Trigonometry
- The two banks of a canal are straight and parallel. A, B, C are three persons of whom A stands on one bank and B and C on the opposite banks. B finds the angle ABC is 30°, while C finds that the angle ACB 60°. If B and C are 100 metres apart, the breadth of the canal is
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BD = x metre (let)
∴ CD = (100 – x) metre
AD ⊥ BC; AD = y metre
From ∆ ABD,tan 30° = AD BD ⇒ 1 = y √3 x
⇒ x = √3 y
From ∆ACD ....(i)tan 60° = y 100 - x ⇒ √3 = y 100 - x
⇒ y = 100 √3 - √3 x
⇒ y = 100 √3 - √3 × √3 y
⇒ y = 100 √3 - 3y
7rArr; 4y = 100 √3
⇒ y = 25 √3 metreCorrect Option: C
BD = x metre (let)
∴ CD = (100 – x) metre
AD ⊥ BC; AD = y metre
From ∆ ABD,tan 30° = AD BD ⇒ 1 = y √3 x
⇒ x = √3 y
From ∆ACD ....(i)tan 60° = y 100 - x ⇒ √3 = y 100 - x
⇒ y = 100 √3 - √3 x
⇒ y = 100 √3 - √3 × √3 y
⇒ y = 100 √3 - 3y
7rArr; 4y = 100 √3
⇒ y = 25 √3 metre
- The base of a triangle is 12 √3 cm and two angles at the base are 30° and 60° respectively. The altitude of the triangle is
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2√3cm.
BD = x cm. (let)
∴ CD = (12√3 - x) cm.
∠ADB = ∠ADC = 90°
From ∆ ABD,tan 30° = AD BD ⇒ 1 = AD √3 BD ⇒ AD = x ....(i) √3
From ∆ ACD,tan 60 ° = AD CD ⇒ √3 = AD 12√3 - x
⇒ AD = √3 (12 √3 - x)
= 36 - √3 x ....(ii)∴ x = 36 -√3 x √3
⇒ x = 36 √3 * 3x
⇒ 4x = 36 √3⇒ x = 36 √3 = 9 √3 4 ∴ AD = x = 9√3 = 9 cm. √3 √3
Correct Option: D
2√3cm.
BD = x cm. (let)
∴ CD = (12√3 - x) cm.
∠ADB = ∠ADC = 90°
From ∆ ABD,tan 30° = AD BD ⇒ 1 = AD √3 BD ⇒ AD = x ....(i) √3
From ∆ ACD,tan 60 ° = AD CD ⇒ √3 = AD 12√3 - x
⇒ AD = √3 (12 √3 - x)
= 36 - √3 x ....(ii)∴ x = 36 -√3 x √3
⇒ x = 36 √3 * 3x
⇒ 4x = 36 √3⇒ x = 36 √3 = 9 √3 4 ∴ AD = x = 9√3 = 9 cm. √3 √3
- A pole stands vertically, inside a scalene triangular park ABC. If the angle of elevation of the top of the pole from each corner of the park is same, then in ∆ ABC, the foot of the pole is at the
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AP = CP = BP
It is possible only when
OA = OB = OC i.e. radii of circum circle. or, (circumference)Correct Option: B
AP = CP = BP
It is possible only when
OA = OB = OC i.e. radii of circum circle. or, (circumference)
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If sin θ + cos θ = 5 , the value of tan² θ + 1 is sin θ - cos θ 4 tan² θ - 1
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= sin θ + cos θ = 5 sin θ - cos θ 4 
⇒ tan θ + 1 = 5 tan θ - 1 4
⇒ 4 tanθ + 4 = 5 tanθ – 5
⇒ tanθ = 9⇒ 2tan θ = 5 + 4 2 5 - 4
(By componendo and dividendo)∴ tan²θ + 1 = (9)² + 1 = 81 + 1 tan²θ - 1 (9)² - 1 81 - 1 = 82 = 41 80 40
Correct Option: C
= sin θ + cos θ = 5 sin θ - cos θ 4 ⇒ tan θ + 1 = 5 tan θ - 1 4
⇒ 4 tanθ + 4 = 5 tanθ – 5
⇒ tanθ = 9⇒ 2tan θ = 5 + 4 2 5 - 4
(By componendo and dividendo)∴ tan²θ + 1 = (9)² + 1 = 81 + 1 tan²θ - 1 (9)² - 1 81 - 1 = 82 = 41 80 40
- The numerical value of
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Expression
= 1 + sin θ . 1 - sinθ cosθ cosθ = 1 - sin² θ = cos² θ = 1 cos² θ cos² θ
Correct Option: C
Expression
= 1 + sin θ . 1 - sinθ cosθ cosθ = 1 - sin² θ = cos² θ = 1 cos² θ cos² θ
