Cos(A-B) = cosAcosB + sinAsinB
<span>
cos(</span>π/2 - θ) = cos(π/2)cosθ + sin(π/2)sinθ
π/2 = 90°
cos(π/2) = cos90° = 0. sin(π/2) = sin90° = 1
cos(π/2 - θ) = cos(π/2)cosθ + sin(π/2)sin<span>θ
</span>
= 0*cosθ + 1*sin<span>θ = </span>sin<span>θ
Therefore </span>cos(π/2 - θ) = sin<span>θ
QED </span>
You do 53 can go into 66 1 time so put
If your looking for Standard Form Rewrite it as... 
Answer:
90°
Step-by-step explanation:
90
Thales Theorem
