Trigonometry
- The shadow of a vertical tower becomes 30 metres longer when the altitude of the sun changes from 60° to 45°. Find the height of the tower.
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AB = Tower = h metre
DC = 30 metre
BD = x metre
From ∆ABC,tan 45° = AB ⇒ 1 = h ⇒ h = x + 30 ......(i) BC x + 30
From ∆ABD,tan 60° = AB ⇒ √3 = h BD x
h = √3x⇒ x = h .........(ii) √3
∴ h = x + 30⇒ h = h + 30 √3
⇒ (√3 + 1)h = 30√3⇒ h = 30√3 √3 - 1 = 30√3(√3 + 1) (√3 - 1)(√3 + 1)
= 15(3 + √3) metre.Correct Option: B
AB = Tower = h metre
DC = 30 metre
BD = x metre
From ∆ABC,tan 45° = AB ⇒ 1 = h ⇒ h = x + 30 ......(i) BC x + 30
From ∆ABD,tan 60° = AB ⇒ √3 = h BD x
h = √3x⇒ x = h .........(ii) √3
∴ h = x + 30⇒ h = h + 30 √3
⇒ (√3 + 1)h = 30√3⇒ h = 30√3 √3 - 1 = 30√3(√3 + 1) (√3 - 1)(√3 + 1)
= 15(3 + √3) metre.
- If cosθ – sinθ = 2 sinθ, then cosθ + sinθ = ?
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cosθ – sinθ = 2 sinθ ..(i)
cosθ + sinθ = x ....(ii)
On squaring both the equations and adding.
2 (cos²θ + sin²θ) = 2sin²θ + x²
⇒ x² = 2 – 2 sin²θ = 2 (1 – sin²θ) = 2cos²θ
⇒ x = √2cosθCorrect Option: C
cosθ – sinθ = 2 sinθ ..(i)
cosθ + sinθ = x ....(ii)
On squaring both the equations and adding.
2 (cos²θ + sin²θ) = 2sin²θ + x²
⇒ x² = 2 – 2 sin²θ = 2 (1 – sin²θ) = 2cos²θ
⇒ x = √2cosθ
- If cosθ + secθ = 2, then cos5θ + sec5θ= ?
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cosθ + secθ = 2
⇒ cosθ + 1 = 2 cosθ
⇒ cos²θ – 2 cosθ + 1 = 0
⇒ (cosθ – 1)² = 0
⇒ cosθ = 1
∴ cos5θ + sec5θ = 1 + 1 = 2Correct Option: B
cosθ + secθ = 2
⇒ cosθ + 1 = 2 cosθ
⇒ cos²θ – 2 cosθ + 1 = 0
⇒ (cosθ – 1)² = 0
⇒ cosθ = 1
∴ cos5θ + sec5θ = 1 + 1 = 2
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If sin A = 3 and cos B = 12 , what is the value of tanA - tanB 5 13 1 + tanA tanB
; it being given that A and B are acute angles?
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sin A = 3 5
∴ cos A = √1 - sin²A= √ 1 - 9 25 = √ 25 - 9 = √ 16 = 4 25 25 5 cos B = 12 13
sin B = √1 - cos²B= √ 1 - 144 169 = √ 
1 - 12 
² 13 = √ 1 - 144 = √ 169 - 144 = √ 25 = 5 169 169 169 5 ∴ tanA = 1 = sinA = 5 3 cosA 4 4 5 ∴ tanB = 5 = sinB = 13 5 cosB 12 12 13 ∴ 3 - 5 tanB - tanB = 4 12 1 + tanA.tanB 1 + 3 . 5 4 12 = 9 - 5 12 1 + 5 16 = 1 3 21 16 = 1 × 16 = 16 3 21 63 Correct Option: D
sin A = 3 5
∴ cos A = √1 - sin²A= √ 1 - 9 25 = √ 25 - 9 = √ 16 = 4 25 25 5 cos B = 12 13
sin B = √1 - cos²B= √ 1 - 144 169 = √ 1 - 12 ² 13 = √ 1 - 144 = √ 169 - 144 = √ 25 = 5 169 169 169 5 ∴ tanA = 1 = sinA = 5 3 cosA 4 4 5 ∴ tanB = 5 = sinB = 13 5 cosB 12 12 13 ∴ 3 - 5 tanB - tanB = 4 12 1 + tanA.tanB 1 + 3 . 5 4 12 = 9 - 5 12 1 + 5 16 = 1 3 21 16 = 1 × 16 = 16 3 21 63
- Find the value of sin²1° + sin²3° + sin²5°+ ...... sin²87° + sin²89°.
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sin 89° = sin (90° – 1°) = cos 1°
sin 87° = sin (90° – 3°) = cos 3°
∴ sin² 1° + sin² 3° + .... + sin² 45° + .... + sin² 87° + sin² 89°
= (sin² 1° + sin² 89°) + (sin² 3° + sin² 87°) + ..... to 22 terms + sin² 45°= (sin² 1° + cos² 1°) + (sin² 3° +cos² 3°) + ..... to 22 terms + 1 = 22 + 1 = 22 1 2 2 2 Correct Option: B
sin 89° = sin (90° – 1°) = cos 1°
sin 87° = sin (90° – 3°) = cos 3°
∴ sin² 1° + sin² 3° + .... + sin² 45° + .... + sin² 87° + sin² 89°
= (sin² 1° + sin² 89°) + (sin² 3° + sin² 87°) + ..... to 22 terms + sin² 45°= (sin² 1° + cos² 1°) + (sin² 3° +cos² 3°) + ..... to 22 terms + 1 = 22 + 1 = 22 1 2 2 2
