Trigonometry
- The angle of elevation of an aeroplane from a point A on the ground is 60°. After a straight flight of the plane for 30 seconds, the angle of elevation becomes 30°. If the palne flies at a constant height of 3600√3 metre, what is the speed of plane?
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P and Q = Positions of plane
∠PAB = 60°, ∠QAB = 30°, PB = 3600√3 metre
In ∆ABP,tan 60° = BP AB ⇒ √3 = 3600√3 AB
⇒ AB = 3600 metre
In ∆ACQ,tan 30° = CQ AC ⇒ 1 = 3600√3 √3 AC
⇒ AC = 3600 × 3 = 10800 metre
∴ PQ = BC = AC – AB = 10800 – 3600 = 7200 metre
This distance is covered in 30 seconds.∴ Speed of plane = 7200 = 240 m/sec. 30
= 240= 
240 × 18 
5
= 864 kmphCorrect Option: A
P and Q = Positions of plane
∠PAB = 60°, ∠QAB = 30°, PB = 3600√3 metre
In ∆ABP,tan 60° = BP AB ⇒ √3 = 3600√3 AB
⇒ AB = 3600 metre
In ∆ACQ,tan 30° = CQ AC ⇒ 1 = 3600√3 √3 AC
⇒ AC = 3600 × 3 = 10800 metre
∴ PQ = BC = AC – AB = 10800 – 3600 = 7200 metre
This distance is covered in 30 seconds.∴ Speed of plane = 7200 = 240 m/sec. 30
= 240= 240 × 18 5
= 864 kmph
- sin 75° + sin 15° can be expressed as
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sin 75° + sin 15° = 2 sin
= 
75° + 15° . cos 75° - 15° 2 2
∵ sinC + sinD= 2sin C + D . cos C - D 2 2
= 2 sin 45° . cos 30°= 2 . 1 . √3 √2 2
√3/2Correct Option: D
sin 75° + sin 15° = 2 sin
= 75° + 15° . cos 75° - 15° 2 2
∵ sinC + sinD= 2sin C + D . cos C - D 2 2
= 2 sin 45° . cos 30°= 2 . 1 . √3 √2 2
√3/2
- What will be the value of 2 cos 45° × sin 15°
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We know that,
2 cosA sinB = sin(A + B) – sin(A – B)
⇒ 2 cos 45° sin 15°
= sin (45° + 15°) – sin (45° – 15°)
= sin 60° – sin 30°= √3 - 1 2 2 = √3 - 1 1 Correct Option: C
We know that,
2 cosA sinB = sin(A + B) – sin(A – B)
⇒ 2 cos 45° sin 15°
= sin (45° + 15°) – sin (45° – 15°)
= sin 60° – sin 30°= √3 - 1 2 2 = √3 - 1 1
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The value of sin 22 1° will be 2
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We know that,
cos2A = 1 – 2 sin²A⇒ cosA = 1 – 2 sin² A 2
Let A = 45°⇒ cos 45° = 1 – 2 sin² 45° 2 ⇒ 2 sin² 22 1° = 1 – cos 45° 2 ⇒ 2 sin² 22 1° = 1 – 1 2 √2 ⇒ 2 sin² 22 1° = √2 - 1 2 √2 ⇒ 2 sin² 22 1° = √2 - 1 2 2√2 ⇒ 2 sin² 22 1° = √ √2 - 1 2 2√2 Correct Option: D
We know that,
cos2A = 1 – 2 sin²A⇒ cosA = 1 – 2 sin² A 2
Let A = 45°⇒ cos 45° = 1 – 2 sin² 45° 2 ⇒ 2 sin² 22 1° = 1 – cos 45° 2 ⇒ 2 sin² 22 1° = 1 – 1 2 √2 ⇒ 2 sin² 22 1° = √2 - 1 2 √2 ⇒ 2 sin² 22 1° = √2 - 1 2 2√2 ⇒ 2 sin² 22 1° = √ √2 - 1 2 2√2
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The value of cosα + cosβ will be sinα + sinβ
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cosα + cosβ sinα + sinβ = 2.cos α + β cos α - β 2 2 2.sin α + β sin α - β 2 2 = cos α + β 2 sin α + β 2 = cot α + β 2 Correct Option: B
cosα + cosβ sinα + sinβ = 2.cos α + β cos α - β 2 2 2.sin α + β sin α - β 2 2 = cos α + β 2 sin α + β 2 = cot α + β 2
