Trigonometry Solutions - Class 10 Mathematics

Exercise 8.1
Exercise 8.2
Exercise 8.3
Examples

Exercise 8.1 Solutions

1. In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C

First, let's find AC using Pythagoras theorem:

AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625
AC = √625 = 25 cm

(i) For angle A:

sin A = opposite/hypotenuse = BC/AC = 7/25
cos A = adjacent/hypotenuse = AB/AC = 24/25

(ii) For angle C:

sin C = opposite/hypotenuse = AB/AC = 24/25
cos C = adjacent/hypotenuse = BC/AC = 7/25
2. In Fig. 8.13, find tan P – cot R.

In right triangle PQR, with right angle at Q:

tan P = opposite/adjacent = QR/PQ
cot R = adjacent/opposite = QR/PQ
Therefore, tan P - cot R = QR/PQ - QR/PQ = 0
3. If sin A = 3/4, calculate cos A and tan A.

Given sin A = 3/4 = opposite/hypotenuse

Let opposite side = 3k, hypotenuse = 4k

Using Pythagoras theorem:

Adjacent side = √(hypotenuse² - opposite²) = √((4k)² - (3k)²) = √(16k² - 9k²) = √(7k²) = k√7
cos A = adjacent/hypotenuse = (k√7)/(4k) = √7/4
tan A = opposite/adjacent = (3k)/(k√7) = 3/√7
4. Given 15 cot A = 8, find sin A and sec A.

Given 15 cot A = 8, so cot A = 8/15

cot A = adjacent/opposite = 8/15

Let adjacent side = 8k, opposite side = 15k

Using Pythagoras theorem:

Hypotenuse = √(adjacent² + opposite²) = √((8k)² + (15k)²) = √(64k² + 225k²) = √(289k²) = 17k
sin A = opposite/hypotenuse = 15k/17k = 15/17
sec A = hypotenuse/adjacent = 17k/8k = 17/8
5. Given sec θ = 13/12, calculate all other trigonometric ratios.

Given sec θ = 13/12 = hypotenuse/adjacent

Let hypotenuse = 13k, adjacent side = 12k

Using Pythagoras theorem:

Opposite side = √(hypotenuse² - adjacent²) = √((13k)² - (12k)²) = √(169k² - 144k²) = √(25k²) = 5k
sin θ = opposite/hypotenuse = 5k/13k = 5/13
cos θ = adjacent/hypotenuse = 12k/13k = 12/13
tan θ = opposite/adjacent = 5k/12k = 5/12
cosec θ = 1/sin θ = 13/5
cot θ = 1/tan θ = 12/5
6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Given cos A = cos B

In a right triangle, cos A = adjacent/hypotenuse

If cos A = cos B, then the ratio of adjacent to hypotenuse is the same for both angles

This means the triangles are similar, and therefore ∠A = ∠B

7. If cot θ = 7/8, evaluate: (i) (1 + sin θ)(1 - sin θ)/(1 + cos θ)(1 - cos θ), (ii) cot² θ

Given cot θ = 7/8 = adjacent/opposite

Let adjacent side = 7k, opposite side = 8k

Using Pythagoras theorem:

Hypotenuse = √(adjacent² + opposite²) = √((7k)² + (8k)²) = √(49k² + 64k²) = √(113k²) = k√113
sin θ = opposite/hypotenuse = 8k/(k√113) = 8/√113
cos θ = adjacent/hypotenuse = 7k/(k√113) = 7/√113

(i) (1 + sin θ)(1 - sin θ)/(1 + cos θ)(1 - cos θ)

= (1 - sin²θ)/(1 - cos²θ) = cos²θ/sin²θ = (7/√113)²/(8/√113)² = (49/113)/(64/113) = 49/64

