MAK LABS — Std 9 Maths Part 2 Chapter 8 Trigonometry — Practice Sets 8.1, 8.2 & Problem Set 8 Solutions

1In the Fig. 8.12, ∠R is the right angle of ΔPQR. Write the following ratios. (i) sin P (ii) cos Q (iii) tan P (iv) tan Q

Fig. 8.12 — hypotenuse PQ

Hypotenuse is PQ (opposite the right angle R). For ∠P, opposite side is QR and adjacent side is PR; for ∠Q, opposite side is PR and adjacent side is QR.

sin P = QRPQ
cos Q = QRPQ
tan P = QRPR
tan Q = PRQR

2In the right angled ΔXYZ, ∠XYZ = 90° and a, b, c are the lengths of the sides as shown in the figure. Write the following ratios, (i) sin X (ii) tan Z (iii) cos X (iv) tan X.

Fig. 8.13 — XY = b, YZ = a, XZ = c

Hypotenuse XZ = c. For ∠X: opposite side YZ = a, adjacent side XY = b. For ∠Z: opposite side XY = b, adjacent side YZ = a.

sin X = ac
tan Z = ba
cos X = bc
tan X = ab

3In right angled ΔLMN, ∠LMN = 90°, ∠L = 50° and ∠N = 40°, write the following ratios. (i) sin 50° (ii) cos 50° (iii) tan 40° (iv) cos 40°

Fig. 8.14 — hypotenuse LN

sin 50° and cos 50° are ratios of ∠L; tan 40° and cos 40° are ratios of ∠N.

sin 50° = MNLN
cos 50° = LMLN
tan 40° = LMMN
cos 40° = MNLN

4In the figure 8.15, ∠PQR = 90°, ∠PQS = 90°, ∠PRQ = α and ∠QPS = θ. Write the following trigonometric ratios. (i) sin α, cos α, tan α (ii) sin θ, cos θ, tan θ

Fig. 8.15

(i) α is an angle of right angled ΔPQR (∠Q = 90°, hypotenuse PR).

sin α = PQPR, cos α = QRPR, tan α = PQQR

(ii) θ is an angle of right angled ΔPQS (∠Q = 90°, hypotenuse PS).

sin θ = QSPS, cos θ = PQPS, tan θ = QSPQ

1In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table.

Use a right angled triangle with an acute angle θ: take the two known sides in the given ratio (as k multiples) and find the third side by Pythagoras theorem.

(1) cos θ = 3537: adjacent 35k, hypotenuse 37k ⇒ opposite = √37² − 35²k = √144k = 12k. sin θ = 1237, tan θ = 1235

(2) sin θ = 1161: opposite 11k, hypotenuse 61k ⇒ adjacent = √3721 − 121k = 60k. cos θ = 6061, tan θ = 1160

(3) tan θ = 1: opposite = adjacent = k ⇒ hypotenuse = √2k. sin θ = cos θ = 1√2 (θ = 45°)

(4) sin θ = 12: θ = 30° ⇒ cos θ = √32, tan θ = 1√3

(5) cos θ = 1√3: adjacent k, hypotenuse √3k ⇒ opposite = √3 − 1k = √2k. sin θ = √2√3, tan θ = √2

(6) tan θ = 2120: opposite 21k, adjacent 20k ⇒ hypotenuse = √441 + 400k = 29k. sin θ = 2129, cos θ = 2029

(7) tan θ = 815: opposite 8k, adjacent 15k ⇒ hypotenuse = √64 + 225k = 17k. sin θ = 817, cos θ = 1517

(8) sin θ = 35: opposite 3k, hypotenuse 5k ⇒ adjacent = 4k. cos θ = 45, tan θ = 34

(9) tan θ = 12√2: opposite k, adjacent 2√2k ⇒ hypotenuse = √1 + 8k = 3k. sin θ = 13, cos θ = 2√23

Completed table (answers found are highlighted):

Ratio	(1)	(2)	(3)	(4)	(5)	(6)	(7)	(8)	(9)
sin θ	1237	1161	1√2	12	√2√3	2129	817	35	13
cos θ	3537	6061	1√2	√32	1√3	2029	1517	45	2√23
tan θ	1235	1160	1	1√3	√2	2120	815	34	12√2

2Find the values of —

Use sin 30° = 12, cos 30° = √32, tan 45° = 1, sin 60° = √32, cos 60° = 12, tan 60° = √3, cos 45° = 1√2, sin 90° = cos 0° = 1.

