JKBOSE 10th Class Maths Solutions chapter – 8 Introduction to Trigonometry

1. If a student is looking at the top of the Minar, a right angled can be imagined to be made, as shown below.

2. Suppose a girl is sitting on the balcony of her house located on the bank of a river. She is looking down at a flower pot placed on a stair of a temple situated nearby on the other bank of a river. A right triangle is made in this situation as shown below.

In all the situations given above, the distances or heights can be found by using some mathematical techniques, which come under a branch of mathematics called ‘trigonometry’. The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides), and ‘metron’ (meaning measure). In fact, trigonometry is the study of relationships between the sides and angles of a triangle.

In this Chapter, we will study some ratios of the sides of a right triangle with respect to its acute angles called trigonometric ratios of the angle. We will also define the trigonometric ratios for angles of measure 0° and 90°. We will I calculate trigonometric ratios for some specific angles and establish some identities involving these ratios called trigonometric identities.

Solution. Given that,

AB = 24 cm; BC = 7 cm; ∠B = 90°

By using Pythagoras Theorem,

AC² = AB² + BC²

AC² = (24)² + (7)²

AC² = 576 + 49 = 625

Solution. Hyp. PR = 13 cm

By using Pythagoras Theorem,

PR² = PQ² + QR²

or               (13)² = (12)² + QR²

or                  169 = 144 + (QR)²

or        169 – 144 = (QR)²

or                    25 = (QR)²

or                  QR = ±√25

or                  QR = 5, – 5.

But               QR = 5 cm.

[QR ≠ – 5, because side cannot be negative]

Solution. Let ABC be any triangle with right angle at B.

Therefore, if BC = 3K, then AC 4K, where K is a positive number.

By using Pythagoras Theorem,

AC² = AB² + BC²

or              (4K)² = (AB)² + (3K)²

or               16K² = AB² + 9K²

or    16K² – 9K² = AB²

or                 7K² =AB²

Solution. Let ABC be any right angled triangle where A is an acute angle with right angle at B.

Solution. Let ABC be any right angled triangle with right angle at B.

Therefore, if AC = 13k, then AB = 12k,

where k is a positive number.

By using Pythagoras Theorem,

AC² = (AB)² + (BC)²

or             (13k)² = (12k)² + (BC)²

or             169k² = 144k² + (BC)²

or 169k² – 144k² = (BC)²

or         (BC)² = 25k²

Solution. Let ABC be any triangle,

where ∠A and ∠B are acute angles. To find cos A and cos B.

Draw CM ⊥ AB

∠AMC = ∠BMC = 90°

In right angled ΔAMC,

Solution. (i) ∠ABC = θ.

In right angled triangle ABC with right angle at C.

Therefore, if BC = 7k, then AC = 8k

where k is a positive number

By using Pythagoras Theorem,

AB² = (BC)² + (AC)²

or   (AB)² = (7k)² + (8k)²

or   (AB)² = 49k² + 64k²

or   (AB)² = 113k²

Solution. Let ABC be a right angled triangle with right angled at B.

⇒ If AB = 4k then, BC = 3k where k is a positive number

By using Pythagoras Theorem,

(AC)² = (AB)² + (BC)²

(AC)² = (4k)² + (3k)²

(AC)² = 16k² + 9k²

(AC)² = 25k²

Solution. (i) Given: ΔABC with right angled at B

If BC = k then AB = √3k

where k is a positive number.

In right angled triangle ABC,

By using Pythagoras Theorem,

(AC)² = (AB)² + (BC)²

Solution. Given: ΔPQR, right angled at Q

PR + QR = 25 cm

PQ = 5 cm

In right angled triangle PQR

By using Pythagoras Theorem,

(PR)² = (PQ)² + (RQ)²

or      (PR²) = (5)² + (RQ)²

[∵ PR + QR = 25

    QR = 25 – PR]

or      (PR)² = 25 + [25 – PR]²

or      (PR)² = 25 + (25)² + (PR)² – 2 × 25 × PR

or      (PR)² = 25 + 625 + (PR)² – 50 PR

or      (PR)² – (PR)² + 50 PR = 650

or      50 PR = 650

Solution. (i) False;

∵ tan 60° = √3 =1 · 732 > 1.