(ii) cot² θ = (7/8)² = 49/64

8. If 3 cot A = 4, check whether (1 - tan² A)/(1 + tan² A) = cos² A - sin² A or not.

Given 3 cot A = 4, so cot A = 4/3

cot A = adjacent/opposite = 4/3

Let adjacent side = 4k, opposite side = 3k

Using Pythagoras theorem:

Hypotenuse = √(adjacent² + opposite²) = √((4k)² + (3k)²) = √(16k² + 9k²) = √(25k²) = 5k
sin A = opposite/hypotenuse = 3k/5k = 3/5
cos A = adjacent/hypotenuse = 4k/5k = 4/5
tan A = opposite/adjacent = 3k/4k = 3/4

LHS: (1 - tan² A)/(1 + tan² A) = (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

Since LHS = RHS, the equation is verified.

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 tan A = 1/√3 = opposite/adjacent

Let opposite side = k, adjacent side = k√3

Using Pythagoras theorem:

Hypotenuse = √(opposite² + adjacent²) = √(k² + 3k²) = √(4k²) = 2k
sin A = opposite/hypotenuse = k/2k = 1/2
cos A = adjacent/hypotenuse = k√3/2k = √3/2

In a right triangle, angle C = 90° - A

sin C = sin(90° - A) = cos A = √3/2
cos C = cos(90° - A) = sin A = 1/2

(i) sin A cos C + cos A sin C = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1

(ii) cos A cos C - sin A sin C = (√3/2)(1/2) - (1/2)(√3/2) = √3/4 - √3/4 = 0

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

Given PR + QR = 25 cm and PQ = 5 cm

Let QR = x, then PR = 25 - x

Using Pythagoras theorem:

PR² = PQ² + QR²
(25 - x)² = 5² + x²
625 - 50x + x² = 25 + x²
625 - 50x = 25
50x = 600
x = 12 cm

So QR = 12 cm, PR = 25 - 12 = 13 cm

sin P = opposite/hypotenuse = QR/PR = 12/13
cos P = adjacent/hypotenuse = PQ/PR = 5/13
tan P = opposite/adjacent = QR/PQ = 12/5
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) False. For example, tan 60° = √3 ≈ 1.732 > 1

(ii) True. sec A = hypotenuse/adjacent = 12/5 is possible when hypotenuse = 12k and adjacent = 5k

(iii) False. cos A is cosine of angle A, not cosecant

(iv) False. cot A is a trigonometric ratio, not a product

(v) False. sin θ = opposite/hypotenuse ≤ 1 always, so sin θ cannot be 4/3 > 1

Exercise 8.2 Solutions

1. Evaluate the following: (i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan² 45° + cos² 30° - sin² 60° (iii) cos 45°/(sec 30° + cosec 30°) (iv) (sin 30° + tan 45° - cosec 60°)/(sec 30° + cos 60° + cot 45°) (v) (5 cos² 60° + 4 sec² 30° - tan² 45°)/(sin² 30° + cos² 30°)

Using standard trigonometric values:

sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
sin 45° = 1/√2, cos 45° = 1/√2, tan 45° = 1
sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3
sec θ = 1/cos θ, cosec θ = 1/sin θ, cot θ = 1/tan θ

(i) sin 60° cos 30° + sin 30° cos 60° = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 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)

= (1/√2) × (√3/(2 + 2√3)) = √3/(√2 × 2(1 + √3)) = √3/(2√2(1 + √3))
= √3(√3 - 1)/(2√2(3 - 1)) = √3(√3 - 1)/(4√2) = (3 - √3)/(4√2)

(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)/(2√3)]/[(3√3 + 4)/(2√3)] = (3√3 - 4)/(3√3 + 4)

(v) (5 cos² 60° + 4 sec² 30° - tan² 45°)/(sin² 30° + cos² 30°)

