Trigonometric Ratios of 30°, 45°, 60° Explained Geometrically with MCQS, Solved Examples

ANAND CLASSES Study Material and Notes to learn the geometric derivation of trigonometric ratios for 30°, 45°, and 60° with diagrams, formulas, and step-by-step explanations. Ideal for Class 10, Class 11, JEE, and NEET preparation.

📚 Trigonometric Ratios of 45°, 60°, and 30° (Geometrically Explained) for Class 10 Math

Trigonometric ratios are the relationships between the sides of a right-angled triangle and its angles. Let’s derive the exact values of trigonometric ratios for 45°, 60°, and 30° using basic geometry and Pythagoras’ Theorem.

🔺 TRIGONOMETRIC RATIOS OF 45° for Class 10 Math

🔧 Construction:

Take a right-angled triangle ΔABC with:

• ∠B = 90°
• ∠A = ∠C = 45° (Since the sum of angles in triangle is 180°)

As ∠A = ∠C, the triangle is isosceles with:

• AB = BC = a units (equal sides)
• AC = hypotenuse

Using Pythagoras’ Theorem:

$$AC = \sqrt{AB^2 + BC^2} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}$$

🧮 T-Ratios of 45°:

$$\sin 45^\circ = \frac{BC}{AC} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}} $$

$$\cos 45^\circ = \frac{AB}{AC} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}}$$

$$\tan 45^\circ = \frac{BC}{AB} = \frac{a}{a} = 1 $$

$$\sec 45^\circ = \frac{1}{\cos 45^\circ} = \sqrt{2} $$

$$\mathrm{cosec} \:45^\circ = \frac{1}{\sin 45^\circ} = \sqrt{2} $$

$$\cot 45^\circ = \frac{1}{\tan 45^\circ} = 1$$

🔺 TRIGONOMETRIC RATIOS OF 60° AND 30° for Class 10 Math

🔧 Construction:

Consider an equilateral triangle ΔABC:

• Each side = 2a
• Each angle = 60°

Draw a perpendicular AD from vertex A to base BC.

Then:

• AD ⊥ BC
• BD = DC = a (since it bisects BC)
• ∠ADB = 90°
• ∠BAD = 30°, ∠DAC = 30°, ∠ADB = 90°, ∠DAB = 60°

Using Pythagoras in ΔADB:

$$AD = \sqrt{AB^2 – BD^2} = \sqrt{(2a)^2 – a^2} = \sqrt{4a^2 – a^2} = \sqrt{3a^2} = a\sqrt{3}$$

✅ T-Ratios of 60° (from ΔADB):

• Base = BD = a
• Height = AD = $a\sqrt{3}$
• Hypotenuse = AB = 2a

$$\sin 60^\circ = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \frac{\sqrt{3}}{2} $$

$$\cos 60^\circ = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2} $$

$$\tan 60^\circ = \frac{AD}{BD} = \frac{a\sqrt{3}}{a} = \sqrt{3} $$

$$\sec 60^\circ = \frac{1}{\cos 60^\circ} = 2 $$

$$\mathrm{cosec} \: 60^\circ = \frac{1}{\sin 60^\circ} = \frac{2}{\sqrt{3}} $$

$$\cot 60^\circ = \frac{1}{\tan 60^\circ} = \frac{1}{\sqrt{3}}$$

✅ T-Ratios of 30° (from the same ΔADB):

• Base = AD = $a\sqrt{3}$
• Height = BD = a
• Hypotenuse = AB = 2a

$$\sin 30^\circ = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2}$$

$$\cos 30^\circ = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \frac{\sqrt{3}}{2}$$

$$\tan 30^\circ = \frac{BD}{AD} = \frac{a}{a\sqrt{3}} = \frac{1}{\sqrt{3}}$$

$$\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{2}{\sqrt{3}}$$

$$\mathrm{cosec} \: 30^\circ = \frac{1}{\sin 30^\circ} = 2 $$

$$\cot 30^\circ = \frac{1}{\tan 30^\circ} = \sqrt{3}$$

🧠 Axioms of Trigonometric Ratios

📍 For 0°:

We define:

• sin 0° = 0
• cos 0° = 1
• tan 0° = 0
• sec 0° = 1
  ❌ cosec 0° and cot 0° are undefined

📍 For 90°:

We define:

• sin 90° = 1
• cos 90° = 0
• cosec 90° = 1
• cot 90° = 0
  ❌ tan 90° and sec 90° are undefined

All Values of Trigonometric Ratios [Some Specific Angles]

Some of the common values of trigonometric ratios are listed in the following table:

MCQs on Trigonometric Ratios of 30°, 45°, and 60° for Class 10 Math

Multiple Choice Questions (MCQs) based on the geometric explanation of Trigonometric Ratios of 30°, 45°, and 60°, perfect for Class 10, Class 11, JEE, and NEET-level preparation

