Hello friend !! Trigonometry ke part-9 aur part-10 me humne aapko trigonometric ratios 30 degree aur 45 degree ki value find karna full details me sikhaya tha. Agar aapne nahi padha toh nichey humne link diya hai. Aap yaha se padh kar sikh lijiye.
❒ Trigonometric Ratios of 45 degree in hindi
❒ Trigonometric ratios of 30 degree
Aaj ka ye lessons trigonometry part - 11 hai. Aaj hum aapko trigonometric ratios 60 degree ki value find karna sikhane jaa rahe hai. Aaj hum aapko batayenge ki, sin60°, cos60°, tan60°, cosec60°, sec60° aur cot60° ki value kya hoti hai? aur inn sabhi ka maan (value) nikalne ka main concept kya hai. Toh chaliye aaj ka lesson start karte hai-:
❒ Similar triangles Defination, Examples and properties in Hindi
❒ Fraction ko percent me kaise convert karte hai? | 3 best methods
❒ Aapke send kiye gaye question ka answer yaha hai part-3
Hume ummeed hai ki, aapko aaj ka lesson Trigonometric ratios of 60 degree ki value find karna achhe se samjh aaya hoga. Agar aapko koi bhi doubt yaa koi problem ho toh aap comment karke puchh sakte hai.
Aap chahe toh humare Facebook page par apne question ki image yaa massage bhi send kar sakte hai. Hum aapka doubt yaa problem dur karne ki puri koshish karenge.
Ab aap hume bataye ki, "Next lesson me aap math ka kon-sa topic sikhana chahte hai?" Abhi comment ya massage karke bataye. Hum uss topic ko full details me aapko sikhayenge
Last me Agar aapko aaj ka topic helpful laga ho toh please ek share jarur kare.
❒ Trigonometric Ratios of 45 degree in hindi
❒ Trigonometric ratios of 30 degree
Aaj ka ye lessons trigonometry part - 11 hai. Aaj hum aapko trigonometric ratios 60 degree ki value find karna sikhane jaa rahe hai. Aaj hum aapko batayenge ki, sin60°, cos60°, tan60°, cosec60°, sec60° aur cot60° ki value kya hoti hai? aur inn sabhi ka maan (value) nikalne ka main concept kya hai. Toh chaliye aaj ka lesson start karte hai-:
Trigonometric ratios of 60 degree ki value find karna in hindi
Trigonometric ratios 60° ki value nikaalane ke liye hume ek aisa triangle banana hoga jiske angle 60° ke ho. Toh isske liye hum ek sambaahu tribhuj (equilateral triangle) ABC lenge kyuki iss triangle ke sabhi angle 60° hote hai aur hum ye bhi jaante hai ki iss triangle ke all sides equal bhi hote hai.
Maana nichey (figure-1) ek sambaau tribhuj ΔABC hai, jiske side AB = BC = CA = x aur ∠A =∠B =∠C = 60°
Ab hum point A se side BC ke point D par ek perpendicular line (lamb rekha) AD banate hai. (Figure-2 dekhiye)
Ab ∆ABD aur ∆ACD ko dekhne par hum paate hai ki,
(i). ∠ADB =∠ADC = 90° 【∵ AD perpendicular hai】
(ii). ∠B = ∠C = 60°
(iii). Side AD = AD 【dono triangle me common hai.】
Toh "Angle-Angle-Side congruency" (kon-kon-bhuja ke samrupata niyam) se, ∆ABD ≅ ∆ACD
∴Hum kah sakte hai ki, BD = CD
Ab hum dekh sakte hai ki, ∆ABD ek "right angled triangle (samkon tribhuj)" hai. Jisme ∠D right angle (90°) aur ∠ABD = 60° hai. Nichey diye Figure-3 ko dekhiye-:
Trigonometric ratios 60° ki value nikalne ke liye hume ∆ABD ke side AD aur BD ki value maalum karna hoga. Toh chaliye ab sabse pahle side BD ki value nikalte hai.
Toh Figure-1 ∆ABC se,
side BC = x
side BD = 1 BC
2
= 1 x = x/2
2
Ab hume side AD ki value nikalani hai. Isske liye hum Pythagoras theorem ka use karenge.
Toh Hum jaante hai ki, Pythagoras theorem ke anusar ∆ABD me -:
AB² = AD²+BD²
AD² = AB²-BD²
= x²-(x/2)²
= x²-(x²/4)
= (4x²-x²)/4
= 3x²/4
AD = √(3x²/4)
= x√3/2
Toh ab hume Figure-3 me ∆ABD ke sabhi side ki value maalum ho gaye hai. Matlab -:
1. Side AB = x
2. Side BD = x/2
3. Side AD = x√3/2
Ab hum, sabhi trigonometric ratios ki 60° ki value nikalte hai.