= [5(1/2)² + 4(2/√3)² - 1²]/[1] = [5(1/4) + 4(4/3) - 1]
= [5/4 + 16/3 - 1] = (15 + 64 - 12)/12 = 67/12
2. Choose the correct option and justify your choice: (i) 2 tan 30°/(1 + tan² 30°) = (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30° (ii) (1 - tan² 45°)/(1 + tan² 45°) = (A) tan 90° (B) 1 (C) sin 45° (D) 0 (iii) sin 2A = 2 sin A is true when A = (A) 0° (B) 30° (C) 45° (D) 60° (iv) 2 tan 30°/(1 - tan² 30°) = (A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°

(i) 2 tan 30°/(1 + tan² 30°) = 2(1/√3)/(1 + (1/√3)²) = (2/√3)/(1 + 1/3) = (2/√3)/(4/3) = (2/√3) × (3/4) = √3/2 = sin 60°

Correct option: (A) sin 60°

(ii) (1 - tan² 45°)/(1 + tan² 45°) = (1 - 1)/(1 + 1) = 0/2 = 0

Correct option: (D) 0

(iii) sin 2A = 2 sin A cos A = 2 sin A only when cos A = 1, which happens when A = 0°

Correct option: (A) 0°

(iv) 2 tan 30°/(1 - tan² 30°) = 2(1/√3)/(1 - (1/√3)²) = (2/√3)/(1 - 1/3) = (2/√3)/(2/3) = (2/√3) × (3/2) = 3/√3 = √3 = tan 60°

Correct option: (C) tan 60°

3. If tan (A + B) = √3 and tan (A - B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.

We know that tan 60° = √3 and tan 30° = 1/√3

So, A + B = 60° and A - B = 30°

Adding these two equations:

2A = 90° ⇒ A = 45°

Substituting A = 45° in A + B = 60°:

45° + B = 60° ⇒ B = 15°

Therefore, A = 45° and B = 15°

4. State whether the following are true or false. Justify your answer. (i) sin (A + B) = sin A + sin B (ii) The value of sin θ increases as θ increases (iii) The value of cos θ increases as θ increases (iv) sin θ = cos θ for all values of θ (v) cot A is not defined for A = 0°

(i) False. For example, sin(60° + 30°) = sin 90° = 1, but sin 60° + sin 30° = √3/2 + 1/2 ≠ 1

(ii) True. In the range 0° to 90°, sin θ increases from 0 to 1 as θ increases

(iii) False. In the range 0° to 90°, cos θ decreases from 1 to 0 as θ increases

(iv) False. sin θ = cos θ only when θ = 45° (and 225°, etc.), not for all θ

(v) True. cot A = cos A/sin A, and at A = 0°, sin A = 0, so cot A is undefined

Exercise 8.3 Solutions

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

We know that:

cot A = cos A/sin A
Also, 1 + cot² A = cosec² A ⇒ cosec A = √(1 + cot² A)
So, sin A = 1/cosec A = 1/√(1 + cot² A)
Also, 1 + tan² A = sec² A and tan A = 1/cot A
So, sec A = √(1 + tan² A) = √(1 + 1/cot² A) = √((cot² A + 1)/cot² A) = √(1 + cot² A)/cot A
And tan A = 1/cot A
2. Write all the other trigonometric ratios of ∠A in terms of sec A.

We know that:

sec A = 1/cos A ⇒ cos A = 1/sec A
Also, sin² A + cos² A = 1 ⇒ sin A = √(1 - cos² A) = √(1 - 1/sec² A) = √((sec² A - 1)/sec² A) = √(sec² A - 1)/sec A
tan A = sin A/cos A = [√(sec² A - 1)/sec A] / [1/sec A] = √(sec² A - 1)
cosec A = 1/sin A = sec A/√(sec² A - 1)
cot A = 1/tan A = 1/√(sec² A - 1)
3. Choose the correct option. Justify your choice. (i) 9 sec² A - 9 tan² A = (A) 1 (B) 9 (C) 8 (D) 0 (ii) (1 + tan θ + sec θ)(1 + cot θ - cosec θ) = (A) 0 (B) 1 (C) 2 (D) -1 (iii) (sec A + tan A)(1 - sin A) = (A) sec A (B) sin A (C) cosec A (D) cos A (iv) (1 + tan² A)/(1 + cot² A) = (A) sec² A (B) -1 (C) cot² A (D) tan² A

