Important questions of Chapter 8 – Introduction to Trigonometry of Class 10 are given here for students who want to score high marks in their board exams. These questions are as per the latest CBSE syllabus and are designed according to NCERT book for the academic session 2020-2021. The questions are formulated after analysing the previous year questions papers, exam trend and latest sample papers 2021. Solving these questions will help students to get prepared for the final exam. Students can also getÂ important questions for all the chapters of 10th standard Maths. Solve them to get acquainted with the types of questions to be asked from each chapter of the Maths subject.

In Chapter 8, students will be introduced to the Trigonometry concept, which states the relationship between angles and sides of a triangle. They will come across many trigonometric formulas which will be used to solve numerical problems. The six primary trigonometric ratios are sine, cosine, tangent, secant, cosecant and cotangent. The whole trigonometry concept revolves around these ratios or sometimes also called functions.

## Class 10 Maths Chapter 8 Important Questions

Below are the important questions for Class 10 students of CBSE board based on Chapter 8-Introduction to Trigonometry. Solve these questions and for any queries, please drop your comments at the end of the section. Students can also refer to the solutions when they get stuck while solving any problem.

**Question. 1 :Â In âˆ† ABC, right-angled at B, AB = 24 cm, BC = 7 cm.**

**Determine:**

**(i) sin A, cos A**

**(ii) sin C, cos C**

Solution:

In a given triangle ABC, right-angled at B = âˆ B = 90Â°

Given: AB = 24 cm and BC = 7 cm

That means, AC = Hypotenuse

According to the Pythagoras Theorem,

In a right-angled triangle, the squares of the hypotenuse side are equal to the sum of the squares of the other two sides.

By applying Pythagoras theorem, we get

AC^{2Â }= AB^{2Â }+ BC^{2}

AC^{2}Â = (24)^{2Â }+ 7^{2}

AC^{2}Â = (576 + 49)

AC^{2}Â =Â 625 cm^{2}

Therefore, AC = 25 cm

(i) We need to find Sin A and Cos A.

As we know, sine of the angle is equal to the ratio of the length of the opposite side and hypotenuse side. Therefore,

Sin A = BC/AC = 7/25

Again, the cosine of an angle is equal to the ratio of the adjacent side and hypotenuse side. Therefore,

cos A = AB/AC = 24/25

(ii) We need to find Sin C and Cos C.

Sin C = AB/AC = 24/25

Cos C = BC/AC = 7/25

**Question 2: If Sin A = 3/4, Calculate cos A and tan A.**

Solution: Let us say, ABC is a right-angled triangle, right-angled at B.

Sin A = 3/4

As we know,

Sin A = Opposite Side/Hypotenuse Side = 3/4

Now, let BC be 3k and AC will be 4k.

where k is the positive real number.

As per the Pythagoras theorem, we know;

Hypotenuse^{2} = Perpendicular^{2}+ Base^{2}

AC^{2} = AB^{2} + BC^{2}

Substitute the value of AC and BC in the above expression to get;

(4k)^{2} = (AB)^{2} + (3k)^{2}

16k^{2} – 9k^{2} = AB^{2}

AB^{2} = 7k^{2}

Hence, AB =Â âˆš7 kÂ

Now, as per the question, we need to find the value of cos A and tan A.

cos A = Adjacent Side/Hypotenuse side = AB/AC

cos A =Â âˆš7 k/4k =Â âˆš7/4

And,

tan A = Opposite side/Adjacent side = BC/AB

tan A = 3k/âˆš7 k = 3/âˆš7

**Question.3:Â If âˆ A and âˆ B are acute angles such that cos A = cos B, then show that âˆ A = âˆ B.**

Solution:

Suppose a triangle ABC, right-angled at C.

Now, we know the trigonometric ratios,

cos A = AC/AB

cos B = BC/AB

Since, it is given,

cos A = cos B

AC/AB = BC/AB

AC = BC

We know that by isosceles triangle theorem, the angles opposite to the equal sides are equal.