Maana nichey (figure-1) ek sambaau tribhuj ΔABC hai, jiske side AB = BC = CA = x aur ∠A =∠B =∠C = 60°
 |
| Figure-1 |
Ab hum point A se side BC ke point D par ek perpendicular line (lamb rekha) AD banate hai. (Figure-2 dekhiye)
 |
| Figure-2 |
Ab ∆ABD aur ∆ACD ko dekhne par hum paate hai ki,
(i). ∠ADB =∠ADC = 90° 【∵ AD perpendicular hai】
(ii). ∠B = ∠C = 60°
(iii). Side AD = AD 【dono triangle me common hai.】
Toh "Angle-Angle-Side congruency" (kon-kon-bhuja ke samrupata niyam) se, ∆ABD ≅ ∆ACD
∴Hum kah sakte hai ki, BD = CD
Ab hum dekh sakte hai ki, ∆ABD ek "right angled triangle (samkon tribhuj)" hai. Jisme ∠D right angle (90°) aur ∠ABD = 60° hai. Nichey diye Figure-3 ko dekhiye-:
 |
| Figure-3 |
Toh Figure-1 ∆ABC se,
side BC = x
side BD = 1 BC
2
= 1 x = x/2
2
Ab hume side AD ki value nikalani hai. Isske liye hum Pythagoras theorem ka use karenge.
Toh Hum jaante hai ki, Pythagoras theorem ke anusar ∆ABD me -:
AB² = AD²+BD²
AD² = AB²-BD²
= x²-(x/2)²
= x²-(x²/4)
= (4x²-x²)/4
= 3x²/4
AD = √(3x²/4)
= x√3/2
Toh ab hume Figure-3 me ∆ABD ke sabhi side ki value maalum ho gaye hai. Matlab -:
1. Side AB = x
2. Side BD = x/2
3. Side AD = x√3/2
Ab hum, sabhi trigonometric ratios ki 60° ki value nikalte hai.
Sabhi trigonometric ratios ke liye 60° ka maan (value)
Toh ab hume samkon ∆ADB ke sabhi sides ki value jaan chuke hai. Ab hum ek-ek karke sabhi trigonometric ratios ki value nikalna start karte hai.
⇒ sin60° = AD
AB
= x√3/2
x
= x√3 × 1
2 x
= √3
2
∴ sin60° = √3
2
⇒ cos60° = BD
AB
= x/2
X
= x × 1
2 x
= 1
2
∴ cos60° = 1
2
⇒ tan60° = AD
BD
= x√3/2
x/2
= x√3 × 2
2 x
= √3
∴ tan60° = √3
Upar ke ratios sin60°, cos60° aur tan60° ki tarah se hum cosec60°, sec60° aur cot60° ki value bhi aashani se nikal sakte hai. Lekin hum cosec60°, sec60° aur cot60° ki value nikalane ke liye trigonometric ratios ke Reciprocal relation ka use karenge.
⇒ cosec60° = 1
sin60°
= 1
√3/2
= 1× 2
√3
= 2
√3
∴ cosec60° = 2
√3
⇒ sec60° = 1
cos60°
= 1
1/2
= 1 × 2
1
= 2
∴ sec60° = 2
⇒ cot60° = 1
tan60°
= 1
√3
∴ cot60° = 1
√3
Toh ab hum sabhi trigonometric ratios ki 60° ki value nikalna sikh gaye hai. Aayiye ab hum iss se related kuch question aur unka solution sikh lete hai.
Solution-: sin60° = √3/2, cos30° = √3/2, cos60° = 1/2 aur sin30° = 1/2
∴ sin60°cos30°+cos60°sin30°
= √3 × √3 + 1 × 1
2 2 2 2
= 3 + 1
4 4
= 4 = 1 Ans.
4
Ques-2. ∆ABC me, jisme ∠B = 90° hai. AB = 8 cm aur AC = 16 cm hai. ∠BAC bataye. (Nichey diye Figure-4 ko dekhiye.)
Solution-: Diya hai, samkon ∆ABC me ∠B = 90°,Base (aadhar) AB = 8 cm aur hypotenuse (karn) AC = 16 aur ∠BAC = ??
Toh hume aisa trigonometric ratio lena padega jisme aadhar aur karn diya ho. Toh isske liye hum cosA ya secA lenge.
Toh cos∠BAC = AB/AC
= 8/16
= 1/2
cos∠BAC = cos60° 【∵cos60° = 1/2】
∴ ∠BAC = 60° Ans.
Ques-3. cos²60°+tan²60°
Solution-: cos²60°+tan²60
= (1/2)²+(√3)²
= 1/4 + 3
= (1+12)/4
= 13/4 Ans.
Ques-4. 2tan²45°+cos²30°-sin²60
Solution-: 2tan²45°+cos²30°-sin²60
= 2×(1)²+(√3/2)²-(√3/2)²
= 2+3/2-3/2
= 2 Ans.
