Trigonometry
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In a triangle ABC, ∠ABC = 75° and ∠ACB = πc . The circular measure of ∠BAC is 4
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Using Rule 1,
∠ABC = 75°
[∵ 180° = π radian or πc ]75° = π × 75 = 5 π radian 180 12 ∴ ∠BAC = π - π - 5 π 4 12 = 12π - 3π - 5π - 4 π 12 12 = π radian 3
Correct Option: B
Using Rule 1,
∠ABC = 75°
[∵ 180° = π radian or πc ]75° = π × 75 = 5 π radian 180 12 ∴ ∠BAC = π - π - 5 π 4 12 = 12π - 3π - 5π - 4 π 12 12 = π radian 3
- In circular measure, the value of the angle 11°15' is
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Using Rule 1,
11°15'= 11 ° + 15° 60 = 11° + 1 = 45° 4 4
[∵ 180° = π c]∴ 45° = π × 45 = πc 4 180 4 16
Correct Option: A
Using Rule 1,
11°15'= 11 ° + 15° 60 = 11° + 1 = 45° 4 4
[∵ 180° = π c]∴ 45° = π × 45 = πc 4 180 4 16
- If 3 cosθ + 4 sinθ = 5, then tanθ = ?
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3 cosθ + 4 sinθ = 5
Dividing by cosθ,
3 + 4 tanθ = 5 secθ
On squaring,
9 + 16 tan²θ + 24 tanθ = 25 (1 + tan²θ)
⇒ 9 tan²θ – 24 tanθ + 16 = 0
⇒ (3 tanθ – 4)2 = 0
⇒ 3 tanθ = 4⇒ tanθ = 4 3 Correct Option: A
3 cosθ + 4 sinθ = 5
Dividing by cosθ,
3 + 4 tanθ = 5 secθ
On squaring,
9 + 16 tan²θ + 24 tanθ = 25 (1 + tan²θ)
⇒ 9 tan²θ – 24 tanθ + 16 = 0
⇒ (3 tanθ – 4)2 = 0
⇒ 3 tanθ = 4⇒ tanθ = 4 3
- Find the simplest numerical value of 3 (sin x – cos x)4 + 4 (sin6x + cos6x) + 6(sinx + cosx)²
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3(sin x – cos x)4
= 3 (sin²x + cos²x – 2 sin x . cos x)²
= 3 (1 – 2sinx. cosx)²
= 3 (1 + 4 sin²x . cos²x – 4 sin x . cos x) 4 (sin6x + cos6x)
= 4 [(sin²x + cos²x)³ – 3 sin²x.cos²x (sin²x + cos²x) [a³ + b³ = (a+b)³ – 3ab (a+b)]
= 4 (1 – 3 sin²x . cos²x)
= 6 (sin x + cos x)²
= 6 (sin²x + cos²x+2 sin x. cos x)
= 6 (1 + 2 sin x . cos x)
∴ Expression = 3 (1 + 4sin²x . cos²x – 4sin x . cos x) + 4 (1 – 3sin²x . cos²x) + 6 (1 + 2sinx . cos x)
= 3 + 4 + 6 = 13Correct Option: D
3(sin x – cos x)4
= 3 (sin²x + cos²x – 2 sin x . cos x)²
= 3 (1 – 2sinx. cosx)²
= 3 (1 + 4 sin²x . cos²x – 4 sin x . cos x) 4 (sin6x + cos6x)
= 4 [(sin²x + cos²x)³ – 3 sin²x.cos²x (sin²x + cos²x) [a³ + b³ = (a+b)³ – 3ab (a+b)]
= 4 (1 – 3 sin²x . cos²x)
= 6 (sin x + cos x)²
= 6 (sin²x + cos²x+2 sin x. cos x)
= 6 (1 + 2 sin x . cos x)
∴ Expression = 3 (1 + 4sin²x . cos²x – 4sin x . cos x) + 4 (1 – 3sin²x . cos²x) + 6 (1 + 2sinx . cos x)
= 3 + 4 + 6 = 13
- A telegraph post is bent at a point above the ground due to starm. Its top just touches the ground at a distance of 10 √3 metre from its foot and makes an angle of 30° with the horizontal. Then height (in metres) of the telegraph post is
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AB = Telegraph post
AC = CD = bent part
BD = 10√3 metre
In ∆BCD,tan 30° = BC BD ⇒ 1 ⇒ BC √3 10√3 ⇒ BC = 1 × 10√3 √3
= 10 metre
Again,sin 30° = BC CD ⇒ 1 = 10 2 CD
⇒ CD = 20 metre
∴ AB = BC + CD = (10 + 20) metre
= 30 metreCorrect Option: A
AB = Telegraph post
AC = CD = bent part
BD = 10√3 metre
In ∆BCD,tan 30° = BC BD ⇒ 1 ⇒ BC √3 10√3 ⇒ BC = 1 × 10√3 √3
= 10 metre
Again,sin 30° = BC CD ⇒ 1 = 10 2 CD
⇒ CD = 20 metre
∴ AB = BC + CD = (10 + 20) metre
= 30 metre