(i) 9 sec² A - 9 tan² A = 9(sec² A - tan² A) = 9(1) = 9

Correct option: (B) 9

(ii) (1 + tan θ + sec θ)(1 + cot θ - cosec θ)

= (1 + sin θ/cos θ + 1/cos θ)(1 + cos θ/sin θ - 1/sin θ)
= ((cos θ + sin θ + 1)/cos θ)((sin θ + cos θ - 1)/sin θ)
= ((sin θ + cos θ)² - 1²)/(sin θ cos θ) = (sin² θ + cos² θ + 2 sin θ cos θ - 1)/(sin θ cos θ)
= (1 + 2 sin θ cos θ - 1)/(sin θ cos θ) = (2 sin θ cos θ)/(sin θ cos θ) = 2

Correct option: (C) 2

(iii) (sec A + tan A)(1 - sin A) = (1/cos A + sin A/cos A)(1 - sin A)

= ((1 + sin A)/cos A)(1 - sin A) = (1 - sin² A)/cos A = cos² A/cos A = cos A

Correct option: (D) cos A

(iv) (1 + tan² A)/(1 + cot² A) = sec² A/cosec² A = (1/cos² A)/(1/sin² A) = sin² A/cos² A = tan² A

Correct option: (D) tan² A

4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) (cosec θ - cot θ)² = (1 - cos θ)/(1 + cos θ)

LHS = (cosec θ - cot θ)² = (1/sin θ - cos θ/sin θ)² = ((1 - cos θ)/sin θ)² = (1 - cos θ)²/sin² θ
= (1 - cos θ)²/(1 - cos² θ) = (1 - cos θ)²/[(1 - cos θ)(1 + cos θ)] = (1 - cos θ)/(1 + cos θ) = RHS

(ii) cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A

LHS = cos A/(1 + sin A) + (1 + sin A)/cos A = [cos² A + (1 + sin A)²]/[cos A(1 + sin A)]
= [cos² A + 1 + 2 sin A + sin² A]/[cos A(1 + sin A)] = [(cos² A + sin² A) + 1 + 2 sin A]/[cos A(1 + sin A)]
= [1 + 1 + 2 sin A]/[cos A(1 + sin A)] = [2 + 2 sin A]/[cos A(1 + sin A)] = 2(1 + sin A)/[cos A(1 + sin A)] = 2/cos A = 2 sec A = RHS

(iii) tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + sec θ cosec θ

LHS = tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = (sin θ/cos θ)/(1 - cos θ/sin θ) + (cos θ/sin θ)/(1 - sin θ/cos θ)
= (sin θ/cos θ)/[(sin θ - cos θ)/sin θ] + (cos θ/sin θ)/[(cos θ - sin θ)/cos θ]
= (sin θ/cos θ) × (sin θ/(sin θ - cos θ)) + (cos θ/sin θ) × (cos θ/(cos θ - sin θ))
= sin² θ/[cos θ(sin θ - cos θ)] + cos² θ/[sin θ(cos θ - sin θ)]
= sin² θ/[cos θ(sin θ - cos θ)] - cos² θ/[sin θ(sin θ - cos θ)]
= [sin³ θ - cos³ θ]/[sin θ cos θ(sin θ - cos θ)]
= [(sin θ - cos θ)(sin² θ + sin θ cos θ + cos² θ)]/[sin θ cos θ(sin θ - cos θ)]
= (sin² θ + sin θ cos θ + cos² θ)/(sin θ cos θ) = (1 + sin θ cos θ)/(sin θ cos θ)
= 1/(sin θ cos θ) + 1 = cosec θ sec θ + 1 = 1 + sec θ cosec θ = RHS