Ques-5. sin30°+tan45°-cosec60°
sec30°+cos60°+cot45°
Solution-: sin30°+tan45°-cosec60°
sec30°+cos60°+cot45°
1 + 1 - 2
= 2 √3
2 + 1 + 1
√3 2
√3+2√3-4
= 2√3
4+√3+2√3
2√3
3√3-4
= 2√3
4+3√3
2√3
= 3√3-4 × 2√3
2√3 4+3√3
= 3√3-4
3√3+4
= 3√3-4 × 3√3-4
3√3+4 3√3-4
= (3√3-4)²
(3√3)²-4²
= (3√3)²+4²-2×3√3×4
27-16
= 27+16-24√3
11
= 43-24√3 Ans.
11
Ye bhi padhe-:⇒ sin60° = AD
AB
= x√3/2
x
= x√3 × 1
2 x
= √3
2
∴ sin60° = √3
2
⇒ cos60° = BD
AB
= x/2
X
= x × 1
2 x
= 1
2
∴ cos60° = 1
2
⇒ tan60° = AD
BD
= x√3/2
x/2
= x√3 × 2
2 x
= √3
∴ tan60° = √3
Upar ke ratios sin60°, cos60° aur tan60° ki tarah se hum cosec60°, sec60° aur cot60° ki value bhi aashani se nikal sakte hai. Lekin hum cosec60°, sec60° aur cot60° ki value nikalane ke liye trigonometric ratios ke Reciprocal relation ka use karenge.
⇒ cosec60° = 1
sin60°
= 1
√3/2
= 1× 2
√3
= 2
√3
∴ cosec60° = 2
√3
⇒ sec60° = 1
cos60°
= 1
1/2
= 1 × 2
1
= 2
∴ sec60° = 2
⇒ cot60° = 1
tan60°
= 1
√3
∴ cot60° = 1
√3
Toh ab hum sabhi trigonometric ratios ki 60° ki value nikalna sikh gaye hai. Aayiye ab hum iss se related kuch question aur unka solution sikh lete hai.
Trigonometric ratios 60° se related kuch Question aur Solution:
Ques-1. sin60°cos30°+cos60°sin30°Solution-: sin60° = √3/2, cos30° = √3/2, cos60° = 1/2 aur sin30° = 1/2
∴ sin60°cos30°+cos60°sin30°
= √3 × √3 + 1 × 1
2 2 2 2
= 3 + 1
4 4
= 4 = 1 Ans.
4
Ques-2. ∆ABC me, jisme ∠B = 90° hai. AB = 8 cm aur AC = 16 cm hai. ∠BAC bataye. (Nichey diye Figure-4 ko dekhiye.)
 |
| Figure-4 |
Solution-: Diya hai, samkon ∆ABC me ∠B = 90°,Base (aadhar) AB = 8 cm aur hypotenuse (karn) AC = 16 aur ∠BAC = ??
Toh hume aisa trigonometric ratio lena padega jisme aadhar aur karn diya ho. Toh isske liye hum cosA ya secA lenge.
Toh cos∠BAC = AB/AC
= 8/16
= 1/2
cos∠BAC = cos60° 【∵cos60° = 1/2】
∴ ∠BAC = 60° Ans.
Ques-3. cos²60°+tan²60°
Solution-: cos²60°+tan²60
= (1/2)²+(√3)²
= 1/4 + 3
= (1+12)/4
= 13/4 Ans.
Ques-4. 2tan²45°+cos²30°-sin²60
Solution-: 2tan²45°+cos²30°-sin²60
= 2×(1)²+(√3/2)²-(√3/2)²
= 2+3/2-3/2
= 2 Ans.
Ques-5. sin30°+tan45°-cosec60°
sec30°+cos60°+cot45°
Solution-: sin30°+tan45°-cosec60°
sec30°+cos60°+cot45°
1 + 1 - 2
= 2 √3
2 + 1 + 1
√3 2
√3+2√3-4
= 2√3
4+√3+2√3
2√3
3√3-4
= 2√3
4+3√3
2√3
= 3√3-4 × 2√3
2√3 4+3√3
= 3√3-4
3√3+4
= 3√3-4 × 3√3-4
3√3+4 3√3-4
= (3√3-4)²
(3√3)²-4²
= (3√3)²+4²-2×3√3×4
27-16
= 27+16-24√3
11
= 43-24√3 Ans.
11
❒ Similar triangles Defination, Examples and properties in Hindi
❒ Fraction ko percent me kaise convert karte hai? | 3 best methods
❒ Aapke send kiye gaye question ka answer yaha hai part-3
Hume ummeed hai ki, aapko aaj ka lesson Trigonometric ratios of 60 degree ki value find karna achhe se samjh aaya hoga. Agar aapko koi bhi doubt yaa koi problem ho toh aap comment karke puchh sakte hai.
Aap chahe toh humare Facebook page par apne question ki image yaa massage bhi send kar sakte hai. Hum aapka doubt yaa problem dur karne ki puri koshish karenge.
Ab aap hume bataye ki, "Next lesson me aap math ka kon-sa topic sikhana chahte hai?" Abhi comment ya massage karke bataye. Hum uss topic ko full details me aapko sikhayenge
Last me Agar aapko aaj ka topic helpful laga ho toh please ek share jarur kare.
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