Q1. If in a triangle $\angle A = 30^\circ$, AB = 10 cm, and $\angle B = 90^\circ$, find the length of side AC.
A. 10 cm
B. 5√3 cm
C. 20 cm
D. 10√3 cm

✅ Correct Answer: D. $10\sqrt{3} cm$
🧠 Explanation:
In right triangle with $\angle A = 30^\circ$ :

$\cos 30^\circ = \frac{AB}{AC} \Rightarrow \frac{\sqrt{3}}{2} = \frac{10}{AC} \Rightarrow AC = \frac{10 \times 2}{\sqrt{3}} = \frac{20}{\sqrt{3}} = 10\sqrt{3}$

Q2. $\tan \theta = \frac{1}{\sqrt{3}}$, and $0^\circ < \theta < 90^\circ$, what is the value of $\cos \theta$ ?
A. $\frac{1}{\sqrt{2}}$
B. $\frac{1}{2}$
C. $\frac{\sqrt{3}}{2}$
D. 1

✅ Correct Answer: C. $\frac{\sqrt{3}}{2}$
🧠 Explanation:
If $\tan \theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = 30^\circ \Rightarrow \cos 30^\circ = \frac{\sqrt{3}}{2}$

Q3. If $\sin \theta = \cos \theta$, then the value of $\theta$ is:
A. 30°
B. 45°
C. 60°
D. 90°

✅ Correct Answer: B. 45°
🧠 Explanation:
Only at 45°, $\sin \theta = \cos \theta = \frac{1}{\sqrt{2}}$

Q4. In a right triangle, the ratio of adjacent side to hypotenuse is $\frac{1}{2}$. The angle opposite to the perpendicular is:
A. 30°
B. 45°
C. 60°
D. 90°

✅ Correct Answer: C. 60°
🧠 Explanation:

$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{2} \Rightarrow \theta = 60^\circ$

Q5. If $\sin \theta = \frac{1}{2}$ and $0^\circ < \theta < 90^\circ$, then what is the value of $\tan \theta + \cot \theta$?
A. $\sqrt{3} + \frac{1}{\sqrt{3}}$
B. 2
C. $\frac{1}{\sqrt{3}} + \sqrt{3}$
D. 3

✅ Correct Answer: A. $\sqrt{3} + \frac{1}{\sqrt{3}}$
🧠 Explanation:
$\sin \theta = \frac{1}{2} \Rightarrow \theta = 30^\circ$

$\tan 30^\circ = \frac{1}{\sqrt{3}}, \quad \cot 30^\circ = \sqrt{3} \Rightarrow \text{Sum} = \frac{1}{\sqrt{3}} + \sqrt{3}$

Solved Examples on Trigonometric Ratios of 30°, 45°, and 60° for Class 10 Math

Find the value of: sin 30^o cos 30^o

Given sin 30^o cos 30^o

By substituting the values, we get

sin 30^o cos 30^o = ½ (√3/2) = √3/4

Find the value of: tan 30^o tan 60^o

Given tan 30^o tan 60^o

By substituting the values, we get

tan 30^o tan 60^o = 1/√3 (√3) = 1

Find the value of: cos² 60^o + sin² 30^o

Given cos² 60^o + sin² 30^o

By substituting the values, we get

cos² 60^o + sin² 30^o = (½)² +(½)² = ¼ + ¼ = ½

Find the value of: cosec² 60^o – tan² 30^o

Given cosec² 60^o – tan² 30^o

By substituting the values, we get

cosec² 60^o – tan² 30^o = (2/√3)² – (1/√3)²

= 4/3 – 1/3 = 1

Find the value of: sin² 30^o + cos² 30^o + cot² 45^o

Given sin² 30^o + cos² 30^o + cot² 45^o

By substituting the values, we get

sin² 30^o + cos² 30^o + cot² 45^o = (½)² + (√3/2)² + 1²

= ¼ + ¾ + 1 = 2

Find the value of: cos² 60^o + sec² 30^o + tan² 45^o

Given cos² 60^o + sec² 30^o + tan² 45^o

By substituting the values, we get

cos² 60^o + sec² 30^o + tan² 45^o = (½)² + (2/√3)² + 1²

= ¼ + 4/3 + 1 = 31/12

Find the value of: tan² 30^o + tan² 45^o + tan² 60^o

Given tan² 30^o + tan² 45^o + tan² 60^o

By substituting the values, we get

tan² 30^o + tan² 45^o + tan² 60^o = (1//√3)² + 1² + (/√3)²

= 1/3 + 1 + 3 = 13/3 = 4 1/3

Find the value of: 3 sin² 30^o + 2 tan² 60^o – 5 cos² 45^o.