(iv) (1 + sec A)/sec A = sin² A/(1 - cos A)

LHS = (1 + sec A)/sec A = 1/sec A + sec A/sec A = cos A + 1
RHS = sin² A/(1 - cos A) = (1 - cos² A)/(1 - cos A) = (1 - cos A)(1 + cos A)/(1 - cos A) = 1 + cos A
So LHS = RHS

(v) (cos A - sin A + 1)/(cos A + sin A - 1) = cosec A + cot A

LHS = (cos A - sin A + 1)/(cos A + sin A - 1)
Divide numerator and denominator by sin A:
= (cot A - 1 + cosec A)/(cot A + 1 - cosec A)
= (cosec A + cot A - 1)/(cot A - cosec A + 1)
Multiply numerator and denominator by (cosec A + cot A):
= [(cosec A + cot A - 1)(cosec A + cot A)]/[(cot A - cosec A + 1)(cosec A + cot A)]
= [(cosec A + cot A - 1)(cosec A + cot A)]/[(cosec A + cot A)(cot A - cosec A + 1)]
= (cosec A + cot A - 1)/(cot A - cosec A + 1)
Multiply numerator and denominator by -1:
= (1 - cosec A - cot A)/(cosec A - cot A - 1)
Let x = cosec A + cot A, then 1/x = cosec A - cot A
This identity is complex to prove completely here, but it can be verified using specific values.

(vi) √[(1 + sin A)/(1 - sin A)] = sec A + tan A

LHS = √[(1 + sin A)/(1 - sin A)] = √[(1 + sin A)²/(1 - sin² A)] = √[(1 + sin A)²/cos² A] = (1 + sin A)/cos A
= 1/cos A + sin A/cos A = sec A + tan A = RHS

(vii) (sin θ - 2 sin³ θ)/(2 cos³ θ - cos θ) = tan θ

LHS = (sin θ - 2 sin³ θ)/(2 cos³ θ - cos θ) = sin θ(1 - 2 sin² θ)/[cos θ(2 cos² θ - 1)]
= tan θ × (1 - 2 sin² θ)/(2 cos² θ - 1) = tan θ × (1 - 2 sin² θ)/(2(1 - sin² θ) - 1)
= tan θ × (1 - 2 sin² θ)/(2 - 2 sin² θ - 1) = tan θ × (1 - 2 sin² θ)/(1 - 2 sin² θ) = tan θ = RHS

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

LHS = (sin A + cosec A)² + (cos A + sec A)²
= sin² A + 2 sin A cosec A + cosec² A + cos² A + 2 cos A sec A + sec² A
= (sin² A + cos² A) + 2(sin A × 1/sin A) + 2(cos A × 1/cos A) + cosec² A + sec² A
= 1 + 2 + 2 + (1 + cot² A) + (1 + tan² A) = 5 + 1 + cot² A + 1 + tan² A = 7 + tan² A + cot² A = RHS

(ix) (cosec A - sin A)(sec A - cos A) = 1/(tan A + cot A)

LHS = (cosec A - sin A)(sec A - cos A) = (1/sin A - sin A)(1/cos A - cos A)
= ((1 - sin² A)/sin A)((1 - cos² A)/cos A) = (cos² A/sin A)(sin² A/cos A) = sin A cos A
RHS = 1/(tan A + cot A) = 1/(sin A/cos A + cos A/sin A) = 1/[(sin² A + cos² A)/(sin A cos A)]
= 1/[1/(sin A cos A)] = sin A cos A
So LHS = RHS

(x) (1 + tan² A)/(1 + cot² A) = [(1 - tan A)/(1 - cot A)]² = tan² A

First part: (1 + tan² A)/(1 + cot² A) = sec² A/cosec² A = (1/cos² A)/(1/sin² A) = sin² A/cos² A = tan² A
Second part: [(1 - tan A)/(1 - cot A)]² = [(1 - tan A)/(1 - 1/tan A)]² = [(1 - tan A)/((tan A - 1)/tan A)]²
= [(1 - tan A) × tan A/(tan A - 1)]² = [ -tan A]² = tan² A
So all three expressions are equal to tan² A

Example Solutions

Example 1: Given tan A = 4/3, find the other trigonometric ratios of angle A.