Therefore, âˆ A =Â âˆ B

**Question 4:Â If 3 cot A = 4, check whether (1 – tan ^{2}A)/(1 + tan^{2}A) = cos^{2} A – sin^{2} A or not.**

Solution:

Let us consider a triangle ABC, right-angled at B.

Given,

3 cot A = 4

cot A = 4/3

Since, tan A = 1/cot A

tan A = 1/(4/3) = 3/4

BC/AB = 3/4

Let BC = 3k and AB = 4k

By using Pythagoras theorem, we get;

Hypotenuse^{2} = Perpendicular^{2} + Base^{2}

AC^{2} = AB^{2} + BC^{2}

AC^{2} = (4k)^{2} + (3k)^{2}

AC^{2} = 16k^{2Â }+ 9k^{2}

AC = âˆš25k^{2Â }= 5k

sin A = Opposite side/Hypotenuse

= BC/AC

=3k/5k

=3/5

In the same way,

cos A = Adjacent side/hypotenuse

= AB/AC

= 4k/5k

= 4/5

To check:Â (1-tan^{2}A)/(1+tan^{2}A) = cos^{2} A – sin^{2} A or not

Let us take L.H.S. first;

(1-tan^{2}A)/(1+tan^{2}A) = [1 – (3/4)^{2}]/ [1 + (3/4)^{2}]

= [1 – (9/16)]/[1 + (9/16)] = 7/25

R.H.S. =Â cos^{2} A – sin^{2} A = (4/5)^{2Â }– (3/5)^{2}

= (16/25) – (9/25) = 7/25

Since,

L.H.S. = R.H.S.

Hence, proved.

**Question 5: In triangle PQR, right-angled at Q,Â PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.**

Solution: Given,

In triangle PQR,

PQ = 5 cm

PR + QR = 25 cm

Let us say, QR = x

Then, PR = 25 – QR = 25 – x

Using Pythagoras theorem:

PR^{2} = PQ^{2} + QR^{2}

Now, substituting the value of PR, PQ and QR, we get;

(25 – x)^{2} = (5)^{2} + (x)^{2}

25^{2Â }+ x^{2Â }– 50x = 25 + x^{2}

625 – 50x = 25

50x = 600

x = 12

So, QR = 12 cm

PR = 25 – QR = 25 – 12 = 13 cm

Therefore,

sin P = QR/PR = 12/13

cos P = PQ/PR = 5/13

tan P = QR/PQ = 12/5

**Question 6: Evaluate 2 tan ^{2} 45Â° + cos^{2} 30Â° – sin^{2} 60Â°.**

Solution: Since we know,

tan 45Â° = 1

cos 30Â° = âˆš3/2

sin 60Â° = âˆš3/2

Therefore, putting these values in the given equation:

2(1)^{2} + (âˆš3/2)^{2} – (âˆš3/2)^{2}

= 2 + 0

= 2

**Question 7:Â If tan (A + B) =âˆš3 and tan (A â€“ B) =1/âˆš3, 0Â° < A + B â‰¤ 90Â°; A > B, find A and B.**

Solution: Given,

tan (A + B) = âˆš3

As we know, tan 60Â° =Â âˆš3

Thus, we can write;

â‡’ tan (A + B) = tan 60Â°

â‡’(A + B) = 60Â° â€¦… (i)

Now again given;

tan (A â€“ B) = 1/âˆš3

Since, tan 30Â° =Â 1/âˆš3

Thus, we can write;

â‡’ tan (A â€“ B) = tan 30Â°

â‡’(A â€“ B) = 30Â° â€¦.. (ii)

Adding the equation (i) and (ii), we get;

A + B + A â€“ B = 60Â° + 30Â°

2A = 90Â°

A= 45Â°

Now, put the value of A in eq. (i) to find the value of B;

45Â° + B = 60Â°

B = 60Â° â€“ 45Â°

B = 15Â°

Therefore A = 45Â° and B = 15Â°

**Question 8:Â Show that :**

**(i) tan 48Â° tan 23Â° tan 42Â° tan 67Â° = 1**

**(ii) cos 38Â° cos 52Â° â€“ sin 38Â° sin 52Â° = 0**

Solution:

