Answer:
idk
Step-by-step explanation:
Answer:
∠A = ∠C = 60°
∠B = ∠D = 120°
Step-by-step explanation:
In quad. DPBQ, by angle sum property we have
∠PDQ + ∠DPB + ∠B + ∠BQD = 360°
60° + 90° + ∠B + 90° = 360°
∠B = 360° – 240°
Therefore, ∠B = 120°
But ∠B = ∠D = 120° opposite angles of parallelogram
As, AB || CD opposite sides of a parallelogram
∠B + ∠C = 180° sum of adjacent interior angles is 180°
120° + ∠C = 180°
∠C = 180° – 120° = 60°
Hence ∠A = ∠C = 60° Opposite angles of parallelogram are equal
The answer is C. The easiest way to find this is by setting the top two angles added together to 180 because they divide a straight line. The just plug the numbers in and you get C
It’s Negative below the lines the touching points do not meet
Answer: 2 - 2*sin³(θ) - √1 -sin²(θ)
Step-by-step explanation: In the expression
cos(theta)*sin2(theta) − cos(theta)
sin (2θ) = 2 sin(θ)*cos(θ) ⇒ cos(θ)*2sin(θ)cos(θ) - cos(θ)
2cos²(θ)sin(θ) - cos(θ) if we use cos²(θ) = 1-sin²(θ)
2 [ (1 - sin²(θ))*sin(θ)] - cos(θ)
2 - 2sin²(θ)sin(θ) - cos(θ) ⇒ 2-2sin³(θ)-cos(θ) ; cos(θ) = √1 -sin²(θ)
2 - 2*sin³(θ) - √1 -sin²(θ)
