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Introduction to trigonometry

Lesson 46 of 113

1. Ye topic hai kya?

Right triangle mein ek acute angle ke hisaab se sides ke ratio — sin, cos, tan. Paper: sin 30° ki value, sin A = 21/29 ho to cos A, sin²θ + cos²θ, tan(90°−θ). Angle of elevation se height nahi — woh agla chapter.

Yahan lock

Opposite / adjacent / hypotenuse (angle ke relative), 6 ratios, reciprocal, teen identities, standard angles, complementary, given-one-ratio → baaki.

Yahan kya nahi

Tower 30°, depression, shadow-height word (Applications). DE ∥ BC. Distance formula (x,y). Clock hands 5:20. Compound sin(A+B) Class-XI.

2. Angle ke relative — opp, adj, hyp

Hamesha right triangle. Hypotenuse = right angle ke saamne, sabse lamba — yeh angle pe depend nahi. Opposite aur adjacent us acute angle pe depend: jis angle pe kaam, uske saamne wali side opposite, uske bagal (jo hyp nahi) adjacent.

C A B ∠A adj 20 opp 21 hyp 29

∠C = 90°. ∠A ke liye: opp = BC = 21, adj = AC = 20, hyp = AB = 29. (20²+21²=400+441=841=29².)

Same triangle, ∠B pe kaam: opposite aur adjacent swap. Hyp same. Isliye sin A ≠ sin B generally. Pehle likho: “kis angle ka?”

3. Chhe ratios

sin θ = opp/hyp   cos θ = adj/hyp   tan θ = opp/adj
Reciprocal: cosec θ = 1/sin θ = hyp/opp,   sec θ = 1/cos θ = hyp/adj,   cot θ = 1/tan θ = adj/opp.
Bonus: tan θ = sin θ / cos θ,   cot θ = cos θ / sin θ.
Example

Upar wale figure, ∠A. sin A, cos A, tan A?

Solution

  1. Names

    opp=21, adj=20, hyp=29.

  2. Ratios

    sin A = 21/29, cos A = 20/29, tan A = 21/20.

Answer21/29, 20/29, 21/20

cosec A = 29/21, sec A = 29/20, cot A = 20/21. Geometry chapter ka “hyp dhoondho 8,15” alag skill — yahan sides di hain, ratio naam dena hai.

4. Identities — hamesha (jahan defined)

Right triangle: opp² + adj² = hyp². Divide by hyp² → (opp/hyp)² + (adj/hyp)² = 1.

sin²θ + cos²θ = 1
1 + tan²θ = sec²θ   (divide hyp² wali equation by adj²)
1 + cot²θ = cosec²θ
Inse: cos²θ = 1 − sin²θ, sin²θ = 1 − cos²θ.
Example

sin θ = 21/29, θ acute. cos θ aur tan θ?

Solution

  1. Identity

    cos²θ = 1 − (21/29)² = (841 − 441)/841 = 400/841. cos θ = 20/29 (acute, +).

  2. tan

    tan θ = sin/cos = 21/20. Triangle bana ke 21, 20, 29 bhi same.

Answercos θ = 20/29, tan θ = 21/20
Class-X acute: sin, cos, tan sab ≥ 0. Negative root mat lo. 90° pe kuch ratios defined nahi — identity tab apply mat karo jab denominator 0.

5. Standard angles — 0°, 30°, 45°, 60°, 90°

45°: isosceles right, legs 1, 1, hyp √2. sin 45° = cos 45° = 1/√2, tan 45° = 1.

30°–60°: equilateral side 2, beech se altitude. Half base 1, altitude √3, hyp 2. 30° ke saamne 1, 60° ke saamne √3.

90° 60° 30° 1 √3 2 45° 45° 1 1 √2

Left: 30°–60°–90° (1, √3, 2). Right: 45°–45°–90° (1, 1, √2). Tower height nahi — sirf ratios nikalne ka figure.

θ0°30°45°60°90°
sin01/21/√2√3/21
cos1√3/21/√21/20
tan01/√31√3not defined
cosecnot defined2√22/√31
sec12/√3√22not defined
cotnot defined√311/√30

Pattern: sin 0→90 badhta 0 se 1; cos girta 1 se 0. tan 45° = 1 checkpoint. 30° aur 60° swap: sin 30 = cos 60 = 1/2, sin 60 = cos 30 = √3/2.

tan 90° number nahi — “∞” MCQ pe bhi Class-X mein not defined. sin 0 = 0, cos 0 = 1 ulta mat yaad. 1/√2 ko √2/2 likh sakte, value same.

6. Complementary — 90° − θ

Right triangle ke do acute angles ka sum 90°. Isliye ek ka sin = dusre ka cos.

sin(90°−θ) = cos θ   cos(90°−θ) = sin θ
tan(90°−θ) = cot θ   cot(90°−θ) = tan θ
sec(90°−θ) = cosec θ   cosec(90°−θ) = sec θ

sin 37° = cos 53°. tan 35° = cot 55°. Agar sin θ = cos 40°, to θ = 50° (acute): sin θ = sin(90°−40°) ⇒ θ = 50° (0°–90° mein).

7. Evaluate — table + identity, tower nahi

Example

2 sin 30° + tan 45° − cos 60° ?

Solution

  1. Plug

    2·(1/2) + 1 − 1/2 = 1 + 1 − 1/2 = 3/2.

Answer3/2
Example

sin 60° cos 30° + cos 60° sin 30° ?