(i) tan 48Â° tan 23Â° tan 42Â° tan 67Â°

We can also write the above given tan functions in terms of cot functions, such as;

tan 48Â° = tan (90Â° â€“ 42Â°) = cot 42Â°

tan 23Â° = tan (90Â° â€“ 67Â°) = cot 67Â°

Hence, substituting these values, we get

= cot 42Â° cot 67Â° tan 42Â° tan 67Â°

= (cot 42Â° tan 42Â°) (cot 67Â° tan 67Â°)

= 1 Ã— 1Â [since cot A.tan A = 1]

= 1

(ii) cos 38Â° cos 52Â° â€“ sin 38Â° sin 52Â°

We can also write the given cos functions in terms of sin functions.

cos 38Â° = cos (90Â° â€“ 52Â°) = sin 52Â°

cos 52Â°= cos (90Â° – 38Â°) = sin 38Â°

Hence, putting these values in the given equation, we get;

sin 52Â° sin 38Â° â€“ sin 38Â° sin 52Â° = 0

**Question 9: If tan 2A = cot (A â€“ 18Â°), where 2A is an acute angle, find the value of A.**

Solution: Given,

tan 2A = cot (A – 18Â°)

As we know by trigonemetric identities,

tan 2A = cot (90Â° â€“ 2A)

Substituting the above equation in the given equation, we get;

â‡’ cot (90Â° â€“ 2A) = cot (A – 18Â°)

Therefore,

â‡’ 90Â° â€“ 2A = A – 18Â°

â‡’ 108Â° = 3A

A = 108Â° / 3

Hence, the value of A = 36Â°

**Question 10:Â If A, B and C are interior angles of a triangle ABC, then show that sin [(B + C)/2] = cos A/2.**

Solution:

As we know, for any given triangle, the sum of all its interior angles is equals to 180Â°.

Thus,

A + B + C = 180Â° â€¦.(1)

Now we can write the above equation as;

â‡’ B + C = 180Â° â€“ A

Dividing by 2 on both the sides;

â‡’ (B + C)/2 = (180Â° – A)/2

â‡’ (B + C)/2 = 90Â° – A/2

Now, put sin function on both sides.

â‡’ sin (B + C)/2 = sin (90Â° – A/2)

Since,

sin (90Â° – A/2) = cos A/2

Therefore,

sin (B + C)/2 = cos A/2

**Question 11: Prove the identities:**

**(i)Â âˆš[1 + sinA/1 – sinA] = sec A + tan A**

**(ii)Â (1 + tan ^{2}A/1 + cot^{2}A) = (1 – tan A/1 – cot A)^{2} = tan^{2}A**

Solution:

(i) Given:âˆš[1 + sinA/1 – sinA] = sec A + tan A

(ii) Given: (1 + tan^{2}A/1 + cot^{2}A) = (1 – tan A/1 – cot A)^{2} = tan^{2}A

LHS:

= 1+tanÂ²A / 1+cotÂ²A

Using the trigonometric identities we know that 1+tanÂ²A= SecÂ²A and 1+cotÂ²A= CosecÂ²A

= SecÂ²A/ CosecÂ²A

On taking the reciprocals we get

= SinÂ²A/CosÂ²A

= tanÂ²A

RHS:

=(1-tanA)Â²/(1-cotA)Â²

Substituting the reciprocal value of tan A and cot A we get,

=(1-sinA/cosA)Â²/(1-cosA/sinA)Â²

=[(cosA-sinA)/cosA]Â²/ [(sinA-cos)/sinA)Â²

=(cosA-sinA)Â²Ã—sinÂ²A /CosÂ²A. /(sinA-cosA)Â²

=1Ã—sinÂ²A/CosÂ²AÃ—1.

=tan^{2}A

The values of LHS and RHS are the same.

Hence proved.

We hope students must have found this information onÂ Important Questions Class 10 Maths Chapter 8 Introduction to Trigonometry helpful for their board exam preparation. They must keep visiting BYJU’S for the latest update on CBSE/ICSE/State Board/ Competitive exams.

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