Important Questions for Class 10 Maths Chapter 8- Introduction to Trigonometry

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 : In ∆ 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 = ∠B = 90°

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

AC2=AB2+BC2

AC2 (24)2+72

AC2 =(576+49)

AC2 = 625cm2

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;

Hypotenuse2 = Perpendicular2+Base2

AC2 = AB2 + BC2

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

(4k)2 = (AB)2 + (3k)2

16k2 – 9k2 = AB2

AB2 = 7k2

Hence, AB = √7 k

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 = √7 k/4k = √7/4

And,

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

tan A = 3k/√7 k = 3/√7

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

Question 4: If 3 cot A = 4, check whether (1-tan2A)/(1+tan2A) = cos2 A – sin2 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,

AC2 = AB2 + BC2

Putting the values of AB and BC, we get;

AC2 = (4k)2 + (3k)2

AC2 = 16k+ 9k2

AC = 25k= 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: (1-tan2A)/(1+tan2A) = cos2 A – sin2 A or not

Let us take L.H.S. first;

(1-tan2A)/(1+tan2A) = 1-(3/4)2/1+(3/4)2

= [1-9/16]/[1+9/16] = 7/25

R.H.S. = cos2 A – sin2 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, 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:

PR2 = PQ2 + QR2

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

(25-x)2 = (5)2 + (x)2

252+x2-50x = 25 + x2

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 2tan2 45 + cos2 30 – sin2 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 + (3/2)2 – (3/2)2

2+0

= 2

Question 7: If tan (A + B) =√3 and tan (A – B) =1/√3,0° < A + B ≤ 90°; A > B, find A and B.

Solution: Given,

tan (A + B) = √3

As we know, tan 60° = √3

Thus, we can write;

⇒ tan (A + B) = tan 60°

⇒(A + B) = 60° …… (i)

Now again given;

tan (A – B) = 1/√3

Since, tan 30° = 1/√3

Thus, we can write;

⇒ tan (A – B) = tan 30°

⇒(A – B) = 30° … equation (ii)

Now on adding the equation (i) and (ii), we get;

A + B + A – B = 60° + 30°

2A = 90°

A= 45°

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

45° + B = 60°

B = 60° – 45°

B = 15°

Therefore A = 45° and B = 15°

Question 8: Show that :

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

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Solution:

(i) tan 48° tan 23° tan 42° tan 67°

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

tan 48° = tan (90° – 42°) = cot 42°

tan 23° = tan (90° – 67°) = cot 67°

Hence, substituting the values, we get

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1        [cot A.tan A = 1]

(ii) cos 38° cos 52° – sin 38° sin 52°

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

cos 38° = cos (90° – 52°) = sin 52°

cos 52°= cos (90°-38°) = sin 38°

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

sin 52° sin 38° – sin 38° sin 52° = 0

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

Solution: Given,

tan 2A = cot (A- 18°)

As we know by trigonemetric identities,

tan 2A = cot (90° – 2A)

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

⇒ cot (90° – 2A) = cot (A -18°)

Therefore,

⇒ 90° – 2A = A- 18°

⇒ 108° = 3A

A = 108° / 3

Hence, the value of A = 36°

Question 10:  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 180°

Thus,

A + B + C = 180° ….(1)

Now we can write the above equation as;

⇒ B + C = 180° – A

Dividing by 2 both the sides;

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = 90°-A/2

Now, put sin function on both sides.

⇒ sin (B+C)/2 = sin (90°-A/2)

Since,

sin (90°-A/2) = = cos A/2

Therefore,

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

Hence proved.

Question 11: Prove the identities:

(i) √[1+sinA/1-sinA] = sec A + tan A

(ii) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A
Solution:
(i) Given:√[1+sinA/1-sinA] = sec A + tan A

(ii) Given: (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A