Given 3 sin² 30^o + 2 tan² 60^o – 5 cos² 45^o.

By substituting the values, we get

3 sin² 30^o + 2 tan² 60^o – 5 cos² 45^o. = 3 (½)² + 2 (√3)² + 5 (1/√3)²

= ¾ + 6 – 5/2

= (3 + 24 – 10)/4 = 4 ¼

Prove that:

(i) sin 60^o cos 30^o + cos 60^o. sin 30^o = 1

(ii) cos 30^o. cos 60^o – sin 30^o. sin 60^o = 0

(iii) cosec² 45^o – cot² 45^o = 1

(iv) cos² 30^o – sin² 30^o = cos 60^o.

(v) 3 cosec² 60^o – 2 cot² 30^o + sec² 45^o = 0.

Solution :

(i) Given sin 60^o cos 30^o + cos 60^o. sin 30^o

LHS = sin 60^o cos 30^o + cos 60^o. sin 30^o

Now we have to prove that RHS = 1

= (√3/2) (√3/2) + ½ ½ = ¾ + ¼ = 1 = RHS

(ii) Given cos 30^o. cos 60^o – sin 30^o. sin 60^o = 0

LHS = cos 30^o. cos 60^o – sin 30^o. sin 60^o

= (√3/2) ½ – ½ (√3/2) = (√3/4) – (√3/4) = 0 = RHS

(iii) Given cosec² 45^o – cot² 45^o = 1

LHS = cosec² 45^o – cot² 45^o = 1

= (√2)² – 1² = 2 – 1 = 1 = RHS

(iv) Given cos² 30^o – sin² 30^o = cos 60^o.

LHS = cos² 30^o – sin² 30^o = cos 60^o.

= (√3/2)² – (½)²

= ¾ – ¼ = ½

= cos 60^o = RHS

(v) Given 3 cosec² 60^o – 2 cot² 30^o + sec² 45^o = 0.

LHS = 3 cosec² 60^o – 2 cot² 30^o + sec² 45^o = 0.

= 3 (2/√3)² – 2 (√3)² + (√2)²

= 4 – 6 + 2 = 0 = RHS

Prove that:

(i) sin 60^o = 2 sin 30^o cos 30^o.

(ii) 4 (sin⁴ 30^o + cos⁴ 60^o) – 3 (cos² 45^o – sin² 90^o) = 2

Solution:

(i) LHS = sin 60^o = √3/2

RHS = 2 sin 30^o cos 30^o = 2 (√3/2) (½) = √3/2

Therefore LHS = RHS

(ii) LHS = 4 (sin⁴ 30^o + cos⁴ 60^o) – 3 (cos² 45^o – sin² 90^o)

Now by substituting the values we get

= 4[(½)⁴ + (½)⁴] – 3 [(1/√2)² + 1⁴]

= 4(1/16 + 1/16) – 3 (½ – 1)

= 8/16 + 3/2 = 2

LHS = RHS

Solve the given Problems, according to given statement

(i) If sin x = cos x and x is acute, state the value of x.

(ii) If sec A = cosec A and 0^o ≤ A ≤ 90^o, state the value of A.

(iii) If tan θ= cot θ and 0^o ≤ θ ≤ 90^o, state the value of θ.

(iv) If sin x = cos y; write the relation between x and y, if both the angles x and y are acute.

Solution:

(i) The angle, x is acute and hence we have, 0 < x

We know that

Cos²x + sin² x = 1

Since cos x = sin x

Above equation will become

2 sin² x = 1

Sin x = 1/√2

Therefore, x = 45^o

(ii) sec A = cosec A

Cos A = sin A

Cos² A = sin² A

Cos²x + sin² x = 1

Above equation will become

Cos² A = 1 – cos² A

2 cos² A = 1

Cos A = 1/√2

A = 45^o

(iii) tan θ = cot θ

tan θ = 1/tan θ

tan² θ = 1

tan θ = 1

tan θ = tan 45^o

θ = 45^o

(iv) sin x = cos y = sin (90^o – y)

If x and y are acute angles

x = 90^o – y

which implies,

x + y = 90^o

hence x and y are complementary angles.

State True or False the given statements

(i) If sin x = cos y, then x + y = 45^o; write true of false.

(ii) sec θ. Cot θ = cosec θ; write true or false.

(iii) For any angle θ, state the value of: Sin² θ + cos² θ.

Solution:

(i) sin x = cos y = sin (π/2 – y)

If x and y acute angles,

x = (π/2 – y)

x + y = π/2

x + y = 45^o is false

(ii) sec θ. Cot θ = 1/ cos θ. cos θ/ sin θ

= cosec θ

sec θ. Cot θ = cosec θ

is true.

