Important Questions Class 10 Maths Chapter 8: Introduction to Trigonometry

Trigonometry might sound like a complex topic, but with the right guidance and practice, it becomes much more approachable. In Class 10 Maths, Chapter 8 introduces the basics of trigonometry, laying the groundwork for understanding angles, triangles, and their relationships.

In this article, we'll explore some of the most crucial questions from the "Introduction to Trigonometry" chapter, explained in simple and easy-to-understand language.

What is Trigonometry?

Trigonometry is a branch of mathematics that studies the relationships between the angles and sides of triangles. The word "trigonometry" comes from the Greek words "trigonon" (meaning triangle) and "metron" (meaning measure).

Class 10 Maths Chapter 8 Important Questions

Q1: If 1 + sin2 θ = 3sin θ cos θ , then prove that tan θ = 1 or ½.

Solution:

Given: 3 sin θ cos θ = 1+sin 2 θ

LHS and RHS equations are divided using sin 2 θ .

We obtain,

(1+sin 2 θ )/sin 2 θ = 3 sin θ cos θ / sin 2 θ
⇒ (1/sin 2 θ ) + 1 = 3cos θ / sin θ
⇒ cosec 2 θ + 1 = 3 cot θ

Since,

cosec 2 θ – cot 2 θ = 1
⇒ cosec 2 θ = cot 2 θ +1
⇒ cot 2 θ +1+1 = 3 cot θ
⇒ cot 2 θ +2 = 3 cot θ
⇒ cot 2 θ –3 cot θ +2 = 0

After dividing the middle term and resolving the equation,

⇒ cot 2 θ – cot θ –2 cot θ +2 = 0
⇒ cot θ (cot θ -1)–2(cot θ +1) = 0
⇒ (cot θ – 1)(cot θ – 2) = 0
⇒ cot θ = 1, 2

Since,
tan θ = 1/cot θ
Thus, tan θ = 1, 1/2

Q2: If sin A = 3/4, so calculate cos A and tan A.

Solution:

Assuming a right-angled triangle ABC with a right angle at B

As stated: Sin A = 3/4
As we know, Sin A = Hypotenuse side / Opposite side = 3/4

Let BC be 3k and let AC be 4k.
in which k is a real positive value.

The squares on the hypotenuse side of a right angle triangle equal the sum of the squares on the other two sides, according to Pythagoras' theorem. As a result, we can see,

AC²=Ab² + BC²
Substitute the value of AC and BC
⇒ (4k)²=Ab² + (3k)²
⇒ 16k²−9k² =Ab²
⇒ Ab²=7k²

Hence, AB = √7k

Next, we need to calculate the values of tan A and cos A.

Cos (A) = Adjacent side/Hypotenuse side
AB/AC = √7k/4k = √7/4

Hence, cos (A) = √7/4

tan(A) = Opposite side/Adjacent side
BC/AB = 3k/√7k = 3/√7

Hence, tan A = 3/√7

Q3: In triangle ABC, right-angled at B, when tan A = 1/√3 find out the value :

• sin A cos C + cos A sin C
• cos A cos C – sin A sin C

Solution:

Consider ΔABC, where ∠B=90°.

tan A = BC/AB = 1/√3

Let’s BC = 1k and AB = √3 k,
Here, k represents the problem's positive real number.

Using ΔABC and the Pythagorean theorem, we obtain:

AC²=Ab²+BC²
⇒ AC²=(√3 k)²+(k)²
⇒ AC²=3k²+k²
⇒ AC²=4k²
⇒ AC = 2k

Then find the values of cos A, Sin A

• Sin A = BC/AC = 1/2
• Cos A = AB/AC = √3/2

now, find the values of cos C and sin C

• Sin C = AB/AC = √3/2
• Cos C = BC/AC = 1/2

then, substitute the values in the given problems

• sin A cos C + cos A sin C = (1/2) ×(1/2 )+ √3/2 ×√3/2 = 1/4 + 3/4 = 1
• cos A cos C – sin A sin C = (√3/2 )(1/2) – (1/2) (√3/2 ) = 0

Q4: Prove that (1 − cos A)/sin⁡ A = sin⁡ A/(1+cos⁡ A).

Solution:

Starting with the LHS, or left side, we have:

LHS = (1−cos⁡A)/sin⁡A
⇒ LHS = (1−cos⁡A)(1+cosA)/sin⁡A(1+cosA)
⇒ LHS = (1 − cos²⁡A)/sin⁡A(1+cosA)
⇒ LHS = sin²⁡A/sin⁡A(1+cosA). [As 1 - cos²A = sin²⁡A]
⇒ LHS = sin⁡A/1+cosA

This proves the equation since it equals the right-hand side (RHS).

Q5: Prove that (sin θ −cos θ )/(sin θ +cos θ )= (1−tan θ )/(1+tan θ ).

