Important questions of Chapter 8- Introduction to Trigonometry of Class 10 is given here for students who want to get score good marks in their board exams. These questions are in accordance with the latest CBSE syllabus (NCERT book) for academic session 2019-2020 and are formulated based on previous year questions papers. Solving these questions will help students to get prepared for the final exam. Get the important questions for all the chapters of 10th standard Maths here.

In Chapter 8, students will be introduced with 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 major 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

Here are the important questions for class 10th 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.

**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

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

AC^{2}Â (24)^{2}+7^{2}

AC^{2}Â =(576+49)

AC^{2}Â =Â 625cm^{2}

Therefore, AC = 25 cm

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

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

Sin A = BC/AC = 7/25

Again, cosine of an angle is equal to the ratio of 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

As we know, the angles opposite to the equal sides are equal, by isosceles triangle theorem.

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.

As per the given question,

cot A = 4/3 = AB/BC

So, AB = 4k and BC = 3k, where k is any positive integer.

Using Pythagoras theorem,

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

Putting the values of AB and BC, we get;

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

AC^{2} = 16k^{2Â }+ 9k^{2}

AC = âˆš25k^{2Â }= 5k

Now we have all the sides.

As we know,

tan A = 1/cot A

tan A = 3/4

sin A = BC/AC = 3/5

cos A = AB/AC = 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)^{2}-(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 = 5cm

PR + QR = 25cm

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^{2}+x^{2}-50x = 25 + x^{2}

625 – 50 x = 25

50 x = 600

x = 12

So, QR = 12

PR = 25 – QR = 25 – 12 = 13

Therefore,

sin P = QR/PR = 12/13

cos P = PQ/pR = 5/13

tan P = QR/PQ = 12/5

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

Solution: Since we know,

tan 45 = 1

cos 30 = 3/2

sin 60 = 3/2

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€¦ equation (ii)

Now on 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 the values, we get

= cot 42Â° cot 67Â° tan 42Â° tan 67Â°

= (cot 42Â° tan 42Â°) (cot 67Â° tan 67Â°) = 1Ã—1 = 1Â Â Â Â [cot A.tan A = 1]

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

We can also write the above given cos functions in terms of sin functions, such as;

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 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 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

Hence proved.

**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

Nice work by byjus ..