(iii) Sin² θ + cos² θ = Sin² θ + 1 – sin² θ.

= 1

State for any acute angle θ whether:

(i) sin θ increases or decreases as θ increases:

(ii) cos θ increases or decreases as θ increases.

(iii) tan θ increases or decreases as θ decreases.

Solution:

(i) For acute angles, remember what sine means: opposite over hypotenuse. If we increase the angle, then the opposite side gets larger. That means “opposite/hypotenuse” gets larger or increases.

(ii)For acute angles, remember what cosine means: base over hypotenuse. If we increase the angle, then the hypotenuse side gets larger. That means “base/hypotenuse” gets smaller or decreases.

(iii)For acute angles, remember what tangent means: opposite over base. If we decrease the angle, then the opposite side gets smaller. That means “opposite /base” gets decreases.

If √3 = 1.732, find (correct to two decimal place) the value of each of the following: (i) sin 60^o  (ii) 2/ tan 30^o

Solution:

(i) sin 60^o = √3 /2

= 1.732/2 = 0.87

(ii) 2/ tan 30^o = 2/ (1/√3)

= 2√3  = 2 (1.732) = 3.46

Given A = 60^o and B = 30^o, prove that:

(i) sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos A cos B – sin A sin B

(iii) cos (A – B) = cos A cos B + sin A sin B

(iv) tan(A-B) = (tan A – tan B)/(1 + tan A tan B)

Solution:

(i) Given A = 60^o and B = 30^o

LHS = sin (A + B)

= sin (60^o + 30^o) = sin 90^o = 1

RHS = sin A cos B + cos A sin B

= sin 60^o cos 30^o + cos 60^o sin 30^o

= √3/2 (√3/2) + ½ ½ = ¾ + ¼ = 1

LHS = RHS

(ii) LHS = cos (A + B)

= cos (60^o + 30^o) = cos 90^o = 0

RHS = cos A cos B – sin A sin B

RHS = cos 60^o cos 30^o – sin 60^o sin 30^o

= ½ (√3/2) – (√3/2) ½ = √¾ – √3/4 = 0

LHS = RHS

(iii) LHS = cos (A – B)

= cos (60^o – 30^o) = cos 30^o = √3/2

RHS = cos A cos B + sin A sin B

RHS = cos 60^o cos 30^o + sin 60^o sin 30^o

= ½ (√3/2) + (√3/2) ½ = √¾ + √3/4 = √3/2

LHS = RHS

(iv) LHS = tan (A – B)

= tan (60^o – 30^o) = tan 30^o = 1/√3

RHS = (tan A – tan B)/(1 + tan A tan B)

(tan 60^o – tan 30^o)/(1 + tan 60^o tan 30^o)

= (√3 – 1/√3)/ 1 + √3 (1/√3)

= 2/ 2 √3 = 1/√3

Therefore, LHS = RHS

If tan (A+ B) =√3, tan (A-B) = 1/√3, then find A and B. [Given that  0° <A+B ≤ 90°; A>B ]

Solution:

Given that 

tan (A+B) = √3.

We know that tan 60 = √3.

Thus, tan (A+B) = tan 60° = √3.

Hence A+B= 60° …(1)

Similarly, given that,

tan (A-B) = 1/√3.

We know that tan 30° = 1/√3.

Thus, tan (A-B) = tan 30° = 1/√3.

Hence, A-B = 30° …(2)

Now, adding the equations (1) and (2), we get

A+B+A-B = 60° + 30°

2A = 90°

A = 45°.

Now, substitute A = 45° in equation (1), we get

45° +B = 60°

B = 60°- 45°

B = 15°

Hence, A = 45 and B = 15°.

If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1.

Solution:

Given,

sin θ + cos θ = √3

Squaring on both sides,

(sin θ + cos θ)² = (√3)²

sin²θ + cos²θ + 2 sin θ cos θ = 3

Using the identity sin²A + cos²A = 1,

1 + 2 sin θ cos θ = 3

2 sin θ cos θ = 3 – 1

2 sin θ cos θ = 2

sin θ cos θ = 1

sin θ cos θ = sin²θ + cos²θ

⇒ (sin²θ + cos²θ)/(sin θ cos θ) = 1

⇒ [sin²θ/(sin θ cos θ)] + [cos²θ/(sin θ cos θ)] = 1

⇒ (sin θ/cos θ) + (cos θ/sin θ) = 1

⇒ tan θ + cot θ = 1

Hence proved.

💡 Do You Know?

• The values of trigonometric ratios are exact for 0°, 30°, 45°, 60°, and 90°.
• These ratios are frequently used in geometry, physics, astronomy, and engineering.
• The identities like sin²θ + cos²θ = 1 help derive all other formulas in trigonometry.

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