If 7 – 2sin2θ – 11cosθ = 0, where 0° < θ < 90° then find the

1.	If 7 – 2sin2θ – 11cosθ = 0, where 0° < θ < 90° then find the value of θ 1. 90° 2. 60° 3. 30° 4. 45°
Answer» Correct Answer - Option 2 : 60° Given: Our given expression is 7 – 2sin²θ – 11cosθ = 0 Formula used: cos²θ + sin²θ = 1 Calculation: Our given expression is 7 – 2sin²θ – 11cosθ = 0 ⇒ 7 – 2(1 – cos²θ) – 11cosθ = 0 ⇒ 7 – 2 + 2cos²θ – 11cosθ = 0 Put cosθ = t, then our equation becomes ⇒ 2t² – 11t + 5 = 0 ⇒ (t – 5)(t – 1/2) = 0 ⇒ t = 5 or t = ½ Here t = 5 is not possible, so t = ½ ⇒ cosθ = ½ ∴ The value of θ is 60°

If 7 – 2sin2θ – 11cosθ = 0, where 0° < θ < 90° then find the value of θ 1. 90° 2. 60° 3. 30° 4. 45°

Answer» Correct Answer - Option 2 : 60°

Given:

Our given expression is 7 – 2sin²θ – 11cosθ = 0

Formula used:

cos²θ + sin²θ = 1

Calculation:

Our given expression is 7 – 2sin²θ – 11cosθ = 0

⇒ 7 – 2(1 – cos²θ) – 11cosθ = 0

⇒ 7 – 2 + 2cos²θ – 11cosθ = 0

Put cosθ = t, then our equation becomes

⇒ 2t² – 11t + 5 = 0

⇒ (t – 5)(t – 1/2) = 0

⇒ t = 5 or t = ½

Here t = 5 is not possible, so t = ½

⇒ cosθ = ½

∴ The value of θ is 60°