Trigonometry
- The ratio of the length of a rod and its shadow is 1 : √3 . The angle of elevation of the sun is :
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AB = 1 BC √3 tanθ = AB = 1 BC √3
⇒ tanθ = tan 30°
⇒ θ = 30°Correct Option: B
AB = 1 BC √3 tanθ = AB = 1 BC √3
⇒ tanθ = tan 30°
⇒ θ = 30°
- If the angle of elevation of the sun changes from 45° to 60°, then the length of the shadow of a pillar decreases by 10 m. The height of the pillar is :
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AB = Height of pillar = h metre (let)
CD = 10 metre
∠ACB = 45°
∠ADB = 60°
BD = x metre (let)
From ∆ABCtan 45° = AB BC ⇒ 1 = h x + 10
⇒ h = (x + 10) metre (i)
From ∆ABDtan 60° = AB BD ⇒ √3 = h x ⇒ x = h metre (ii) √3
From equation (i),h = h + 10 √3 ⇒ h - h = 10 √3 ⇒ √3h - h = 10 √3
⇒ h(√3 - 1) = 10√3⇒ h = 10√3 √3- 1 = 10√3(√3+ 1) (√3- 1)(√3+ 1) = 10√3(√3+ 1) 3 - 1
= 5√3 (√3 + 1)
= 5 (3 + √3)metreCorrect Option: D
AB = Height of pillar = h metre (let)
CD = 10 metre
∠ACB = 45°
∠ADB = 60°
BD = x metre (let)
From ∆ABCtan 45° = AB BC ⇒ 1 = h x + 10
⇒ h = (x + 10) metre (i)
From ∆ABDtan 60° = AB BD ⇒ √3 = h x ⇒ x = h metre (ii) √3
From equation (i),h = h + 10 √3 ⇒ h - h = 10 √3 ⇒ √3h - h = 10 √3
⇒ h(√3 - 1) = 10√3⇒ h = 10√3 √3- 1 = 10√3(√3+ 1) (√3- 1)(√3+ 1) = 10√3(√3+ 1) 3 - 1
= 5√3 (√3 + 1)
= 5 (3 + √3)metre
- TF is a tower with F on the ground. The angle of elevation of T from A is x° such that tan x° = (2 / 5) and AF = 200m. The angle of elevation of T from a nearer point B is y° with BF = 80m. The value of y° is
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TF = Tower = h metre
∠TAF = x° ; ∠TBF = y°,
BF = 80 metre
In ∆AFT,tan x° = TF AF ⇒ 2 = h 5 200 ⇒ h = 2 × 200 5
= 80 metre
In ∆BFTtan y° = TF FB ⇒ tan y° = 80 = 1 80
⇒ tan y° = tan 45°
⇒ y = 45°Correct Option: D
TF = Tower = h metre
∠TAF = x° ; ∠TBF = y°,
BF = 80 metre
In ∆AFT,tan x° = TF AF ⇒ 2 = h 5 200 ⇒ h = 2 × 200 5
= 80 metre
In ∆BFTtan y° = TF FB ⇒ tan y° = 80 = 1 80
⇒ tan y° = tan 45°
⇒ y = 45°
- If sin A – cos A = (√3 - 1 / 2) , then the value of sin A . cos A is
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sinA – cosA = √3 - 1 2
On squaring both sides,
sin²A + cos²A – 2 sinA . cosA
⇒ 1 – 2 sinA cosA= 1 (4 - 2√3) 4 ⇒ 1 – 2 sinA cosA = 2 - √3 2
⇒ 2 – 4 sinA cosA = 2 – √3
⇒ 4 sinA . cosA = 2–2+ √3 = √3⇒ sinA.cosA = √3 4
Correct Option: C
sinA – cosA = √3 - 1 2
On squaring both sides,
sin²A + cos²A – 2 sinA . cosA
⇒ 1 – 2 sinA cosA= 1 (4 - 2√3) 4 ⇒ 1 – 2 sinA cosA = 2 - √3 2
⇒ 2 – 4 sinA cosA = 2 – √3
⇒ 4 sinA . cosA = 2–2+ √3 = √3⇒ sinA.cosA = √3 4
- The value of x in the equation
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tan² π - cos² π 4 3 = x sin π . cos π . tan π 4 4 3 
⇒ 1 - 1 = x × √3 4 2 ⇒ 4 - 1 = x × √3 4 2 ⇒ x = 3 × 2 = √3 4 √3 2
Correct Option: D
tan² π - cos² π 4 3 = x sin π . cos π . tan π 4 4 3 ⇒ 1 - 1 = x × √3 4 2 ⇒ 4 - 1 = x × √3 4 2 ⇒ x = 3 × 2 = √3 4 √3 2