(i) 5 sin 30° + 3 tan 45° = 5 × 12 + 3 × 1 = 52 + 3 = 112

(ii) 45 tan² 60° + 3 sin² 60° = 45 × 3 + 3 × 34 = 125 + 94 = 4820 + 4520 = 9320

(iii) 2 sin 30° + cos 0° + 3 sin 90° = 2 × 12 + 1 + 3 × 1 = 1 + 1 + 3 = 5

(iv) tan 60°sin 60° + cos 60° = √3√32 + 12 = √3√3 + 12 = 2√3√3 + 1
Multiply numerator and denominator by (√3 − 1): 2√3(√3 − 1)3 − 1 = √3(√3 − 1) = 3 − √3

(v) cos² 45° + sin² 30° = (1√2)² + (12)² = 12 + 14 = 34

(vi) cos 60° × cos 30° + sin 60° × sin 30° = 12 × √32 + √32 × 12 = √34 + √34 = √32

3If sin θ = 4/5 then find cos θ.

sin² θ + cos² θ = 1

(45)² + cos² θ = 1 ∴ cos² θ = 1 − 1625 = 925

θ is acute, so cos θ is positive ∴ cos θ = 35

∴ cos θ = 35

4If cos θ = 15/17 then find sin θ.

sin² θ + cos² θ = 1

sin² θ = 1 − (1517)² = 1 − 225289 = 64289

∴ sin θ = 817

1Choose the correct alternative answer for following multiple choice questions.

(i) Which of the following statements is true? — (A) sin θ = cos (90 − θ)
Sine of an angle equals cosine of its complement.
(ii) Which of the following is the value of sin 90°? — (D) 1
sin 90° = 1.
(iii) 2 tan 45° + cos 45° − sin 45° = ? — (C) 2
2 × 1 + 1√2 − 1√2 = 2.
(iv) cos 28°sin 62° = ? — (D) 1
28° + 62° = 90°, so sin 62° = cos 28°.

2In right angled ΔTSU, TS = 5, ∠S = 90°, SU = 12 then find sin T, cos T, tan T. Similarly find sin U, cos U, tan U.

Fig. 8.26

TU² = TS² + SU² = 5² + 12² = 25 + 144 = 169 ∴ TU = 13 ...... Pythagoras theorem

sin T = SUTU = 1213, cos T = TSTU = 513, tan T = SUTS = 125

sin U = TSTU = 513, cos U = SUTU = 1213, tan U = TSSU = 512

∴ sin T = 1213, cos T = 513, tan T = 125

∴ sin U = 513, cos U = 1213, tan U = 512

3In right angled ΔYXZ, ∠X = 90°, XZ = 8 cm, YZ = 17 cm, find sin Y, cos Y, tan Y, sin Z, cos Z, tan Z.

Fig. 8.27

XY² = YZ² − XZ² = 17² − 8² = 289 − 64 = 225 ∴ XY = 15 cm ...... Pythagoras theorem

sin Y = XZYZ = 817, cos Y = XYYZ = 1517, tan Y = XZXY = 815

sin Z = XYYZ = 1517, cos Z = XZYZ = 817, tan Z = XYXZ = 158

∴ sin Y = 817, cos Y = 1517, tan Y = 815

∴ sin Z = 1517, cos Z = 817, tan Z = 158

4In right angled ΔLMN, if ∠N = θ, ∠M = 90°, cos θ = 24/25, find sin θ and tan θ. Similarly, find (sin² θ) and (cos² θ).

Fig. 8.28

cos θ = MNLN = 2425. Let MN = 24k and LN = 25k.

LM² = LN² − MN² = 625k² − 576k² = 49k² ∴ LM = 7k

sin θ = LMLN = 7k25k = 725; tan θ = LMMN = 7k24k = 724

sin² θ = (725)² = 49625; cos² θ = (2425)² = 576625

Check: sin² θ + cos² θ = 625625 = 1

∴ sin θ = 725, tan θ = 724, sin² θ = 49625, cos² θ = 576625

5Fill in the blanks. (i) sin 20° = cos __° (ii) tan 30° × tan __° = 1 (iii) cos 40° = sin __°

Use sin θ = cos (90° − θ), cos θ = sin (90° − θ) and tan θ × tan (90° − θ) = 1.

sin 20° = cos (90° − 20°) = cos 70°
tan 30° × tan 60° = 1 (since 30° + 60° = 90°)
cos 40° = sin (90° − 40°) = sin 50°

8. Trigonometry

Practice Set 8.1

Practice Set 8.2

Problem Set 8