Solution:

Starting with the RHS, or right side:

RHS= (1−tan θ )/(1+tan θ)
⇒ RHS = (1 - sin θ /cos θ )/(1 + sin θ /cos θ ) {As we know tan θ = sin θ /cos θ}
⇒ RHS = (cos θ - sin θ )/cos θ /(cos θ + sin θ )/cos θ
⇒ RHS = (cos θ - sin θ )/ (cos θ + sin θ )
⇒ RHS = (sin θ −cos θ )/(sin θ +cos θ )

Hence Prove.

Q6: Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°

Solution:

To answer the following question, we shall use the trigonometric ratios of complementary angles: cos (90° - θ ) = sin θ

Given that: sin 67° + cos 75°
sin 67° + cos 75° = sin (90° - 23°) + cos (90° - 15°)
= cos 23° + sin 15°

As a result, there are trigonometric ratios of angles between 0° and 45° in the statement cos 23° + sin 15°.

Q7: Evaluate:

• (sin²63° + sin²27°) / (cos²17° + cos²73°)
• sin 25° cos 65° + cos 25° sin 65°

Solution:

Given: sin² A + cos² A = 1

• sin (90° - θ )= cos θ
• cos (90° - θ ) = sin θ

(i) (sin²63° + sin²27°) / (cos²17° + cos²73°)
= [sin(90° - 27°)]²+ sin²27°/ [cos(90° - 73°)]² + cos²73°
= (cos²27° + sin²27°) / (sin²73° + cos²73°) [ Since sin (90° - θ ) = cos θ and cos (90° - θ ) = sin θ ]
= 1/1 (By using the identity sin² A + cos² A = 1)
= 1

(ii) sin25° cos 65° + cos 25° sin 65°
= sin 25° [cos(90° - 25°)] + cos 25° [sin (90° - 25°)]
= sin 25° sin 25° + cos 25° cos 25° [Since sin (90° - θ ) = cos θ & cos (90° - θ ) = sin θ ]
= sin² 25° + cos² 25°
= 1 (By using the identity sin² A + cos² A = 1)

Q8: If tan (A + B) = √3 and tan (A - B) = 1/√3; 0° < (A + B) ≤ 90° ; A > B, find A and B

Solution:

To solve the problem, we make use of the trigonometric identities and the trigonometric table.

Given that, tan (A + B) = √3 and, tan (A - B) = 1/√3

Since, tan 60° = √3 and tan 30° = 1/√3
tan (A + B) = tan 60°
(A + B) = 60° . . . (i)

tan (A - B) = tan 30°
(A - B) = 30° . . .(ii)

On adding both equations (i) and (ii), we obtain:

A + B + A - B = 60° + 30°
2A = 90°
A = 45°

By substituting the value of A in equation (i) we obtain

A + B = 60°
45° + B = 60°
B = 60° - 45° = 15°

Therefore, ∠A = 45° and ∠B = 15° (A > B).

Practice Questions for Introduction to Trigonometry

Q1: Find the value of sin⁡ θ , cos⁡ θ , and tan⁡ θ if in a right-angled triangle, the opposite side is 5 cm and the adjacent side is 12 cm.

Q2: In a right-angled triangle, if one acute angle is 30° and the hypotenuse is 10 cm, find the lengths of the other two sides.

Q3: Prove that sin⁡² θ + cos⁡² θ = 1 using a right-angled triangle.

Q4: Calculate the height of a tree if its shadow is 30 meters long and the angle of elevation of the sun is 60°.

Q5: If tan⁡ θ = 3/4​, find the value of sin⁡ θ and cos⁡ θ .

Q6: A ladder is leaning against a wall, making an angle of 75° with the ground. If the ladder is 10 meters long, find the height reached on the wall.

Q7: In a right-angled triangle, if the hypotenuse is 13 cm and one side is 5 cm, find the other side and all trigonometric ratios of the given angle.

Q8: A flagpole is standing straight. From a point 15 meters away from its base, the angle of elevation of the top is 45°. Find the height of the flagpole.

Q9: If cos⁡ θ = 12/13, find sin⁡ θ and tan⁡ θ .

Answer Key

1.

• sin ⁡ θ = 5/13
• cos⁡ θ = 12/13
• tan⁡ θ = 5/12

2.

• Opposite side = 5 cm
• Adjacent side = 5√3 cm

4. Height =30√3 meters

5.

• sin ⁡ θ = 3/5
• cos ⁡ θ = 4/5

6. Height reached ≈ 9.659 meters

7.

• Other side = 12 cm
• sin⁡ θ = 5/13
• cos ⁡ θ = 12/13
• tan ⁡ θ = 5/12

8. Height = 15 meters

9.

• sin⁡ θ = 5/13
• tan⁡ θ = 5/12

Read More,

• Trigonometry in Maths
• Trigonometry Table
• Trigonometric Identities
• Difference Between Trigonometry and Geometry