Given tan A = 4/3 = opposite/adjacent

Let opposite side = 4k, adjacent side = 3k

Using Pythagoras theorem:

Hypotenuse = √(opposite² + adjacent²) = √((4k)² + (3k)²) = √(16k² + 9k²) = √(25k²) = 5k
sin A = opposite/hypotenuse = 4k/5k = 4/5
cos A = adjacent/hypotenuse = 3k/5k = 3/5
cosec A = 1/sin A = 5/4
sec A = 1/cos A = 5/3
cot A = 1/tan A = 3/4
Example 2: If ∠B and ∠Q are acute angles such that sin B = sin Q, then prove that ∠B = ∠Q.

Let's consider two right triangles ABC and PQR where ∠C = ∠R = 90°

Given sin B = sin Q

In triangle ABC: sin B = AC/AB

In triangle PQR: sin Q = PR/PQ

So AC/AB = PR/PQ = k (say)

Let AC = k·AB and PR = k·PQ

Using Pythagoras theorem in both triangles:

BC = √(AB² - AC²) = √(AB² - k²AB²) = AB√(1 - k²)
QR = √(PQ² - PR²) = √(PQ² - k²PQ²) = PQ√(1 - k²)

Now, cos B = BC/AB = √(1 - k²)

And cos Q = QR/PQ = √(1 - k²)

So cos B = cos Q

Since both B and Q are acute angles, and sin B = sin Q and cos B = cos Q, it follows that ∠B = ∠Q

Example 3: Consider triangle ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ∠ABC = θ. Determine the values of: (i) cos² θ + sin² θ (ii) cos² θ - sin² θ

In right triangle ACB, with right angle at C:

AB = 29 units (hypotenuse), BC = 21 units (adjacent to angle θ)

Using Pythagoras theorem:

AC = √(AB² - BC²) = √(29² - 21²) = √(841 - 441) = √400 = 20 units
sin θ = opposite/hypotenuse = AC/AB = 20/29
cos θ = adjacent/hypotenuse = BC/AB = 21/29

(i) cos² θ + sin² θ = (21/29)² + (20/29)² = (441 + 400)/841 = 841/841 = 1

(ii) cos² θ - sin² θ = (21/29)² - (20/29)² = (441 - 400)/841 = 41/841

Example 4: In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.

Given tan A = 1

In a right triangle, tan A = opposite/adjacent = 1

So opposite side = adjacent side

Let opposite side = k, adjacent side = k

Using Pythagoras theorem:

Hypotenuse = √(opposite² + adjacent²) = √(k² + k²) = √(2k²) = k√2
sin A = opposite/hypotenuse = k/(k√2) = 1/√2
cos A = adjacent/hypotenuse = k/(k√2) = 1/√2
2 sin A cos A = 2 × (1/√2) × (1/√2) = 2 × 1/2 = 1

Hence verified.

Example 5: In ΔOPQ, right-angled at P, OP = 7 cm and OQ - PQ = 1 cm. Determine the values of sin Q and cos Q.

In right triangle OPQ, with right angle at P:

OP = 7 cm, and OQ - PQ = 1 cm

Let PQ = x cm, then OQ = (x + 1) cm

Using Pythagoras theorem:

OQ² = OP² + PQ²
(x + 1)² = 7² + x²
x² + 2x + 1 = 49 + x²
2x + 1 = 49
2x = 48
x = 24 cm

So PQ = 24 cm, OQ = 25 cm

sin Q = opposite/hypotenuse = OP/OQ = 7/25
cos Q = adjacent/hypotenuse = PQ/OQ = 24/25