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Exercise - 8.1

Introduction to Trigonometry

**Question-1 :-** In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm.

Determine : (i) sin A, cos A (ii) sin C, cos C

Given : In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm, AC = ? By Pythagoras Theorem : AC² = AB² + BC² = (24)² + (7)² = 576 + 49 = 625 AC = 25 cm (i) sin A, cos A sin A = BC/AC = 7/25 cos A = AB/AC = 24/25 (ii) sin C, cos C sin C = AB/AC = 24/25 cos C = BC/AC = 7/25

**Question-2 :-** In Figure, find tan P – cot R.

Given : In Δ PQR, right-angled at Q, PQ = 12 cm, PR = 13 cm, QR = ? By Pythagoras Theorem : QR² = PR² - PQ² = (13)² - (12)² = 169 - 144 = 25 QR = 5 cm tan P = QR/PQ = 5/12 cot R = QR/PQ = 5/12 tan P - cot R = 5/12 - 5/12 = 0

**Question-3 :-** If sin A = 3/4, calculate cos A and tan A.

Given : In Δ ABC, right-angled at B, sin A = 3/4 = BC/AC AC = 4 cm, BC = 3 cm, AB = ? By Pythagoras Theorem : AB² = AC² - BC² = (4)² - (3)² = 16 - 9 = 7 AB = √7 cm cos A = AB/AC = √7/4 tan A = BC/AB = 3/√7

**Question-4 :-** Given 15 cot A = 8, find sin A and sec A.

Given : In Δ ABC, right-angled at B, 15 cot A = 8 cot A = 8/15 = AB/BC AB = 8 cm, BC = 15 cm, AC = ? By Pythagoras Theorem : AC² = AB² + BC² = (8)² + (15)² = 64 + 225 = 289 AC = 17 cm sin A = BC/AC = 15/17 sec A = AC/AB = 17/8

**Question-5 :-** Given sec θ = 13/12 calculate all other trigonometric ratios.

Given : In Δ ABC, right-angled at B, sec θ = 13/12 = AC/AB AB = 12 cm, BC = ?, AC = 13 cm By Pythagoras Theorem : BC² = AC² - AB² = (13)² - (12)² = 169 - 144 = 25 BC = 5 cm sin A = BC/AC = 5/13 cos A = AB/AC = 12/13 tan A = BC/AB = 5/12 cot A = AB/BC = 12/5 cosec A = AC/BC = 13/5

**Question-6 :-** If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Let us consider CD Perpendicular to AB Given : cos A = cos B AD/AC = BD/BC Let AD/BD = AC/BC = k AD = k BD .....(i) AC = k BC .....(ii) By using Pythagoras Theorem for Δ CAD and Δ CBD, we obtain : CD² = AC² - AD² .....(iii) CD² = BC² - BD² .....(iv) from equation (iii) & (iv), we obtain AC² - AD² = BC² - BD² (k BC)² - (k BD)² = BC² - BD² k² BC² - k² BD² = BC² - BD² k²(BC² - BD²) = BC² - BD² k² = 1 k = ±1 K = 1 Putting this value in equation (ii), we obtain AC = BC So, ∠ A = ∠ B (Angles opposite to equal sides of a triangle)

**Question-7 :-** If cot θ = 7/8, evaluate :

(i)

(ii) cot² θ

Given : In Δ ABC, right-angled at B, cot θ = 7/8 = AB/BC AB = 7 cm, BC = 8 cm, AC = ? By Pythagoras Theorem : AC² = AB² + BC² = (7)² + (8)² = 49 + 64 = 113 AC = √113 cm sin θ = BC/AC = 8/√113 cos θ = AB/Ac = 7/√113 (i) (ii) cot² θ = (cos θ/sin θ)² = (7/√113 x √113/8)² = (7/8)² = 49/64

**Question-8 :-** If 3 cot A = 4, check whether
or Not.

Given : In Δ ABC, right-angled at B, 3 cot A = 4 cot A = 4/3 = AB/BC AB = 4 cm, BC = 3 cm, AC = ? By Pythagoras Theorem : AC² = AB² + BC² = (4)² + (3)² = 16 + 9 = 25 AC = 5 cm sin A = BC/AC = 3/5 cos A = AB/AC = 4/5 tan A = BC/AB = 3/4 Now, L.H.S R.H.S cos² A - sin² A = (4/5)² - (3/5)² = 16/25 - 9/25 = 7/25 L.H.S = R.H.S

**Question-9 :-** 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

Given : In Δ ABC, right-angled at B, tan A = 1/√3 = BC/AB AB = √3 cm, BC = 1 cm, AC = ? By Pythagoras Theorem : AC² = AB² + BC² = (√3)² + (1)² = 3 + 1 = 4 AC = 2 cm sin A = BC/AC = 1/2 cos A = AB/AC = √3/2 sin C = AB/AC = √3/2 cos C = BC/AC = 1/2 (i) sin A cos C + cos A sin C = 1/2 x 1/2 + √3/2 x √3/2 = 1/4 + 3/4 = 4/4 = 1 (ii) cos A cos C – sin A sin C = √3/2 x 1/2 - 1/2 x √3/2 = √3/4 - √3/4 = 0

**Question-10 :-** In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Given : In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm PR = 25 - QR By Pythagoras Theorem : PR² = PQ² + QR² (25 - QR)² = (5)² + QR² 25² + QR² - 50 QR = 25 + QR² 625 - 50 QR = 25 -50 QR = 25 - 625 -50 QR = -600 QR = 600/50 QR = 12 cm Now, PR = 25 - QR = 25 - 12 = 13 cm sin P = QR/PR = 12/13 cos P = PQ/PR = 5/13 tan P = QR/PQ = 12/5

**Question-11 :-** State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) sec A = 12/5, for some value of angle A.

(iii) cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sin θ = 4/3 for some angle θ.

(i) Consider a Δ ABC, right-angled at B. tan A = 10/5 But 10/5 > 1 tan A > 1 So, tan A < 1 in not always true. Hence, the given statement is false.

(ii) sec A = 12/5 AC/AB = 12/5 In Δ ABC, right-angled at B, AC = 12 cm, AB = 5 cm, BC = ? By Pythagoras Theorem : BC² = AC² - AB² = (12)² - (5)² = 144 - 25 = 119 BC = 10.9 It can be observe that for give two sides AC = 12 and AB = 5 BC should not be such that, (AC - AB) < BC < (AC + AB) (12 - 5) < BC < (12 + 5) 7 < BC < 17 However, BC = 10.9. Clearly, such a triangle is possible and hence, such value of sec A is possible. Hence the given statement is true.

(iii) Abbreviation used for cosecant of angle A is cosec A and cos A is the abbreviation used for cosine of angle A. Hence, the given statement is false.

(iv) cot A is not a product of cot and A. It is the cotangent of angle A. Hence, the given statement is false.

(v) sin θ = 4/3 We know that in a Right angled Triangle. sin θ = Perpendicular/Hypotaneouss In a right angled triangle, hypotaneous is always greater than the remaining two sides. Therefore, such value of sin θ is not possible. Hence, the given statement is false.

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