Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π → C = π - (A + B)
→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))
→ sin C = sin (A + B) cos C = - cos(A + B)
Use the following Sum to Product Identity:
sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]
cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]
Use the following Double Angle Identity:
sin 2A = 2 sin A · cos A
<u>Proof LHS → RHS</u>
LHS: (sin 2A + sin 2B) + sin 2C




![\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cqquad%20%5Cqquad%20%5Cqquad%202%5Csin%20C%5Ccdot%20%5B%5Ccos%20%28A-B%29%2B%5Ccos%20%28A%2BB%29%5D)


LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C 
X > -8
The line would be on -8 going to the left, the circle would not be shaded
Answer:
90°
Step-by-step explanation:
All sides of the given quadrilateral are 4 units, that is they are equal. Hence it is a RHOMBUS.
Diagonals of a rhombus bisects each other at right angles (90°).
x = 90°
(-5, 2 ) means x = -5 and y = 2
so answer is <span>point D</span>