(ii) True; sec A = 12/5 = 2 · 40 > 1.

∵ Sec A is always greater than 1.

(iii) False;

Because cos A is used for cosine A.

(iv) False;

Because cot A is cotangent of the angle A not the product of cot and A.

(v) False; sin θ =4/3 = 1 · 666 > 1

Because sin θ is always less than 1.

Solution. (i) sin 60° cos 30° + sin 30° cos 60°

So, correct answer is (A). Ans.

So, correct answer is (D).

(iii) Here when A = 90°

L.H.S. = sin 2A = sin 0° = 0

and R.H.S. = 2 sin A = 2 sin 0°

= 2 × 0 = 0

∴ Option (A) is correct. Ans.

Option (A) is correct. Ans.

Put value of A = 45° in (1)

45° + B = 60°

B = 60° – 45°

B = 15°

Hence     A = 45° and B = 15°   Ans.

Solution. (i) False.

when A = 60°, B = 30°

L.H.S. = sin (A + B) = sin (60° + 30°)

= sin 90° = 1

R.H.S. = sin A + sin B

= sin 60° + sin 30°

i.e. L.H.S. ≠ R.H.S.

and   sin 90° = 1

i.e. value of sin θ increases as θ increases from 0° to 90°.

(iii) False

(iii) cos 48° – sin 42°

= cos (90° – 42°) – sin 42°

[ ∵ cos (90°- θ) = sin θ]

= sin 42° – sin 42° = 0. Ans.

(iv) cosec 31° – sec 59°.

= cosec 31° – sec (90° – 31°)

= cosec 31°- cosec 31°

[∵ sec (90° – θ) = cosec θ]

= 0. Ans.

Solution. (i) L.H.S.

= tan 48° tan 23° tan 42° tan 67°

= tan 48° × tan 23° × tan (90° – 48°) × tan (90° – 23°)

= tan 48° × tan 23° × cot 48° × cot 23°

= tan 48° × tan 23° × 1/tan 48° × 1/tan 23° = 1

∴ L.H.S. = R.H.S.

(ii) L.H.S.= cos 38° cos 52° – sin 38° sin 52°

= cos 38° × cos (90° – 38°) – sin 38° × sin (90° – 38°)

= cos 38° × sin 38° – sin 38° × cos 38°

= 0

∴L.H.S. = R.H.S.

Solution. Given: tan 2A = cot (A – 18°)

⇒ cot (90° – 2A) = cot (A – 18°)

[∵ cot (90°- θ) = tan θ]

⇒ 90° – 2A = A – 18°

⇒ 3A = 108º

⇒ A = 36⁰.

Solution. Given that: tan A = cot B

⇒ tan A = tan (90°- B)

[∵ tan (90° – θ) = cot θ]

⇒ A = 90° – B

⇒ A + B = 90°.

Solution. Given that : sec 4A = cosec (A – 20°)

⇒ cosec (90°- 4A) = cosec (A -20°)

[∵ cosec (90°- θ) = sec θ]

⇒ 90° – 4A = A – 20°

⇒ 5A = 110°

⇒ A = 22°.

Solution. Since, A, B and C are interior angles of a triangle

∴  A+B+C = 180°

[Sum of three angles of a triangle is 180°]

Taking sin on both sides, we get

Solution. Given that: sin 67° + cos 75°

= sin (90° -23°) + cos (90° -15°)

= cos 23° + sin 15°

[∵ sin (90°- θ) = cos θ and cos (90° – θ) = sin θ]

Solution. By using Identity,

cosec² A – cot² A = 1

⇒      cosec² A = 1 + cot² A

⇒      (cosec A)² = cot² A + 1

We reject negative values of sin A for acute angle A.

Solution. By using Identity,

Solution. (i) Consider, 9 sec² A – 9 tan² A

= 9 (sec² A – tan² A)

= 9 × 1 = 9.

Option (B) is correct. Ans.

(ii) Consider, (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

Option (D) is correct. Ans.

From (1) and (2), it is clear that

L.H.S. = R.H.S.

Hence, (cosec A – sin A) (sec A – cos A)

From (1) and (2), it is clear that