Ch 8: Introduction to Trigonometry (Part 1)
Master the basics of Sin, Cos, Tan ratios and the standard angle table (0°, 30°, 45°, 60°, 90°).
Exercise 8.1 Solutions
💡 Trick: Pandit Badri Prasad Hari Hari Bol (P/H = sin, B/H = cos, P/B = tan)
Q1. In ∆ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C
Solution:
By Pythagoras Theorem in ∆ABC: AC² = AB² + BC²
AC² = (24)² + (7)² = 576 + 49 = 625 ⇒ AC = 25 cm (Hypotenuse)
(i) For Angle A: Base(AB)=24, Perpendicular(BC)=7
sin A = P/H = BC/AC = 7/25
cos A = B/H = AB/AC = 24/25
(ii) For Angle C: Base(BC)=7, Perpendicular(AB)=24
sin C = P/H = AB/AC = 24/25
cos C = B/H = BC/AC = 7/25
Q2. In Fig. 8.13, find tan P - cot R. (Given PQ = 12 cm, PR = 13 cm, right angled at Q)
Solution:
By Pythagoras Theorem: QR² = PR² - PQ² = 13² - 12² = 169 - 144 = 25 ⇒ QR = 5 cm
For tan P: P = QR(5), B = PQ(12) ⇒ tan P = 5/12
For cot R: P = PQ(12), B = QR(5) ⇒ cot R = B/P = 5/12
tan P - cot R = 5/12 - 5/12 = 0
Q3. If sin A = 3/4, calculate cos A and tan A.
Solution:
sin A = P/H = 3k/4k. So, Perpendicular = 3k, Hypotenuse = 4k.
Base² = H² - P² = (4k)² - (3k)² = 16k² - 9k² = 7k² ⇒ Base = √7 k
cos A = B/H = √7 / 4
tan A = P/B = 3 / √7
Q4. Given 15 cot A = 8, find sin A and sec A.
Solution:
cot A = 8/15. (B/P). Base = 8k, P = 15k.
H² = (8k)² + (15k)² = 64k² + 225k² = 289k² ⇒ H = 17k
sin A = P/H = 15/17
sec A = H/B = 17/8
Q5. Given sec θ = 13/12, calculate all other trigonometric ratios.
Solution:
sec θ = H/B = 13k/12k. P² = (13k)² - (12k)² = 169k² - 144k² = 25k² ⇒ P = 5k.
- sin θ = P/H = 5/13
- cos θ = B/H = 12/13
- tan θ = P/B = 5/12
- cosec θ = H/P = 13/5
- cot θ = B/P = 12/5
Q6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
Let ∆ABC be a right-angled triangle at C.
cos A = AC/AB
cos B = BC/AB
Given cos A = cos B ⇒ AC/AB = BC/AB ⇒ AC = BC
Angles opposite to equal sides of a triangle are equal. Therefore, ∠A = ∠B. (Proved)
Q7. If cot θ = 7/8, evaluate: (i) (1 + sin θ)(1 - sin θ) / (1 + cos θ)(1 - cos θ) (ii) cot² θ
Solution:
Using (a+b)(a-b) = a² - b², part (i) becomes:
(1 - sin² θ) / (1 - cos² θ) = cos² θ / sin² θ = cot² θ
Since cot θ = 7/8, cot² θ = (7/8)² = 49/64
Therefore, both (i) and (ii) yield the value 49/64.
Q8. If 3 cot A = 4, check whether (1 - tan² A)/(1 + tan² A) = cos² A - sin² A or not.
Solution:
cot A = 4/3. B = 4k, P = 3k. H² = 16k² + 9k² = 25k² ⇒ H = 5k.
tan A = 3/4, sin A = 3/5, cos A = 4/5.
LHS: [1 - (3/4)²] / [1 + (3/4)²] = [1 - 9/16] / [1 + 9/16] = (7/16) / (25/16) = 7/25
RHS: cos² A - sin² A = (4/5)² - (3/5)² = 16/25 - 9/25 = 7/25
LHS = RHS. Yes, they are equal.
Q9. In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of: (i) sin A cos C + cos A sin C (ii) cos A cos C - sin A sin C.
Solution:
tan A = 1/√3. P = 1k, B = √3k. H² = 1k² + 3k² = 4k² ⇒ H = 2k.
sin A = 1/2, cos A = √3/2.
sin C = √3/2, cos C = 1/2.
