Class 10 Maths (NCERT)
Chapter 8: Introduction to Trigonometry
Exercise 8.1 – Stepwise Solutions
Basic Trigonometric Ratios (Concept)
In a right triangle, for an angle θ:
- sin θ = (Perpendicular) / (Hypotenuse)
- cos θ = (Base) / (Hypotenuse)
- tan θ = (Perpendicular) / (Base)
- cosec θ = 1/sin θ
- sec θ = 1/cos θ
- cot θ = 1/tan θ
1. In ΔABC right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
Step 1: Find AC using Pythagoras theorem
AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625
AC = √625 = 25 cm
(i) sin A, cos A
For angle A: Opposite = BC, Adjacent = AB, Hypotenuse = AC
sin A = BC/AC = 7/25
cos A = AB/AC = 24/25
(ii) sin C, cos C
For angle C: Opposite = AB, Adjacent = BC, Hypotenuse = AC
sin C = AB/AC = 24/25
cos C = BC/AC = 7/25
2. In Fig. 8.13, find tan P − cot R.
From figure: right triangle PQR right-angled at Q
PR = 13 cm (hypotenuse), PQ = 12 cm
Step 1: Find QR
QR² = PR² − PQ² = 13² − 12² = 169 − 144 = 25
QR = 5 cm
Step 2: Find tan P
tan P = QR/PQ = 5/12
Step 3: Find cot R
For angle R: opposite = PQ, adjacent = QR
tan R = PQ/QR = 12/5 ⇒ cot R = 5/12
tan P − cot R = 5/12 − 5/12 = 0
3. If sin A = 3/4, calculate cos A and tan A.
Step 1: Let perpendicular = 3 and hypotenuse = 4
Step 2: Find base using Pythagoras theorem
Base = √(4² − 3²) = √(16 − 9) = √7
Step 3: cos A = Base/Hypotenuse = √7/4
Step 4: tan A = Perpendicular/Base = 3/√7
Rational form: tan A = (3√7)/7
4. Given 15 cot A = 8, find sin A and sec A.
Step 1: 15 cot A = 8 ⇒ cot A = 8/15
So adjacent : opposite = 8 : 15
Step 2: Hypotenuse = √(8² + 15²) = √(64 + 225) = √289 = 17
Step 3: sin A = opposite/hypotenuse = 15/17
Step 4: cos A = adjacent/hypotenuse = 8/17 ⇒ sec A = 17/8
5. Given sec θ = 13/12, calculate all other trigonometric ratios.
Step 1: sec θ = 13/12 ⇒ cos θ = 12/13
Step 2: Take base = 12, hypotenuse = 13
Step 3: perpendicular = √(13² − 12²) = √(169 − 144) = √25 = 5
sin θ = 5/13
cos θ = 12/13
tan θ = 5/12
cot θ = 12/5
sec θ = 13/12
cosec θ = 13/5
6. If ∠A and ∠B are acute angles and cos A = cos B, show that ∠A = ∠B.
Step 1: For acute angles, cosine function is one-to-one (unique value for each angle).
Step 2: Given cos A = cos B and both A, B are acute.
Therefore: ∠A = ∠B.
7. If cot θ = 7/8, evaluate:
(i) (1 + sin θ)(1 − sin θ) / (1 + cos θ)(1 − cos θ)
Step 1: Use identities:
(1 + sin θ)(1 − sin θ) = 1 − sin² θ = cos² θ
(1 + cos θ)(1 − cos θ) = 1 − cos² θ = sin² θ
So expression = cos² θ / sin² θ = cot² θ
cot² θ = (7/8)² = 49/64
(ii) cot² θ
cot² θ = (7/8)² = 49/64
8. If 3 cot A = 4, check whether (1 − tan²A)/(1 + tan²A) = cos²A − sin²A
Step 1: 3 cot A = 4 ⇒ cot A = 4/3 ⇒ tan A = 3/4
LHS: (1 − tan²A)/(1 + tan²A)
= (1 − (3/4)²) / (1 + (3/4)²)
= (1 − 9/16) / (1 + 9/16)
= (7/16) / (25/16) = 7/25
Step 2: From tan A = 3/4 ⇒ take perpendicular=3, base=4, hypotenuse=5
sin A = 3/5, cos A = 4/5
RHS: cos²A − sin²A = (4/5)² − (3/5)² = 16/25 − 9/25 = 7/25
LHS = RHS = 7/25, so verified.
9. In ΔABC right-angled at B, if tan A = 1/√3, find:
tan A = 1/√3 ⇒ A = 30° (acute)
So C = 60°
(i) sin A cos C + cos A sin C
Use identity: sin(A + C) = sin A cos C + cos A sin C
A + C = 30° + 60° = 90°
sin 90° = 1
(ii) cos A cos C − sin A sin C
Use identity: cos(A + C) = cos A cos C − sin A sin C
cos 90° = 0
10. In ΔPQR right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find sin P, cos P and tan P.
Right angle at Q ⇒ PR is hypotenuse
Step 1: Let QR = x, PR = 25 − x
Step 2: Apply Pythagoras theorem
PR² = PQ² + QR²
(25 − x)² = 5² + x²
625 − 50x + x² = 25 + x²
625 − 50x = 25
600 = 50x ⇒ x = 12
So QR = 12 cm and PR = 25 − 12 = 13 cm
Step 3: For angle P: opposite = QR, adjacent = PQ, hypotenuse = PR
sin P = 12/13
cos P = 5/13
tan P = 12/5
11. True/False (Justify)
(i) The value of tan A is always less than 1.
False. Example: tan 60° = √3 > 1.
(ii) sec A = 12/5 for some value of angle A.
True. Because sec A ≥ 1 for acute A, and 12/5 > 1, so possible.
(iii) cos A is the abbreviation used for the cosecant of angle A.
False. cos means cosine. Cosecant is written as cosec.
(iv) cot A is the product of cot and A.
False. cot A means cotangent of angle A, not multiplication.
(v) sin θ = 4/3 for some angle θ.
False. Because sin θ always lies between −1 and 1, but 4/3 > 1.
Final Answers (Quick)
1(i) sinA=7/25, cosA=24/25; (ii) sinC=24/25, cosC=7/25
2) 0
3) cosA=√7/4, tanA=3√7/7
4) sinA=15/17, secA=17/8
5) sin=5/13, cos=12/13, tan=5/12, cot=12/5, sec=13/12, cosec=13/5
6) ∠A=∠B
7(i)=49/64, (ii)=49/64
8) Verified, both sides = 7/25
9(i)=1, (ii)=0
10) sinP=12/13, cosP=5/13, tanP=12/5
11) (i)F (ii)T (iii)F (iv)F (v)F