Solution

  1. Table

    (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1. (Compound formula ki zaroorat nahi — values se.)

Answer1

Agar tan θ = 1 aur 0° < θ < 90°, θ = 45°. Agar sin θ = cos θ, tan θ = 1, phir 45°.

8. Sawal — basic se pro

Q1 · Basic

∠A ke liye sin A?

  1. opp / hyp
  2. adj / hyp
  3. opp / adj
  4. hyp / opp

Solution

  1. Definition

    A. B cos. C tan. D cosec.

AnswerA
Q2 · Basic

Figure: opp 21, adj 20, hyp 29. cos A?

  1. 20/29
  2. 21/29
  3. 21/20
  4. 29/20

Solution

  1. adj/hyp

    20/29. B sin. C tan. D sec.

Answer20/29
Q3 · Basic

sin 30° ?

  1. 1/2
  2. √3/2
  3. 1/√2
  4. 1

Solution

  1. Table

    1/2. B sin 60. Mix 30↔60 common trap.

Answer1/2
Q4 · Basic

tan 45° ?

  1. 1
  2. 0
  3. √3
  4. not defined

Solution

  1. opp=adj

    1. D tan 90. C tan 60.

Answer1
Q5 · Medium

sin² 60° + cos² 60° ?

  1. 1
  2. 3/2
  3. 1/2
  4. 0

Solution

  1. Identity

    sin²θ+cos²θ=1 har θ (defined). Check: (3/4)+(1/4)=1. B sirf sin² 60.

Answer1
Q6 · Medium

sin θ = 21/29, acute. tan θ?

  1. 21/20
  2. 20/21
  3. 21/29
  4. 29/21

Solution

  1. cos

    √(1−441/841)=20/29. tan=sin/cos=21/20. B cot. D cosec.

Answer21/20
Q7 · Medium

2 sin 30° + tan 45° − cos 60° ?

  1. 3/2
  2. 1
  3. 2
  4. 1/2

Solution

  1. 1+1−1/2

    3/2. C: 2·sin30 ko 2+1/2. Tower nahi — sirf values.

Answer3/2
Q8 · Medium

sin(90° − 37°) ?

  1. cos 37°
  2. sin 37°
  3. tan 37°
  4. cos 53°

Solution

  1. sin(90−θ)=cos θ

    cos 37°. D: cos 53° = sin 37°, yeh is expression ke barabar nahi.

Answercos 37°
Q9 · Pro

sin 60° cos 30° + cos 60° sin 30° ?

  1. 1
  2. √3/2
  3. 1/2
  4. √3

Solution

  1. 3/4+1/4

    1. B sirf pehla term. Compound formula yaad rakhne ki zaroorat nahi.

Answer1
Q10 · Pro

Kaun sa defined nahi?

  1. tan 90°
  2. sin 90°
  3. cos 0°
  4. tan 45°

Solution

  1. adj=0

    tan = opp/adj, 90° pe adj 0. sin 90=1, cos 0=1, tan 45=1.

Answertan 90°

Practice

Pehle khud, phir neeche kholo.

12345 678910

P1

sec θ ka reciprocal?

  1. cos θ
  2. sin θ
  3. cosec θ
  4. tan θ

Solution

  1. 1/sec

    cos. C 1/sin.

Answercos θ
P2

cos 60° ?

  1. 1/2
  2. √3/2
  3. 1
  4. 0

Solution

  1. Table

    1/2. B cos 30. D cos 90.

Answer1/2
P3

tan 60° · tan 30° ?

  1. 1
  2. √3
  3. 1/3
  4. 0

Solution

  1. √3 · 1/√3

    1. Complementary: tan(90−θ)=cot θ, product tan·cot=1.

Answer1
P4

cosec 30° ?

  1. 2
  2. 1/2
  3. 2/√3
  4. not defined

Solution

  1. 1/sin30

    1/(1/2)=2. B sin 30. C cosec 60.

Answer2
P5

1 + tan² 45° ?

  1. 2
  2. 1
  3. √2
  4. 0

Solution

  1. sec² 45

    1+1=2. sec 45=√2, (√2)²=2.

Answer2
P6

sin θ = cos θ, 0° < θ < 90°. θ?

  1. 45°
  2. 30°
  3. 60°
  4. 90°

Solution

  1. tan=1

    45°. 30° pe sin < cos.

Answer45°
P7

tan(90° − 35°) ?

  1. cot 35°
  2. tan 35°
  3. cot 55°
  4. sin 35°

Solution

  1. tan(90−θ)=cot θ

    cot 35°. C: cot 55° = tan 35°.

Answercot 35°
P8

Same triangle, ∠B (dusra acute). sin B, agar sin A = 21/29?

  1. 20/29
  2. 21/29
  3. 21/20
  4. 29/21

Solution

  1. Swap

    B ke liye opp = A ka adj = 20. sin B = 20/29 = cos A. Complementary.

Answer20/29
P9

cos 0° − sin 90° ?

  1. 0
  2. 1
  3. −1
  4. not defined

Solution

  1. 1−1

    0. C: sirf −sin90. D nahi — dono defined.

Answer0
P10

sec 60° + cot 45° ?

  1. 3
  2. 2
  3. 1
  4. √3

Solution

  1. 2+1

    sec 60=2, cot 45=1. Sum 3. B sirf sec 60.

Answer3

← Index · Coordinate geometry

Agle topic: Applications of trigonometry.