ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: If θ is a positive acute angle and 4cos²θ - 1 = 0, then the value of tan(θ - 30°) is equal to?

Solution:
Given,
4cos²θ - 1 = 0
⇒ 4cos²θ = 1
⇒ cos²θ = 1/4
⇒ cosθ = 1/2
⇒ cosθ = cos60°
∴ θ = 60°

Now, 
tan(θ - 30°) = tan(60° - 30°)
= tan 30°
= 1/√3

Question: asinθ = 1, acosθ = √3 then the value of √3tanθ - 1 = ?

Solution:
asinθ = 1
acosθ = √3

Now,
asinθ/acosθ = 1/√3
⇒ tanθ = 1/√3
⇒ √3tanθ = 1
∴ √3tanθ - 1 = 0

Question: Which one is incorrect?

Solution: 
ত্রিকোনোমিতিক অনুপাতগুলোর সম্পর্ক:
sin²θ + cos²θ = 1
sec²θ - tan²θ = 1
cosec²θ - cot²θ = 1

Question: If sec(4x - 50°) = cosec(50° - x), then the value of x is?

Solution:
sec(4x - 50°) = cosec(50° - x)
⇒ sec(4x - 50°) = cosec {90° - (40° + x)}
⇒ sec(4x - 50°) = sec(40° + x)
⇒ 4x - 50° = 40° + x
⇒ 3x = 90°
∴ x = 30°

Question: = ? 

Solution:
1/{tanA√(1 - sin²A)} 
= 1/(tanA × √cos²A)
= 1/(tanA × cosA)
= 1/{(sinA/cosA) × cosA}
= 1/sinA
= cosecA

Question: A pole of 66 metre long breaks into two parts without complete separation and makes an angle 30° with the ground. Find the length of the broken part of the pole.

Solution: 


sin30 = x/(66 - x)
⇒ 1/2 = x/(66 - x) 
⇒ 66 - x = 2x 
⇒ 3x = 66
⇒ x = 66/3 = 22

The length of the broken part of the pole = 66 - 22 = 44 m

Question: The value of 1 + {(tan 15° - tan 60°)/(cot 30° - cot 75°)} is -

Solution:
1 + (tan 15° - tan 60°)/(cot 30° - cot 75°)
= 1 + (tan 15° - tan 60°)/cot (90° - 60°) - cot (90° - 15°)
= 1 + (tan 15°- tan 60°)/(tan 60° - tan 15°)
= 1 + (tan 15°- tan 60°)/(- 1)(tan 15° - tan 60°)
= 1 - 1
= 0

Question: The angle of elevation of the top of a tower at a point on the ground 37 m away from the foot of the tower is 45°. What is the height of the tower? 

Solution:

 
Let AB be tower and C is a point on the ground 37 m away
From the foot of tower B
The angle of elevation is 45°

Let h be the height of the tower.
∴ tanθ = AB/BC
⇒ tan45^∘ = AB/37
⇒ 1 = AB/37
⇒ AB = 37 m