle with respect to one of its acute angles. They express how the lengths of the sides are related to each other and remain constant for a given angle, regardless of the size of the triangle. These ratios form the foundation of trigonometry and are widely used to determine unknown sides and angles in right-angled triangles.

The six trigonometric ratios are Sine (sin), Cosine (cos), Tangent (tan), Cosecant (cosec), Secant (sec) and Cotangent (cot). Each ratio is defined using the three sides of a right-angled triangle: the hypotenuse, the opposite side, and the adjacent side. Understanding these ratios is essential before studying trigonometric identities, standard angles, heights and distances, and other applications of trigonometry.

Parts of a Right Angled Triangle

Hypotenuse: The side opposite the right angle. It is always the longest side
Adjacent Side: The side next to the chosen acute angle, excluding the hypotenuse
Opposite Side: The side directly opposite the chosen acute angle

The names opposite and adjacent depend on the reference angle, whereas the hypotenuse always remains the same.

The Six Trigonometric Ratios

sin θ = Opposite ÷ Hypotenuse
cos θ = Adjacent ÷ Hypotenuse
tan θ = Opposite ÷ Adjacent
cosec θ = Hypotenuse ÷ Opposite
sec θ = Hypotenuse ÷ Adjacent
cot θ = Adjacent ÷ Opposite

The first three are called the primary trigonometric ratios, while the remaining three are their reciprocals.

Reciprocal Relationships

Each of the three reciprocal ratios is the multiplicative inverse of a primary ratio.

sin θ × cosec θ = 1
cos θ × sec θ = 1
tan θ × cot θ = 1

Therefore,

cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ

Relationship Among the Ratios

Since tan θ = Opposite ÷ Adjacent and
sin θ = Opposite ÷ Hypotenuse
cos θ = Adjacent ÷ Hypotenuse
it follows that
tan θ = sin θ ÷ cos θ
Similarly,
cot θ = cos θ ÷ sin θ
These relationships allow one trigonometric ratio to be expressed in terms of another.

Range of Trigonometric Ratios

For any acute angle in a right-angled triangle,

0 < sin θ < 1
0 < cos θ < 1
tan θ > 0
cosec θ ≥ 1
sec θ ≥ 1
cot θ > 0

Since the hypotenuse is always the longest side:

sin θ and cos θ are always less than 1.
cosec θ and sec θ are always greater than or equal to 1.
tan θ and cot θ are always positive.

These observations are useful for checking whether a calculated trigonometric ratio is reasonable.

Summary of Trigonometric Ratios

Concept	Summary
Hypotenuse	The side opposite the right angle. It is always the longest side of a right-angled triangle.
Adjacent Side	The side next to the reference angle, excluding the hypotenuse.
Opposite Side	The side directly opposite the reference angle.
sin θ	Opposite ÷ Hypotenuse
cos θ	Adjacent ÷ Hypotenuse
tan θ	Opposite ÷ Adjacent
cosec θ	Hypotenuse ÷ Opposite = 1 ÷ sin θ
sec θ	Hypotenuse ÷ Adjacent = 1 ÷ cos θ
cot θ	Adjacent ÷ Opposite = 1 ÷ tan θ
Important Relationships	tan θ = sin θ ÷ cos θ, cot θ = cos θ ÷ sin θ, sin θ × cosec θ = 1, cos θ × sec θ = 1, tan θ × cot θ = 1.

Common Mistakes

Confusing the Opposite and Adjacent Sides:
Students often identify the opposite and adjacent sides incorrectly. These sides always depend on the reference angle θ, while the hypotenuse is always the side opposite the right angle.
Using the Wrong Trigonometric Ratio:
Students sometimes apply the formula for one trigonometric ratio in place of another. Always identify the required ratio before substituting the side lengths.
Formulae:
sin θ = Opposite ÷ Hypotenuse
cos θ = Adjacent ÷ Hypotenuse
tan θ = Opposite ÷ Adjacent
Ignoring Pythagoras’ Theorem:
Students often try to calculate a trigonometric ratio without first finding the missing side of the triangle. Whenever two sides are known, use Pythagoras’ theorem to determine the third side before evaluating the required ratio.
Formula:
Hypotenuse² = Opposite² + Adjacent²
Confusing Reciprocal Ratios:
Students sometimes assume that sec θ is the reciprocal of sin θ or that cosec θ is the reciprocal of cos θ. Remember that cosec θ is the reciprocal of sin θ, sec θ is the reciprocal of cos θ, and cot θ is the reciprocal of tan θ.
Failing to Simplify the Final Answer:
Students often leave trigonometric ratios in an unsimplified form. Always reduce fractions to their lowest terms and rationalise the denominator wherever required.

Practice Questions

Question 1: A right-angled triangle has an area of 540 cm² and one of its legs is 24 cm. Find all six trigonometric ratios for the angle opposite the shorter leg.

Question 2: If sin θ = 3 cos θ, find all six trigonometric ratios.

Question 3: In a right-angled triangle, sec θ = 17/15 and the difference between the hypotenuse and the adjacent side is 14 cm. Find the lengths of all three sides.

Question 4: In a right-angled triangle, sin θ = 8/17 and the sum of the opposite side and the hypotenuse is 150 cm. Find the lengths of all three sides.

Question 5: A right-angled triangle has an area of 1260 cm² and the difference between its two legs is 7 cm. Find sin θ, cos θ and tan θ for the angle opposite the shorter leg.