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 2021-2022. The questions are formulated after analysing the previous year questions papers, exam trend and latest sample papers 2022. Solving these questions will help students to get prepared for the final exam. Students can also get 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 : 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
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
AC2 = AB2 + BC2
AC2 = (24)2 + 72
AC2 = (576 + 49)
AC2 = 625 cm2
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;
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
We know that by isosceles triangle theorem, the angles opposite to the equal sides are equal.
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.
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;
Hypotenuse2 = Perpendicular2 + Base2
AC2 = AB2 + BC2
AC2 = (4k)2 + (3k)2
AC2 = 16k2 + 9k2
AC = √25k2 = 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: (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 = 5 cm
PR + QR = 25 cm
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 – 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 tan2 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° ….. (ii)
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 these values, we get
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° tan 42°) (cot 67° tan 67°)
= 1 × 1 [since cot A.tan A = 1]
= 1
(ii) cos 38° cos 52° – sin 38° sin 52°
We can also write the given cos functions in terms of sin functions.
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 is equals to 180°.
Thus,
A + B + C = 180° ….(1)
Now we can write the above equation as;
⇒ B + C = 180° – A
Dividing by 2 on 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
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
LHS:
= 1+tan²A / 1+cot²A
Using the trigonometric identities we know that 1+tan²A= Sec²A and 1+cot²A= Cosec²A
= Sec²A/ Cosec²A
On taking the reciprocals we get
= Sin²A/Cos²A
= tan²A
RHS:
=(1-tanA)²/(1-cotA)²
Substituting the reciprocal value of tan A and cot A we get,
=(1-sinA/cosA)²/(1-cosA/sinA)²
=[(cosA-sinA)/cosA]²/ [(sinA-cos)/sinA)²
=(cosA-sinA)²×sin²A /Cos²A. /(sinA-cosA)²
=1×sin²A/Cos²A×1.
=tan2A
The values of LHS and RHS are the same.
Hence proved.