(i) (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 4/4 = 1
(ii) (√3/2)(1/2) - (1/2)(√3/2) = √3/4 - √3/4 = 0
Q10. In ∆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:
Let QR = x. Then PR = 25 - x.
By Pythagoras: PR² = PQ² + QR²
(25 - x)² = (5)² + x²
625 + x² - 50x = 25 + x²
600 = 50x ⇒ x = 12 (So QR = 12 cm)
PR = 25 - 12 = 13 cm. (So H = 13, P = 12, B = 5 for Angle P).
- sin P = P/H = 12/13
- cos P = B/H = 5/13
- tan P = P/B = 12/5
Q11. State whether the following are true or false. Justify.
(i) The value of tan A is always less than 1. (False, tan 60° = √3 > 1)
(ii) sec A = 12/5 for some value of angle A. (True, Hypotenuse is greater than Base)
(iii) cos A is the abbreviation used for the cosecant of angle A. (False, it's cosine)
(iv) cot A is the product of cot and A. (False, cot has no meaning without angle)
(v) sin θ = 4/3 for some angle θ. (False, Perpendicular cannot be greater than Hypotenuse)
Exercise 8.2 (Trigonometric Tables)
Q1. Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
= (√3/2)(√3/2) + (1/2)(1/2)
= 3/4 + 1/4 = 4/4 = 1
(ii) 2 tan² 45° + cos² 30° - sin² 60°
= 2(1)² + (√3/2)² - (√3/2)²
= 2 + 3/4 - 3/4 = 2
(iii) cos 45° / (sec 30° + cosec 30°)
= (1/√2) / (2/√3 + 2)
= (1/√2) / [ (2 + 2√3) / √3 ]
= √3 / [ √2(2 + 2√3) ]
= √3 / (2√2 + 2√6). Rationalizing the denominator gives:
(3√2 - √6) / 8
(iv) (sin 30° + tan 45° - cosec 60°) / (sec 30° + cos 60° + cot 45°)
= (1/2 + 1 - 2/√3) / (2/√3 + 1/2 + 1)
= [ (3/2) - (2/√3) ] / [ (3/2) + (2/√3) ]
= (3√3 - 4) / (3√3 + 4). Rationalizing gives:
(43 - 24√3) / 11
(v) (5 cos² 60° + 4 sec² 30° - tan² 45°) / (sin² 30° + cos² 30°)
= [ 5(1/2)² + 4(2/√3)² - (1)² ] / [ (1/2)² + (√3/2)² ]
= [ 5(1/4) + 4(4/3) - 1 ] / [ 1/4 + 3/4 ]
= [ 5/4 + 16/3 - 1 ] / 1
= (15 + 64 - 12) / 12 = 67 / 12
Q2. Choose the correct option and justify your choice:
(i) 2 tan 30° / (1 + tan² 30°)
= 2(1/√3) / (1 + 1/3) = (2/√3) / (4/3) = √3/2.
Since sin 60° = √3/2. Option (A) sin 60° is correct.
(ii) (1 - tan² 45°) / (1 + tan² 45°)
= (1 - 1) / (1 + 1) = 0/2 = 0.
Option (D) 0 is correct.
(iii) sin 2A = 2 sin A is true when A =
Put A = 0°: sin 0 = 0 and 2 sin 0 = 0. Both sides equal.
Option (A) 0° is correct.
(iv) 2 tan 30° / (1 - tan² 30°)
= (2/√3) / (1 - 1/3) = (2/√3) / (2/3) = √3.
Since tan 60° = √3. Option (C) tan 60° is correct.
Q3. If tan (A + B) = √3 and tan (A - B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.
Solution:
tan (A + B) = √3 ⇒ tan (A + B) = tan 60° ⇒ A + B = 60° --- (Eq 1)
tan (A - B) = 1/√3 ⇒ tan (A - B) = tan 30° ⇒ A - B = 30° --- (Eq 2)
Adding (1) and (2):
2A = 90° ⇒ A = 45°
Put A = 45° in Eq 1: 45° + B = 60° ⇒ B = 15°
Q4. State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B. (False, put A=30, B=60. sin 90=1, but sin 30+sin 60 ≠ 1)
(ii) The value of sin θ increases as θ increases. (True, from 0 to 1)
(iii) The value of cos θ increases as θ increases. (False, it decreases from 1 to 0)
(iv) sin θ = cos θ for all values of θ. (False, it is only true for 45°)
(v) cot A is not defined for A = 0°. (True, cot 0 = cos 0 / sin 0 = 1 / 0 = undefined)